PHYSICS BOOK PART-2

THERMODYNAMICS 321 POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 12.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. 12.2 If the geyser operates on a gas burner, what is the rate of consumption of the fuel if 12.3 its heat of combustion is 4.0 × 104 J/g ? What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 12.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 12.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 12.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 2020-21

322 PHYSICS 12.7 A steam engine delivers 5.4×108J of work per minute and services 3.6 × 109J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is 12.8 wasted per minute? 12.9 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13) Fig. 12.13 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F 12.10 A refrigerator is to maintain eatables kept inside at 90C. If room temperature is 360C, calculate the coefficient of performance. 2020-21

CHAPTER THIRTEEN KINETIC THEORY 13.1 Introduction 13.1 INTRODUCTION 13.2 Molecular nature of matter 13.3 Behaviour of gases Boyle discovered the law named after him in 1661. Boyle, 13.4 Kinetic theory of an ideal gas Newton and several others tried to explain the behaviour of 13.5 Law of equipartition of energy gases by considering that gases are made up of tiny atomic 13.6 Specific heat capacity particles. The actual atomic theory got established more than 13.7 Mean free path 150 years later. Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving Summary atoms or molecules. This is possible as the inter-atomic forces, Points to ponder which are short range forces that are important for solids Exercises and liquids, can be neglected for gases. The kinetic theory Additional exercises was developed in the nineteenth century by Maxwell, Boltzmann and others. It has been remarkably successful. It gives a molecular interpretation of pressure and temperature of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 13.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus 2020-21

324 PHYSICS Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by elements when they combine into compounds. From many observations, in recent times we The first law says that any given compound has, now know that molecules (made up of one or a fixed proportion by mass of its constituents. more atoms) constitute matter. Electron The second law says that when two elements microscopes and scanning tunnelling form more than one compound, for a fixed mass microscopes enable us to even see them. The of one element, the masses of the other elements size of an atom is about an angstrom (10 -10 m). are in ratio of small integers. In solids, which are tightly packed, atoms are spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas 2020-21

KINETIC THEORY 325 is misleading. The gas is full of activity and the for a given sample of the gas. Here T is the equilibrium is a dynamic one. In dynamic equilibrium, molecules collide and change their temperature in kelvin or (absolute) scale. K is a speeds during the collision. Only the average properties are constant. constant for the given sample but varies with Atomic theory is not the end of our quest, but the volume of the gas. If we now bring in the the beginning. We now know that atoms are not indivisible or elementary. They consist of a idea of atoms or molecules, then K is proportional nucleus and electrons. The nucleus itself is made up of protons and neutrons. The protons and to the number of molecules, (say) N in the neutrons are again made up of quarks. Even sample. We can write K = N k . Observation tells quarks may not be the end of the story. There may be string like elementary entities. Nature us that this k is same for all gases. It is called always has surprises for us, but the search for truth is often enjoyable and the discoveries Boltzmann constant and is denoted by k . beautiful. In this chapter, we shall limit ourselves B to understanding the behaviour of gases (and a little bit of solids), as a collection of moving As P1V1 = P2V2 = constant = kB (13.2) molecules in incessant motion. N1T1 N 2T2 if P, V and T are same, then N is also same for all gases. This is Avogadro’s hypothesis, that the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure. The number in 22.4 litres of any gas is 6.02 × 1023. This is known as Avogadro number and is denoted by NA. The mass of 22.4 litres of any gas is equal to its molecular weight 13.3 BEHAVIOUR OF GASES in grams at S.T.P (standard temperature 273 K Properties of gases are easier to understand than and pressure 1 atm). This amount of substance those of solids and liquids. This is mainly because in a gas, molecules are far from each is called a mole (see Chapter 2 for a more precise other and their mutual interactions are negligible except when two molecules collide. definition). Avogadro had guessed the equality of Gases at low pressures and high temperatures numbers in equal volumes of gas at a fixed much above that at which they liquefy (or solidify) approximately satisfy a simple relation temperature and pressure from chemical between their pressure, temperature and volume given by (see Chapter 11) reactions. Kinetic theory justifies this hypothesis. The perfect gas equation can be written as PV = µ RT (13.3) PV = KT (13.1) where µ is the number of moles and R = NA akbB sisolauutentievmerpsearlactounrset.aCnht.oTohsientgemkeplveirnatsucareleTfoisr John Dalton (1766 –1844) He was an English chemist. When different types of atoms combine, they obey certain simple laws. Dalton’s atomic theory explains these laws in a simple way. He also gave a theory of colour blindness. Amedeo Avogadro (1776 – 1856) He made a brilliant guess that equal volumes of gases have equal number of molecules at the same temperature and pressure. This helped in understanding the combination of different gases in a very simple way. It is now called Avogadro’s hypothesis (or law). He also suggested that the smallest constituent of gases like hydrogen, oxygen and nitrogen are not atoms but diatomic molecules. 2020-21

326 PHYSICS absolute temperature, R = 8.314 J mol–1K–1. i.e., keeping temperature constant, pressure of Here a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 13.2 shows µ= M = N (13.4) comparison between experimental P-V curves M0 NA and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is where M is the mass of the gas containing N good at high temperatures and low pressures. Next, if you fix P, Eq. (13.1) shows that V ∝ T molecules, M0 is the molar mass and NA the i.e., for a fixed pressure, the volume of a gas is Avogadro’s number. Using Eqs. (13.4) and (13.3) proportional to its absolute temperature T (Charles’ law). See Fig. 13.3. can also be written as PV = kB NT or P = kB nT J mol–1K –1) pV ( µT P (atm) Fig.13.1 Real gases approach ideal gas behaviour at low pressures and high temperatures. where n is the number density, i.e. number of Fig.13.2 Experimental P-V curves (solid lines) for cmoonlsetcaunletsinpterroduuncitevdoalubmovee..kIBtsisvatlhueeBinolStzImuannitns steam at three temperatures compared is 1.38 × 10–23 J K–1. with Boyle’s law (dotted lines). P is in units of 22 atm and V in units of 0.09 litres. Another useful form of Eq. (13.3) is P = ρRT (13.5) Finally, consider a mixture of non-interacting M0 ideal gases: µ moles of gas 1, µ moles of gas where ρ is the mass density of the gas. 2, etc. in a ves1sel of volume V at t2emperature T A gas that satisfies Eq. (13.3) exactly at all and pressure P. It is then found that the equation of state of the mixture is : pressures and temperatures is defined to be an ideal gas. An ideal gas is a simple theoretical PV = ( µ1 + µ2 +… ) RT (13.7) model of a gas. No real gas is truly ideal. Fig. 13.1 shows departures from ideal gas i.e. P = µ1 RT + µ2 RT + ... (13.8) behaviour for a real gas at three different V V temperatures. Notice that all curves approach the ideal gas behaviour for low pressures and = P1 + P2 + … (13.9) high temperatures. Clearly P1 = µ1 R T/V is the pressure that gas 1 would exert at the same conditions of At low pressures or high temperatures the volume and temperature if no other gases were molecules are far apart and molecular present. This is called the partial pressure of the interactions are negligible. Without interactions gas. Thus, the total pressure of a mixture of ideal the gas behaves like an ideal one. If we fix µ and T in Eq. (13.3), we get gases is the sum of partial pressures. This is PV = constant (13.6) Dalton’s law of partial pressures. 2020-21

KINETIC THEORY 327 Fig. 13.3 Experimental T-V curves (solid lines) for density of water molecule may therefore, be CO2 at three pressures compared with regarded as roughly equal to the density of bulk Charles’ law (dotted lines). T is in units of water = 1000 kg m–3. To estimate the volume of a water molecule, we need to know the mass of 300 K and V in units of 0.13 litres. a single water molecule. We know that 1 mole of water has a mass approximately equal to We next consider some examples which give us information about the volume occupied by (2 + 16)g = 18 g = 0.018 kg. the molecules and the volume of a single Since 1 mole contains about 6 × 1023 molecule. molecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = Example 13.1 The density of water is 1000 3 × 10–26 kg. Therefore, a rough estimate of the kg m–3. The density of water vapour at 100 °C volume of a water molecule is as follows : and 1 atm pressure is 0.6 kg m–3. The Volume of a water molecule volume of a molecule multiplied by the total = (3 × 10–26 kg)/ (1000 kg m–3) number gives ,what is called, molecular = 3 × 10–29 m3 volume. Estimate the ratio (or fraction) of = (4/3) π (Radius)3 the molecular volume to the total volume Hence, Radius ≈ 2 ×10-10 m = 2 Å occupied by the water vapour under the above conditions of temperature and Example 13.3 What is the average pressure. distance between atoms (interatomic distance) in water? Use the data given in Answer For a given mass of water molecules, Examples 13.1 and 13.2. the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 ×10 -4 ) Answer : A given mass of water in vapour state times larger. If densities of bulk water and water has 1.67×103 times the volume of the same mass molecules are same, then the fraction of of water in liquid state (Ex. 13.1). This is also molecular volume to the total volume in liquid the increase in the amount of volume available state is 1. As volume in vapour state has for each molecule of water. When volume increased, the fractional volume is less by the increases by 103 times the radius increases by same amount, i.e. 6×10-4. V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å. Example 13.2 Estimate the volume of a water molecule using the data in Example Example 13.4 A vessel contains two non- 13.1. reactive gases : neon (monatomic) and Answer In the liquid (or solid) phase, the molecules of water are quite closely packed. The oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O2 = 32.0 u. Answer Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed ideal) obeys the gas law. Since V and T are common to tµ(t3oh2/enR2teT)wo,(nogii.gaveaen. nsd(Pe),o1s/x,(µPyw12g/)ee=nµh2(r)µae=1vs/ep3e/µPc221t).Vi.vHe=leyµr.e1S1RinTacneadn(Pd21/PrP2eV2f)e==r 2020-21

328 PHYSICS (i) BWTow1(mfhyhea12een/drrcaMdeeeanff2nNo2id)nr;1w2ieaata,,ilhnnosa(eNddnonr1NedM/µw21NNm1rai2=Aa1tr)nieeas(=dNntµth(d1hMµ1/ee1mN2=n/AAa2uv(µr)ammoe2rag)b1etna/he=dtdMrehr3io1eµorf/)’2smmm2a=n.onao(ullsdNeemscc2eµ/uubs2NlleeaoAr=srf.) (ii) ImMf 2aρ1ssshaeonsud.l(dρB2boatehreemxtp1hraeensmsdeaMdss1i;ndaetshnwesiestlialemsaseofmu12naaitnnsdd). 2 respectively, we have ρ1 = m1 /V = m1 = µ1 ×  M1  Fig. 13.4 Elastic collision of a gas molecule with ρ2 m2 /V m2 µ2  M2  the wall of the container. = 3 × 20.2 = 0.947 p(vlxa,nvey,ofvaz r)eahiAts(=thl2e). planar wall parallel to yz- 2 32.0 Since the collision is elastic, the molecule rebounds with the same velocity; 13.4 KINETIC THEORY OF AN IDEAL GAS its y and z components of velocity do not change Kinetic theory of gases is based on the molecular in the collision but the x-component reverses picture of matter. A given amount of gas is a collection of a large number of molecules sign. That is, the velocity after collision is (typically of the order of Avogadro’s number) that are in incessant random motion. At ordinary (m-voxl,evcyu, lvez ). The change in momentum of the pressure and temperature, the average distance principle is: co–nmsvexrv–at(imonvx)o=f m– o2mmevnxt.umBy, the between molecules is a factor of 10 or more than of the the typical size of a molecule (2 Å). Thus, interaction between molecules is negligible and momentum imparted to the wall in the collision we can assume that they move freely in straight lines according to Newton’s first law. However, = 2Tmovcxa.lculate the force (and pressure) on the occasionally, they come close to each other, experience intermolecular forces and their wall, we need to calculate momentum imparted velocities change. These interactions are called collisions. The molecules collide incessantly to the wall per unit time. In a small time interval against each other or with the walls and change their velocities. The collisions are considered to ∆t, a molecule with iixss-,cwaoilmtlhmpinoonlteehcneutldeoissf twvaeintlochecinitvyxth∆vext be elastic. We can derive an expression for the will hit the wall if it pressure of a gas based on the kinetic theory. from the wall. That We begin with the idea that molecules of a volume Athvxe∆at voenrlaygec,anhahlfitotfhtehwesaell in time ∆t. gas are in incessant random motion, colliding But, on are moving against one another and with the walls of the container. All collisions between molecules towards the wall and the other half away from among themselves or between molecules and the walls are elastic. This implies that total kinetic the wall. Thus, the number of molecules with energy is conserved. The total momentum is conserved as usual. velocity (vx, vy, vz ) hitting the wall in time ∆t is ½A vxu∆nti where n is the number of molecules per n, volume . The total momentum t transferred to the wall by these molecules in time ∆t is : ∆t is Q= f(2ormcvexo) n(½thne Awavlxl ) rate of (13.10) The the momentum transfer Q/∆t and pressure is force per unit area : PAc=tuQall/y(,Aa∆llt)m=olencumlevsx2in a gas do (3.11) not have 13.4.1 Pressure of an Ideal Gas the same velocity; there is a distribution in Consider a gas enclosed in a cube of side l. Take velocities. The above equation, therefore, stands the axes to be parallel to the sides of the cube, as shown in Fig. 13.4. A molecule with velocity for pressure due to the group of molecules with nspuemedbevrx in the x-direction and n stands for the density of that group of molecules. The 2020-21

KINETIC THEORY 329 total pressure is obtained by summing over the the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in contribution due to all groups: the derivation. Though this assumption is P = n m v 2 (13.12) difficult to justify rigorously, we can qualitatively x where v 2 is the average of vx2 . Now the gas see that it will not lead to erroneous results. x is isotropic, i.e. there is no preferred direction The number of molecules hitting the wall in time of velocity of the molecules in the vessel. ∆t was found atondbet½hengAavsx ∆t. Now the collisions are random is in a steady state. Therefore, by symmetry, v 2 = vy2 = vz2 Thus, if a molecule with vdeuloectiotyco(lvlixs,iovny, wvizth) x acquires a different velocity = (1/3) [ v 2 + vy2 + v 2 ] = (1/3) v2 (13.13) some molecule, there will always be some other x z molecule with a different initial velocity which where v is the speed and v2 denotes the mean after a collision saoc,quthireedsitshtreibvueltoiocintyof(vvxe,lvoyc,itvize)s. If this were not of the squared speed. Thus would not remain steady. In any case we are P = (1/3) n m v2 (13.14) finding v 2 . Thus, on the whole, molecular x Some remarks on this derivation. First, collisions (if they are not too frequent and the though we choose the container to be a cube, time spent in a collision is negligible compared the shape of the vessel really is immaterial. For to time between collisions) will not affect the a vessel of arbitrary shape, we can always choose calculation above. a small infinitesimal (planar) area and carry 13.4.2 Kinetic Interpretation of Temperature through the steps above. Notice that both A and Equation (13.14) can be written as ∆t do not appear in the final result. By Pascal’s law, given in Ch. 10, pressure in one portion of PV = (1/3) nV m v2 (13.15a) Founders of Kinetic Theory of Gases James Clerk Maxwell (1831 – 1879), born in Edinburgh, Scotland, was among the greatest physicists of the nineteenth century. He derived the thermal velocity distribution of molecules in a gas and was among the first to obtain reliable estimates of molecular parameters from measurable quantities like viscosity, etc. Maxwell’s greatest achievement was the unification of the laws of electricity and magnetism (discovered by Coulomb, Oersted, Ampere and Faraday) into a consistent set of equations now called Maxwell’s equations. From these he arrived at the most important conclusion that light is an electromagnetic wave. Interestingly, Maxwell did not agree with the idea (strongly suggested by the Faraday’s laws of electrolysis) that electricity was particulate in nature. Ludwig Boltzmann (1844 – 1906) born in Vienna, Austria, worked on the kinetic theory of gases independently of Maxwell. A firm advocate of atomism, that is basic to kinetic theory, Boltzmann provided a statistical interpretation of the Second Law of thermodynamics and the concept of entropy. He is regarded as one of the founders of classical statistical mechanics. The proportionality constant connecting energy and temperature in kinetic theory is known as Boltzmann’s constant in his honour. 2020-21

330 PHYSICS PV = (2/3) N x ½ m v2 (13.15b) m = M N2 = 28 = 4.65 × 10–26 kg. NA 6.02 × 1026 where N (= nV ) is the number of molecules in the sample. The quantity in the bracket is the average v 2 = 3 kB T / m = (516)2 m2s-2 translational kinetic energy of the molecules in The square root of v2 is known as root mean the gas. Since the internal energy E of an ideal square (rms) speed and is denoted by vrms, gas is purely kinetic*, ( We can also write v2 as < v2 >.) E = N × (1/2) m v2 (13.16) vrms = 516 m s-1 The speed is of the order of the speed of sound Equation (13.15) then gives : in air. It follows from Eq. (13.19) that at the same temperature, lighter molecules have greater rms PV = (2/3) E (13.17) speed. We are now ready for a kinetic interpretation of temperature. Combining Eq. (13.17) with the ideal gas Eq. (13.3), we get Example 13.5 A flask contains argon and E = (3/2) kB NT (13.18) chlorine in the ratio of 2:1 by mass. The or E/ N = ½ m v2 = (3/2) kBT (13.19) temperature of the mixture is 27 °C. Obtain i.e., the average kinetic energy of a molecule is the ratio of (i) average kinetic energy per proportional to the absolute temperature of the molecule, and (ii) root mean square speed gas; it is independent of pressure, volume or the nature of the ideal gas. This is a fundamental Avrtmosmoicf the molecules of the two gases. result relating temperature, a macroscopic mass of argon = 39.9 u; Molecular measurable parameter of a gas (a thermodynamic variable as it is called) to a mass of chlorine = 70.9 u. molecular quantity, namely the average kinetic energy of a molecule. The two domains are Answer The important point to remember is that connected by the Boltzmann constant. We note in passing that Eq. (13.18) tells us that internal the average kinetic energy (per molecule) of any energy of an ideal gas depends only on temperature, not on pressure or volume. With (ideal) gas (be it monatomic like argon, diatomic this interpretation of temperature, kinetic theory of an ideal gas is completely consistent with the like chlorine or polyatomic) is always equal to ideal gas equation and the various gas laws based on it. (i3s/in2d) ekpBTen. Idtednetpoefntdhse only on temperature, and nature of the gas. For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas (i) Since argon and chlorine both have the same in the mixture. Equation (13.14) becomes temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1. (ii) oNmfooawlemc½uollmeec=vurlm(e3s2o/f2=t)ha) evkeBgrTaaswg.ehTkehrieneremetfiocirseent,heergmy apsesr P = (1/3) [n1m1 v12 + n2 m2 v22 +… ] (13.20) ( )vr2ms (m )Cl (M )Cl ( )vr2ms (m )Ar (M )Ar In equilibrium, the average kinetic energy of Ar = = = 70.9 =1.77 the molecules of different gases will be equal. Cl 39.9 That is, where M denotes the molecular mass of the gas. ½ m1 v12 = ½ m2 v22 = (3/2) kB T (For argon, a molecule is just an atom of argon.) so that Taking square root of both sides, P = (n1 + n2 +… ) kB T (13.21) ( )vrms Ar ( )vrms Cl which is Dalton’s law of partial pressures. = 1.33 From Eq. (13.19), we can get an idea of the You should note that the composition of the typical speed of molecules in a gas. At a mixture by mass is quite irrelevant to the above temperature T = 300 K, the mean square speed of a molecule in nitrogen gas is : * E denotes the translational part of the internal energy U that may include energies due to other degrees of freedom also. See section 13.5. 2020-21

KINETIC THEORY 331 Maxwell Distribution Function In a given mass of gas, the velocities of all molecules are not the same, even when bulk parameters like pressure, volume and temperature are fixed. Collisions change the direction and the speed of molecules. However in a state of equilibrium, the distribution of speeds is constant or fixed. Distributions are very important and useful when dealing with systems containing large number of objects. As an example consider the ages of different persons in a city. It is not feasible to deal with the age of each individual. We can divide the people into groups: children up to age 20 years, adults between ages of 20 and 60, old people above 60. If we want more detailed information we can choose smaller intervals, 0-1, 1-2,..., 99-100 of age groups. When the size of the interval becomes smaller, say half year, the number of persons in the interval will also reduce, roughly half the original number in the one year interval. The number of persons dN(x) in the age interval x and x+dx is proportional to dx or dN(x) = nx dx. We have used nx to denote the number of persons at the value of x. Maxwell distribution of molecular speeds In a similar way the molecular speed distribution gives the number of molecules between N a3e–bv2 v2 the speeds v and v+ dv. dN(v) = 4p the figure. Tdhve=frnavcdtvio. nTohfisthies called Maxwell distribution. The plot of nv against v is shown in molecules with speeds v and v+dv is equal to the area of the strip shown. The average of any quantity like v2 is defined by the integral <v2> = (1/N ) ∫ v2 dN(v) = ª(3kB T/m) which agrees with the result derived from more elementary considerations. calculation. Any other proportion by mass of mass of the molecule, faster will be the speed. argon and chlorine would give the same answers The ratio of speeds is inversely proportional to to (i) and (ii), provided the temperature remains the square root of the ratio of the masses. The unaltered. masses are 349 and 352 units. So Example 13.6 Uranium has two isotopes v349 / v352 = ( 352/ 349)1/2 = 1.0044 . of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which Hence difference ∆V = 0.44 %. would have the larger average speed ? If V atomic mass of fluorine is 19 units, estimate the percentage difference in [235U is the isotope needed for nuclear fission. speeds at any temperature. To separate it from the more abundant isotope 238U, the mixture is surrounded by a porous Answer At a fixed temperature the average cylinder. The porous cylinder must be thick and energy = ½ m <v2 > is constant. So smaller the narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than 2020-21

332 PHYSICS the slower one and so there is more of the lighter is V + u towards the bat. When the ball rebounds molecule (enrichment) outside the porous (after hitting the massive bat) its speed, relative cylinder (Fig. 13.5). The method is not very to bat, is V + u moving away from the bat. So efficient and has to be repeated several times relative to the wicket the speed of the rebounding for sufficient enrichment.]. ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the When gases diffuse, their rate of diffusion is collision with the bat. The rebound speed will inversely proportional to square root of the be less than u if the bat is not massive. For a masses (see Exercise 13.12 ). Can you guess the molecule this would imply an increase in explanation from the above answer? temperature. You should be able to answer (b) (c) and (d) based on the answer to (a). (Hint: Note the correspondence, piston bat, cylinder wicket, molecule ball.) 13.5 LAW OF EQUIPARTITION OF ENERGY The kinetic energy of a single molecule is εt = 1 mv 2 + 1 mvy2 + 1 mvz2 (13.22) 2 x 2 2 For a gas in thermal equilibrium at temperature T the average value of energy denoted by < εt > is εt = 1 mvx2 + 1 mvy2 + 1 mvz2 = 3 (13.23) 2 2 2 2 kBT Fig. 13.5 Molecules going through a porous wall. Since there is no preferred direction, Eq. (13.23) Example 13.7 (a) When a molecule (or implies an elastic ball) hits a ( massive) wall, it rebounds with the same speed. When a ball 1 mv 2 = 1 1 mvy2 = 1 hits a massive bat held firmly, the same 2 x 2 kBT , 2 2 kBT , thing happens. However, when the bat is moving towards the ball, the ball rebounds 1 mvz2 = 1 (13.24) with a different speed. Does the ball move 2 2 kBT faster or slower? (Ch.6 will refresh your memory on elastic collisions.) A molecule free to move in space needs three coordinates to specify its location. If it is (b) When gas in a cylinder is compressed by pushing in a piston, its temperature constrained to move in a plane it needs two; and rises. Guess at an explanation of this in if constrained to move along a line, it needs just terms of kinetic theory using (a) above. one coordinate to locate it. This can also be (c) What happens when a compressed gas expressed in another way. We say that it has pushes a piston out and expands. What one degree of freedom for motion in a line, two would you observe ? for motion in a plane and three for motion in (d) Sachin Tendulkar used a heavy cricket bat while playing. Did it help him in space. Motion of a body as a whole from one anyway ? point to another is called translation. Thus, a molecule free to move in space has three Answer (a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving translational degrees of freedom. Each towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat translational degree of freedom contributes a term that contains square of some variable of motion, e.g., ½ (1m3v.x224)awnedseseimthilaatrintetrhmersmianl average of each such term is evqy uanilidbrvizu. mIn,, Eq. the ½ kBT . 2020-21

KINETIC THEORY 333 Molecules of a monatomic gas like argon have ε = εt + εr + εv (13.26) only translational degrees of freedom. But what about a diatomic gas such as O2 or N2? A where k is the force constant of the oscillator molecule of O2 has three translational degrees of freedom. But in addition it can also rotate and y the vibrational co-ordinate. about its centre of mass. Figure 13.6 shows the two independent axes of rotation 1 and 2, normal Once again the vibrational energy terms in to the axis joining the two oxygen atoms about which the molecule can rotate*. The molecule Eq. (13.26) contain squared terms of vibrational thus has two rotational degrees of freedom, each of which contributes a term to the total energy variables of motion y and dy/dt . consisting of translational energy εt and rotational energy εr. At this point, notice an important feature in Eq.(13.26). While each translational and rotational degree of freedom has contributed only one ‘squared term’ in Eq.(13.26), one vibrational mode contributes two ‘squared terms’ : kinetic and potential energies. 1 1 1 1 1 Each quadratic term occurring in the 2 2 2 2 2 εt + εr = mv 2 + mvy2 + mvz2 + I1ω12 + I 2 ω22 (13.25) expression for energy is a mode of absorption of x energy by the molecule. We have seen that in thermal equilibrium at absolute temperature T, for each translational mode of motion, the average energy is ½ kBT. The most elegant principle of classical statistical mechanics (first proved by Maxwell) states that this is so for each mode of energy: translational, rotational and vibrational. That is, in equilibrium, the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law of equipartition of energy. Accordingly, each Fig. 13.6 The two independent axes of rotation of a translational and rotational degree of freedom diatomic molecule of a molecule contributes ½ kBT to the energy, while each vibrational frequency contributes where ω1 and ω2 are the angular speeds about 2 × ½ kBT = kBT , since a vibrational mode has the axes 1 and 2 and I1, I2 are the corresponding both kinetic and potential energy modes. moments of inertia. Note that each rotational The proof of the law of equipartition of energy degree of freedom contributes a term to the is beyond the scope of this book. Here, we shall energy that contains square of a rotational apply the law to predict the specific heats of variable of motion. gases theoretically. Later, we shall also discuss Wa e‘rhigaivderaostsautomr’e,di.aeb.,otvheethmatoltehceuOle2 molecule is does not briefly, the application to specific heat of solids. vibrate. This assumption, though found to be 13.6 SPECIFIC HEAT CAPACITY true (at moderate temperatures) for ,Oe2,viesnnaott 13.6.1 Monatomic Gases always valid. Molecules, like CO moderate temperatures have a mode of vibration, The molecule of a monatomic gas has only three i.e., its atoms oscillate along the interatomic axis translational degrees of freedom. Thus, the like a one-dimensional oscillator, and contribute average energy of a molecule at temperature a vibrational energy term εv to the total energy: oTfissu(3c/h2a)kgBaTs. The total internal energy of a mole is 1  dy  2 1 2 dt 2 εv = m + ky 2 * Rotation along the line joining the atoms has very small moment of inertia and does not come into play for quantum mechanical reasons. See end of section 13.6. 2020-21

334 PHYSICS U = 3 × NA = 3 RT (13.27) i.e.,Cv = (3 + f ) R, Cp = (4 + f ) R, 2 kBT 2 γ = (4 + f ) (13.36) (3 + f ) The molar specific heat at constant volume, Note that Cp – Cv = R is true for any ideal Cv, is gas, whether mono, di or polyatomic. Cv (monatomic gas) = dU = 3 RT (13.28) Table 13.1 summarises the theoretical dT 2 predictions for specific heats of gases ignoring any vibrational modes of motion. The values are For an ideal gas, in good agreement with experimental values of specific heats of several gases given in Table 13.2. pwrheesCrsepu–CrepC.ivsT=thhRuesm, olar specific heat at (13.29) Of course, there are discrepancies between constant predicted and actual values of specific heats of several other gases (not shown in the table), such Cp = 5 R (13.30) as Cl2, C2H6 and many other polyatomic gases. 2 Usually, the experimental values for specific heats of these gases are greater than the The ratio of specific heats γ = Cp = 5 (13.31) predicted values as given in Table13.1 suggesting Cv 3 that the agreement can be improved by including vibrational modes of motion in the calculation. 13.6.2 Diatomic Gases The law of equipartition of energy is, thus, well As explained earlier, a diatomic molecule treated Table 13.1 Predicted values of specific heat as a rigid rotator, like a dumbbell, has 5 degrees capacities of gases (ignoring of freedom: 3 translational and 2 rotational. vibrational modes) Using the law of equipartition of energy, the total internal energy of a mole of such a gas is U = 5 × NA = 5 RT (13.32) 2 kBT 2 Nature of Cv Cp The molar specific heats are then given by Gas (J mol-1 K-1) (J mol-1 K-1) Cp - Cv g (J mol-1 K-1) Cv (rigid diatomic) = 5 R, Cp = 7 (13.33) Monatomic 12.5 20.8 8.31 1.67 2 R Diatomic 20.8 29.1 8.31 1.40 2 γ (rigid diatomic) = 7 (13.34) Triatomic 24.93 33.24 8.31 1.33 If the diatomic mole5cule is not rigid but has in addition a vibrational mode Table13.2 Measured values of specific heat capacities of some gases = 5 +kBT  = 7 U  2 k BT N A RT 2 Cv = 7 R, Cp = 9 R, γ = 9 (13.35) 2 2 7R 13.6.3 Polyatomic Gases In general a polyatomic molecule has 3 translational, 3 rotational degrees of freedom and a certain number ( f ) of vibrational modes. According to the law of equipartition of energy, it is easily seen that one mole of such a gas has U=  3 kBT + 3 kBT + f kBT NA 2 2 2020-21

KINETIC THEORY 335 verified experimentally at ordinary As Table 13.3 shows the prediction generally temperatures. agrees with experimental values at ordinary temperature (Carbon is an exception). Example 13.8 A cylinder of fixed capacity 44.8 litres contains helium gas at standard 13.6.5 Specific Heat Capacity of Water temperature and pressure. What is the amount of heat needed to raise the We treat water like a solid. For each atom average temperature of the gas in the cylinder by 15.0 °C ? (R = 8.31 J mo1–1 K–1). etwnoerhgyydisro3gkeBnT.aWndatoernemooxleycguelne.hSaos three atoms, it has Answer Using the gas law PV = µRT, you can U = 3 × 3 kBT × NA = 9 RT easily show that 1 mol of any (ideal) gas at and C = ∆Q/ ∆T =∆ U / ∆T = 9R . standard temperature (273 K) and pressure (1 atm = 1.01 × 105 Pa) occupies a volume of This is the value observed and the agreement 22.4 litres. This universal volume is called molar is very good. In the calorie, gram, degree units, volume. Thus the cylinder in this example water is defined to have unit specific heat. As 1 contains 2 mol of helium. Further, since helium calorie = 4.179 joules and one mole of water is monatomic, its predicted (and observed) molar is 18 grams, the heat capacity per mole is specific heat at constant avtolcuomnset,aCnvt = (3/2) R, ~ 75 J mol-1 K-1 ~ 9R . However with more and molar specific heat pressure, complex molecules like alcohol or acetone the Cthpe= (3/2) R + R = (5/2) R . Since the volume of arguments, based on degrees of freedom, become cylinder is fixed, the heat required is more complicated. determined by =Cvn. oT.heorfemfoorele,s Heat required × molar specific Lastly, we should note an important aspect heat × rise in temperature of the predictions of specific heats, based on the = 2 × 1.5 R × 15.0 = 45 R classical law of equipartition of energy. The = 45 × 8.31 = 374 J. predicted specific heats are independent of temperature. As we go to low temperatures, 13.6.4 Specific Heat Capacity of Solids however, there is a marked departure from this prediction. Specific heats of all substances We can use the law of equipartition of energy to approach zero as T 0. This is related to the determine specific heats of solids. Consider a fact that degrees of freedom get frozen and ineffective at low temperatures. According to solid of N atoms, each vibrating about its mean classical physics, degrees of freedom must position. An oscillation in one dimension has remain unchanged at all times. The behaviour of specific heats at low temperatures shows the damenvimoeerlreeagngyseoiisofennses,rogtlhyideo,af vNe2ra×=ge½eNnAke,BrTgay=niskdB3Ttkh.BIeTn. three inadequacy of classical physics and can be For a explained only by invoking quantum total considerations, as was first shown by Einstein. U = 3 kBT × NA = 3 RT Quantum mechanics requires a minimum, Now at constant pressure ∆Q = ∆U + P∆V non-zero amount of energy before a degree of = ∆U, since for a solid ∆V is negligible. Hence, freedom comes into play. This is also the reason C = ∆Q = ∆U = 3R (13.37) why vibrational degrees of freedom come into play ∆T ∆T only in some cases. Table 13.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure 13.7 MEAN FREE PATH Molecules in a gas have rather large speeds of the order of the speed of sound. Yet a gas leaking from a cylinder in a kitchen takes considerable time to diffuse to the other corners of the room. The top of a cloud of smoke holds together for hours. This happens because molecules in a gas have a finite though small size, so they are bound to undergo collisions. As a result, they cannot 2020-21

336 PHYSICS Seeing is Believing Can one see atoms rushing about. Almost but not quite. One can see pollen grains of a flower being pushed around by molecules of water. The size of the grain is ~ 10-5 m. In 1827, a Scottish botanist Robert Brown, while examining, under a microscope, pollen grains of a flower suspended in water noticed that they continuously moved about in a zigzag, random fashion. Kinetic theory provides a simple explanation of the phenomenon. Any object suspended in water is continuously bombarded from all sides by the water molecules. Since the motion of molecules is random, the number of molecules hitting the object in any direction is about the same as the number hitting in the opposite direction. The small difference between these molecular hits is negligible compared to the total number of hits for an object of ordinary size, and we do not notice any movement of the object. When the object is sufficiently small but still visible under a microscope, the difference in molecular hits from different directions is not altogether negligible, i.e. the impulses and the torques given to the suspended object through continuous bombardment by the molecules of the medium (water or some other fluid) do not exactly sum to zero. There is a net impulse and torque in this or that direction. The suspended object thus, moves about in a zigzag manner and tumbles about randomly. This motion called now ‘Brownian motion’ is a visible proof of molecular activity. In the last 50 years or so molecules have been seen by scanning tunneling and other special microscopes. In 1987 Ahmed Zewail, an Egyptian scientist working in USA was able to observe not only the molecules but also their detailed interactions. He did this by illuminating them with flashes of laser light for very short durations, of the order of tens of femtoseconds and photographing them. ( 1 femto- second = 10-15 s ). One could study even the formation and breaking of chemical bonds. That is really seeing ! move straight unhindered; their paths keep will collide with it (see Fig. 13.7). If n is the getting incessantly deflected. number of molecules per unit volume, the molecule suffers nπd2 <v> ∆t collisions in time ∆t. Thus the rate of collisions is nπd2 <v> or the time between two successive collisions is on the average, τ = 1/(nπ <v> d2 ) (13.38) The average distance between two successive t collisions, called the mean free path l, is : v l = <v> τ = 1/(nπd2) (13.39) d d In this derivation, we imagined the other molecules to be at rest. But actually all molecules are moving and the collision rate is determined by the average relative velocity of the molecules. Thus we need to replace <v> by <v > in Eq. (13.38). A more exact treatment gives r ( )l = 1/ 2 nπd 2 (13.40) Fig. 13.7 The volume swept by a molecule in time ∆t Let us estimate l and τ for air molecules with average speeds <v> = ( 485m/s). At STP in which any molecule will collide with it. ( )0.02 × 1023 Suppose the molecules of a gas are spheres ( )n = 22.4 × 10–3 of diameter d. Focus on a single molecule with the average speed <v>. It will suffer collision with = 2.7 × 10 25 m -3. (13.41) any molecule that comes within a distance d between the centres. In time ∆t, it sweeps a Taking, d = 2 × 10–10 m, volume πd2 <v> ∆t wherein any other molecule τ = 6.1 × 10–10 s and l = 2.9 × 10–7 m ≈ 1500d 2020-21

KINETIC THEORY 337 As expected, the mean free path given by So n = 2.7 × 1025 × 273 = 2 × 1025 m –3 Eq. (13.40) depends inversely on the number 373 density and the size of the molecules. In a highly evacuated tube n is rather small and the mean Hence, mean free path l = 4 × 10–7 m free path can be as large as the length of the tube. Note that the mean free path is 100 times the interatomic distance ~ 40 Å = 4 ×10-9 m calculated Example 13.9 Estimate the mean free path earlier. It is this large value of mean free path that for a water molecule in water vapour at 373 K. leads to the typical gaseous behaviour. Gases can Use information from Exercises 13.1 and not be confined without a container. Eq. (13.41) above. Using, the kinetic theory of gases, the bulk Answer The d for water vapour is same as that measurable properties like viscosity, heat of air. The number density is inversely conductivity and diffusion can be related to the proportional to absolute temperature. microscopic parameters like molecular size. It is through such relations that the molecular sizes were first estimated. SUMMARY 1. The ideal gas equation connecting pressure (P ), volume (V ) and absolute temperature (T ) is where µ is the PV = µ RT =NkisB NT number of molecules. R and kB are universal constants. number of moles and the R = 8.314 J mol–1 K–1, R = 1.38 × 10–23 J K–1 kB = N A Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures. 2. Kinetic theory of an ideal gas gives the relation P = 1 n m v2 3 where n is number density of molecules, m the mass of the molecule and v2 is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature. ( )1 3 3k B T 2 m 2 m v2 = kB T , vrms = v2 1/2 = This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed. 3. The translational kinetic energy E = 3 kB NT. 2 This leads to a relation PV = 2 E 3 4. The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total energy is distributed equally in different energy modes of 2020-21

338 PHYSICS absorption, the energy in each mode being equal to ½ kmB oTd.eEoafcahbtsroarnpstiloantioannadl and rotational degree of freedom corresponds to one energy has wenitehrgcyo½rreksBpTo.nEdaicnhg vibrational frequency has two modes of energy (kinetic and potential) energy equal to 2 × ½ kB T = kB T. 5. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion. 6. The mean free path l is the average distance covered by a molecule between two successive collisions : l= 1 2 n π d2 where n is the number density and d the diameter of the molecule. POINTS TO PONDER 1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer. 2. We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule. 3. The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is ½ k T. Each quadratic term in the total energy expression of a molecule is to be counted aBs a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy 2 × ½ k T = k T. BB 4. Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy ½ mv2 of the molecules. 5. < v2 > is not always equal to ( < v >)2. The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement. 13.1 EXERCISES 13.2 13.3 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å. Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres. Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures. 2020-21

KINETIC THEORY 339 y T1 T2 PTV(J K–1) Px Fig. 13.8 (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.) 13.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and 13.5 a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u). An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ? 13.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure. 13.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star). 13.8 Three vessels of equal capacity have gases at the same temperature and pressure. 13.9 The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ? At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u). 2020-21

340 PHYSICS Additional Exercises 13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ? 13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 s–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3 s–1. Identify the gas. [Hint : Use Graham’s law of diffusion: R1/R2 = ( M2 /M1 )1/2, where R1, R2 are diffusion rates of gases 1 and 2, and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.] 13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2 = n1 exp [ -mg (h2 – h1)/ kBT ] where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n2 = n1 exp [ -mg NA (ρ - ρ′ ) (h2 –h1)/ (ρ RT)] where ρ is the density of the suspended particle, and ρ′ , that of surrounding medium. [pNrAinisciAplveogtoadfirnod’s number, and R the universal gas constant.] [Hint : Use Archimedes the apparent weight of the suspended particle.] 13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms : Substance Atomic Mass (u) Density (103 Kg m-3) Carbon (diamond) 12.01 2.22 Gold 197.00 19.32 Nitrogen (liquid) Lithium 14.01 1.00 Fluorine (liquid) 6.94 0.53 1.14 19.00 [Hint : Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å]. 2020-21

CHAPTER FOURTEEN OSCILLATIONS 14.1 Introduction 14.1 INTRODUCTION 14.2 Periodic and oscillatory In our daily life we come across various kinds of motions. motions You have already learnt about some of them, e.g., rectilinear motion and motion of a projectile. Both these motions are 14.3 Simple harmonic motion non-repetitive. We have also learnt about uniform circular 14.4 Simple harmonic motion motion and orbital motion of planets in the solar system. In these cases, the motion is repeated after a certain interval of and uniform circular time, that is, it is periodic. In your childhood, you must have motion enjoyed rocking in a cradle or swinging on a swing. Both these motions are repetitive in nature but different from the 14.5 Velocity and acceleration periodic motion of a planet. Here, the object moves to and fro about a mean position. The pendulum of a wall clock executes in simple harmonic motion a similar motion. Examples of such periodic to and fro motion abound: a boat tossing up and down in a river, the 14.6 Force law for simple piston in a steam engine going back and forth, etc. Such a motion is termed as oscillatory motion. In this chapter we harmonic motion study this motion. 14.7 Energy in simple harmonic The study of oscillatory motion is basic to physics; its concepts are required for the understanding of many physical motion phenomena. In musical instruments, like the sitar, the guitar or the violin, we come across vibrating strings that produce 14.8 Some systems executing pleasing sounds. The membranes in drums and diaphragms in telephone and speaker systems vibrate to and fro about simple harmonic motion their mean positions. The vibrations of air molecules make the propagation of sound possible. In a solid, the atoms vibrate 14.9 Damped simple harmonic about their equilibrium positions, the average energy of vibrations being proportional to temperature. AC power motion supply give voltage that oscillates alternately going positive and negative about the mean value (zero). 14.10 Forced oscillations and The description of a periodic motion, in general, and resonance oscillatory motion, in particular, requires some fundamental concepts, like period, frequency, displacement, amplitude Summary and phase. These concepts are developed in the next section. Points to ponder Exercises Additional Exercises 2020-21

342 PHYSICS 14.2 PERIODIC AND OSCILLATORY MOTIONS Very often, the body undergoing periodic motion has an equilibrium position somewhere Fig. 14.1 shows some periodic motions. Suppose inside its path. When the body is at this position an insect climbs up a ramp and falls down, it no net external force acts on it. Therefore, if it is comes back to the initial point and repeats the left there at rest, it remains there forever. If the process identically. If you draw a graph of its body is given a small displacement from the height above the ground versus time, it would position, a force comes into play which tries to look something like Fig. 14.1 (a). If a child climbs bring the body back to the equilibrium point, up a step, comes down, and repeats the process giving rise to oscillations or vibrations. For identically, its height above the ground would example, a ball placed in a bowl will be in look like that in Fig. 14.1 (b). When you play the equilibrium at the bottom. If displaced a little game of bouncing a ball off the ground, between from the point, it will perform oscillations in the your palm and the ground, its height versus time bowl. Every oscillatory motion is periodic, but graph would look like the one in Fig. 14.1 (c). every periodic motion need not be oscillatory. Note that both the curved parts in Fig. 14.1 (c) Circular motion is a periodic motion, but it is are sections of a parabola given by the Newton’s not oscillatory. equation of motion (see section 3.6), There is no significant difference between h = ut + 1 gt 2 for downward motion, and oscillations and vibrations. It seems that when the frequency is small, we call it oscillation (like, 2 the oscillation of a branch of a tree), while when the frequency is high, we call it vibration (like, h = ut – 1 gt 2 for upward motion, the vibration of a string of a musical instrument). 2 Simple harmonic motion is the simplest form of oscillatory motion. This motion arises when with different values of u in each case. These the force on the oscillating body is directly proportional to its displacement from the mean are examples of periodic motion. Thus, a motion position, which is also the equilibrium position. Further, at any point in its oscillation, this force that repeats itself at regular intervals of time is is directed towards the mean position. called periodic motion. In practice, oscillating bodies eventually come to rest at their equilibrium positions (a) because of the damping due to friction and other dissipative causes. However, they can be forced (b) to remain oscillating by means of some external periodic agency. We discuss the phenomena of (c) damped and forced oscillations later in the chapter. Fig. 14.1 Examples of periodic motion. The period T is shown in each case. Any material medium can be pictured as a collection of a large number of coupled oscillators. The collective oscillations of the constituents of a medium manifest themselves as waves. Examples of waves include water waves, seismic waves, electromagnetic waves. We shall study the wave phenomenon in the next chapter. 14.2.1 Period and frequency We have seen that any motion that repeats itself at regular intervals of time is called periodic motion. The smallest interval of time after which the motion is repeated is called its period. Let us denote the period by the symbol T. Its SI unit is second. For periodic motions, 2020-21

OSCILLATIONS 343 which are either too fast or too slow on the scale as a displacement variable [see Fig.14.2(b)]. The of seconds, other convenient units of time are term displacement is not always to be referred used. The period of vibrations of a quartz crystal is expressed in units of microseconds (10–6 s) Fig. 14.2(a) A block attached to a spring, the other abbreviated as µs. On the other hand, the orbital end of which is fixed to a rigid wall. The period of the planet Mercury is 88 earth days. block moves on a frictionless surface. The The Halley’s comet appears after every 76 years. motion of the block can be described in terms of its distance or displacement x The reciprocal of T gives the number of from the equilibrium position. repetitions that occur per unit time. This quantity is called the frequency of the periodic motion. It is represented by the symbol ν. The relation between ν and T is ν = 1/T (14.1) The unit of ν is thus s–1. After the discoverer of radio waves, Heinrich Rudolph Hertz (1857–1894), a special name has been given to the unit of frequency. It is called hertz (abbreviated as Hz). Thus, 1 hertz = 1 Hz =1 oscillation per second =1s–1 (14.2) Note, that the frequency, ν, is not necessarily an integer. Example 14.1 On an average, a human heart is found to beat 75 times in a minute. Calculate its frequency and period. Answer The beat frequency of heart = 75/(1 min) Fig.14.2(b) An oscillating simple pendulum; its = 75/(60 s) motion can be described in terms of angular displacement θ from the vertical. = 1.25 s–1 = 1.25 Hz in the context of position only. There can be The time period T = 1/(1.25 s–1) many other kinds of displacement variables. The = 0.8 s voltage across a capacitor, changing with time 14.2.2 Displacement in an A C circuit, is also a displacement variable. In section 4.2, we defined displacement of a In the same way, pressure variations in time in particle as the change in its position vector. In the propagation of sound wave, the changing this chapter, we use the term displacement electric and magnetic fields in a light wave are in a more general sense. It refers to change examples of displacement in different contexts. with time of any physical property under The displacement variable may take both consideration. For example, in case of rectilinear positive and negative values. In experiments on motion of a steel ball on a surface, the distance oscillations, the displacement is measured for from the starting point as a function of time is different times. its position displacement. The choice of origin The displacement can be represented by a is a matter of convenience. Consider a block mathematical function of time. In case of periodic attached to a spring, the other end of the spring motion, this function is periodic in time. One of is fixed to a rigid wall [see Fig.14.2(a)]. Generally, the simplest periodic functions is given by it is convenient to measure displacement of the f (t) = A cos ωt (14.3a) body from its equilibrium position. For an oscillating simple pendulum, the angle from the If the argument of this function, ωt, is increased by an integral multiple of 2π radians, vertical as a function of time may be regarded 2020-21

344 PHYSICS the value of the function remains the same. The = 2 sin [ω (t + 2π/ω) + π/4] function f (t ) is then periodic and its period, T, is given by The periodic time of the function is 2π/ω. T = 2π (14.3b) (ii) This is an example of a periodic motion. It ω can be noted that each term represents a Thus, the function f (t) is periodic with period T, periodic function with a different angular f (t) = f (t+T ) frequency. Since period is the least interval The same result is obviously correct if we of time after which a function repeats its consider a sine function, f (t ) = A sin ωt. Further, phvaealrsuioead,ps2einπr/iωo4dtωhπa=/sωTa0=/pT4e0.r/Tio2hd;eaTpn0=edr2isoπidn/ωo4f;ωtchtoehs fa2isrωsatt a linear combination of sine and cosine functions term is a multiple of the periods of the last like, f (t) = A sin ωt + B cos ωt (14.3c) two terms. Therefore, the smallest interval is also a periodic function with the same period of time after which the sum of the three T. Taking, terms repeats is T0, and thus, the sum is a periodic function with a period 2π/ω. A = D cos φ and B = D sin φ (iii) The function e–ωt is not periodic, it Eq. (14.3c) can be written as, decreases monotonically with increasing f (t) = D sin (ωt + φ ) , (14.3d) time and tends to zero as t → ∞ and thus, Here D and φ are constant given by never repeats its value. D= A2 + B2 and φ= tan–1  B  (iv) The function log (ωt) increases A monotonically with time t. It, therefore, The great importance of periodic sine and never repeats its value and is a non- cosine functions is due to a remarkable result proved by the French mathematician, Jean periodic function. It may be noted that as Baptiste Joseph Fourier (1768 –1830): Any t → ∞, log(ωt) diverges to ∞. It, therefore, periodic function can be expressed as a cannot represent any kind of physical superposition of sine and cosine functions of different time periods with suitable displacement. coefficients. 14.3 SIMPLE HARMONIC MOTION Consider a particle oscillating back and forth about the origin of an x-axis between the limits +A and –A as shown in Fig. 14.3. This oscillatory motion is said to be simple harmonic if the Example 14.2 Which of the following Fig. 14.3 A particle vibrating back and forth about the origin of x-axis, between the limits +A functions of time represent (a) periodic and and –A. (b) non-periodic motion? Give the period for displacement x of the particle from the origin each case of periodic motion [ω is any varies with time as : positive constant]. (i) sin ωt + cos ωt (ii) sin ωt + cos 2 ωt + sin 4 ωt (iii) e–ωt (iv) log (ωt) x (t) = A cos (ω t + φ ) (14.4) Answer where A, ω and φ are constants. (i) sin ωt + cos ωt is a periodic function, it can Thus, simple harmonic motion (SHM) is not also be written as 2 sin (ωt + π/4). any periodic motion but one in which Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π) displacement is a sinusoidal function of time. Fig. 14.4 shows the positions of a particle executing SHM at discrete value of time, each interval of time being T/4, where T is the period 2020-21

OSCILLATIONS 345 The amplitutde A of SHM is the magnitude of maximum displacement of the particle. [Note, A can be taken to be positive without any loss of generality]. As the cosine function of time varies from +1 to –1, the displacement varies between the extremes A and – A. Two simple harmonic motions may have same ω and φ but different amplitudes A and B, as shown in Fig. 14.7 (a). Fig. 14.4 The location of the particle in SHM at the discrete values t = 0, T/4, T/2, 3T/4, T, 5T/4. The time after which motion repeats Fig. 14.7 (a) A plot of displacement as a function of itself is T. T will remain fixed, no matter time as obtained from Eq. (14.4) with what location you choose as the initial (t = φ = 0. The curves 1 and 2 are for two 0) location. The speed is maximum for zero different amplitudes A and B. displacement (at x = 0) and zero at the extremes of motion. While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of motion. Fig. 14.5 plots the graph of x versus t, of the particle at any time t is determined by the argument (ωt + φ) in the cosine function. This which gives the values of displacement as a time-dependent quantity, (ωt + φ) is called the phase of the motion. The value of plase at t = 0 continuous function of time. The quantities A, is φ and is called the phase constant (or phase ω and φ which characterize a given SHM have angle). If the amplitude is known, φ can be standard names, as summarised in Fig. 14.6. determined from the displacement at t = 0. Two simple harmonic motions may have the same A Let us understand these quantities. and ω but different phase angle φ, as shown in Fig. 14.7 (b). Fig. 14.5 Displacement as a continuous function of time for simple harmonic motion. x (t) : displacement x as a function of time t Fig. 14.7 (b) A plot obtained from Eq. (14.4). The A : amplitude curves 3 and 4 are for φ = 0 and -π/4 ω : angular frequency respectively. The amplitude A is same for ωt + φ : phase (time-dependent) φ : phase constant both the plots. Fig. 14.6 The meaning of standard symbols in Eq. (14.4) 2020-21

346 PHYSICS Finally, the quantity ω can be seen to be This function represents a simple harmonic motion having a period T = 2π/ω and a related to the period of motion T. Taking, for phase angle (–π/4) or (7π/4) simplicity, φ = 0 in Eq. (14.4), we have (b) sin2 ωt x(t ) = A cos ωt (14.5) = ½ – ½ cos 2 ωt Since the motion has a period T, x (t) is equal to The function is periodic having a period x (t + T ). That is, T = π/ω. It also represents a harmonic motion with the point of equilibrium A cos ωt = A cos ω (t + T ) (14.6) occurring at ½ instead of zero. Now the cosine function is periodic with period 2π, i.e., it first repeats itself when the argument 14.4 SIMPLE HARMONIC MOTION AND changes by 2π. Therefore, UNIFORM CIRCULAR MOTION ω(t + T ) = ωt + 2π In this section, we show that the projection of uniform circular motion on a diameter of the that is ω = 2π/ T (14.7) circle follows simple harmonic motion. A simple experiment (Fig. 14.9) helps us visualise ω is called the angular frequency of SHM. Its this connection. Tie a ball to the end of a string and make it move in a horizontal plane about S.I. unit is radians per second. Since the a fixed point with a constant angular speed. frequency of oscillations is simply 1/T, ω is 2π The ball would then perform a uniform circular motion in the horizontal plane. Observe the times the frequency of oscillation. Two simple ball sideways or from the front, fixing your harmonic motions may have the same A and φ, attention in the plane of motion. The ball will but different ω, as seen in Fig. 14.8. In this plot appear to execute to and fro motion along a horizontal line with the point of rotation as the curve (b) has half the period and twice the the midpoint. You could alternatively observe the shadow of the ball on a wall which is frequency of the curve (a). perpendicular to the plane of the circle. In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing. Fig. 14.8 Plots of Eq. (14.4) for φ = 0 for two different periods. Example 14.3 Which of the following Fig. 14.9 Circular motion of a ball in a plane viewed functions of time represent (a) simple edge-on is SHM. harmonic motion and (b) periodic but not simple harmonic? Give the period for each Fig. 14.10 describes the same situation case. mathematically. Suppose a particle P is moving (1) sin ωt – cos ωt uniformly on a circle of radius A with angular (2) sin2 ωt speed ω. The sense of rotation is anticlockwise. The initial position vector of the particle, i.e., Answer the vector OP at t = 0 makes an angle of φ with (a) sin ωt – cos ωt the positive direction of x-axis. In time t, it will cover a further angle ωt and its position vector = sin ωt – sin (π/2 – ωt) = 2 cos (π/4) sin (ωt – π/4) = √2 sin (ωt – π/4) 2020-21

OSCILLATIONS 347 Example 14.4 The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. Fig. 14.10 Answer will make an angle of ωt + φ with the +ve (a) At t = 0, OP makes an angle of 45o = π/4 rad x-axis. Next, consider the projection of the with the (positive direction of ) x-axis. After position vector OP on the x-axis. This will be OP′. The position of P′ on the x-axis, as the time t, it covers an angle 2πt in the particle P moves on the circle, is given by T x(t ) = A cos (ωt + φ ) anticlockwise sense, and makes an angle which is the defining equation of SHM. This of 2πt + π with the x-axis. shows that if P moves uniformly on a circle, T 4 its projection P′ on a diameter of the circle executes SHM. The particle P and the circle The projection of OP on the x-axis at time t on which it moves are sometimes referred to is given by, as the reference particle and the reference circle, respectively. x (t) = A cos  2π t + π  T 4 We can take projection of the motion of P on For T = 4 s, any diameter, say the y-axis. In that case, the displacement y(t) of P′ on the y-axis is given by x(t) = A cos  2π t + π  4 4 y = A sin (ωt + φ ) which is a SHM of amplitude A, period 4 s, which is also an SHM of the same amplitude as that of the projection on x-axis, but differing and an initial phase* = π . by a phase of π/2. 4 In spite of this connection between circular motion and SHM, the force acting on a particle in linear simple harmonic motion is very different from the centripetal force needed to keep a particle in uniform circular motion. * The natural unit of angle is radian, defined through the ratio of arc to radius. Angle is a dimensionless quantity. Therefore it is not always necessary to mention the unit ‘radian’ when we use π, its multiples or submultiples. The conversion between radian and degree is not similar to that between metre and centimetre or mile. If the argument of a trigonometric function is stated without units, it is understood that the unit is radian. On the other hand, if degree is to be used as the unit of angle, then it must be shown explicitly. For example, sin(150) means sine of 15 degree, but sin(15) means sine of 15 radians. Hereafter, we will often drop ‘rad’ as the unit, and it should be understood that whenever angle is mentioned as a numerical value, without units, it is to be taken as radians. 2020-21

348 PHYSICS (b) In this case at t = 0, OP makes an angle of where the negative sign shows that v (t) has a direction opposite to the positive direction of 90o = π with the x-axis. After a time t, it x-axis. Eq. (14.9) gives the instantaneous velocity of a particle executing SHM, where 2 2π t in the clockwise displacement is given by Eq. (14.4). We can, of covers an angle of course, obtain this equation without using geometrical argument, directly by differentiating T (Eq. 14.4) with respect of t: sense and makes an angle of  π − 2π t  2 T with the x-axis. The projection of OP on the x-axis at time t is given by v(t) = d x(t ) (14.10) dt x(t) = B cos  π − 2π t  The method of reference circle can be similarly 2 T used for obtaining instantaneous acceleration = B sin  2π t  of a particle undergoing SHM. We know that the T centripetal acceleration of a particle P in uniform For T = 30 s, circular motion has a magnitude v2/A or ω2A,   and it is directed towards the centre i.e., the x(t) = B sin π t direction is along PO. The instantaneous 15 acceleration of the projection particle P′ is then Writing this as x (t ) = B cos 1π5 t − π2 , and (See Fig. 14.12) comparing with Eq. (14.4). We find that this a (t) = –ω2A cos (ωt + φ) represents a SHM of amplitude B, period 30 s, = –ω2x (t) (14.11) and an initial phase of − π . 2 14.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A. v = ωA (14.8) The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the Fig. 14.12 The acceleration, a(t), of the particle P′ is the projection of the acceleration a of the geometry of Fig. 14.11, it is clear that the velocity reference particle P. of the projection particle P′ at time t is Eq. (14.11) gives the acceleration of a particle in SHM. The same equation can again be v(t ) = –ωA sin (ωt + φ ) (14.9) obtained directly by differentiating velocity v(t) given by Eq. (14.9) with respect to time: a(t ) = d v(t ) (14.12) dt Fig. 14.11 The velocity, v (t), of the particle P′ is the We note from Eq. (14.11) the important property that acceleration of a particle in SHM projection of the velocity v of the is proportional to displacement. For x(t) > 0, a(t) < 0 and for x(t) < 0, a(t) > 0. Thus, whatever reference particle, P. 2020-21

OSCILLATIONS 349 the value of x between –A and A, the acceleration (b) Using Eq. (14.9), the speed of the body a(t) is always directed towards the centre. = – (5.0 m)(2π s–1) sin [(2π s–1) ×1.5 s For simplicity, let us put φ = 0 and write the + π/4] expression for x (t), v (t) and a(t) = – (5.0 m)(2π s–1) sin [(3π + π/4)] x(t) = A cos ωt, v(t) = – ω Asin ωt, a(t)=–ω2 A cos ωt = 10 π × 0.707 m s–1 The corresponding plots are shown in Fig. 14.13. All quantities vary sinusoidally with time; only = 22 m s–1 their maxima differ and the different plots differ in phase. x varies between –A to A; v(t) varies (c) Using Eq.(14.10), the acceleration of the from –ωA to ωA and a(t) from –ω2A to ω2A. With respect to displacement plot, velocity plot has a body phase difference of π/2 and acceleration plot = –(2π s–1)2 × displacement has a phase difference of π. = – (2π s–1)2 × (–3.535 m) = 140 m s–2 Fig. 14.13 Displacement, velocity and acceleration of a particle in simple harmonic motion have 14.6 FORCE LAW FOR SIMPLE HARMONIC the same period T, but they differ in phase MOTION Example 14.5 A body oscillates with SHM Using Newton’s second law of motion, and the according to the equation (in SI units), expression for acceleration of a particle undergoing SHM (Eq. 14.11), the force acting x = 5 cos [2π t + π/4]. on a particle of mass m in SHM is At t = 1.5 s, calculate the (a) displacement, (b) speed and (c) acceleration of the body. F (t ) = ma (14.13) Answer The angular frequency ω of the body = –mω2 x (t ) (14.14a) = 2π s–1 and its time period T = 1 s. At t = 1.5 s i.e., F (t ) = –k x (t ) (a) displacement = (5.0 m) cos [(2π s–1) × where k = mω2 1.5 s + π/4] or ω = k (14.14b) = (5.0 m) cos [(3π + π/4)] m = –5.0 × 0.707 m = –3.535 m Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (14.4) for displacement or by Eq. (14.13) that gives its force law. Going from Eq. (14.4) to Eq. (14.13) required us to differentiate two times. Likewise, by integrating the force law Eq. (14.13) two times, we can get back Eq. (14.4). Note that the force in Eq. (14.13) is linearly proportional to x(t). A particle oscillating under such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force may contain small additional terms proportional to x2, x3, etc. These then are called non-linear oscillators. Example 14.6 Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 14.14. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations. 2020-21

350 PHYSICS Fig. 14.14 14.7 ENERGY IN SIMPLE HARMONIC MOTION Answer Let the mass be displaced by a small distance x to the right side of the equilibrium Both kinetic and potential energies of a particle position, as shown in Fig. 14.15. Under this in SHM vary between zero and their maximum situation the spring on the left side gets values. Fig. 14.15 In section14.5 we have seen that the velocity of a particle executing SHM, is a periodic elongated by a length equal to x and that on function of time. It is zero at the extreme positions the right side gets compressed by the same of displacement. Therefore, the kinetic energy (K) length. The forces acting on the mass are of such a particle, which is defined as then, K = 1 mv2 F1 = –k x (force exerted by the spring on 2 the left side, trying to pull the mass towards the mean = 1 m ω2 A2 sin2(ωt + φ ) position) 2 F2 = –k x (force exerted by the spring on = 1 k A2 sin2 (ωt + φ ) (14.15) the right side, trying to push the 2 mass towards the mean position) is also a periodic function of time, being zero when the displacement is maximum and The net force, F, acting on the mass is then maximum when the particle is at the mean given by, position. Note, since the sign of v is immaterial in K, the period of K is T/2. F = –2kx Hence the force acting on the mass is What is the potential energy (U) of a particle proportional to the displacement and is directed executing simple harmonic motion? In towards the mean position; therefore, the motion Chapter 6, we have seen that the concept of executed by the mass is simple harmonic. The potential energy is possible only for conservative time period of oscillations is, forces. The spring force F = –kx is a conservative force, with associated potential energy T = 2π m 2k U = 1k x2 (14.16) 2 Hence the potential energy of a particle executing simple harmonic motion is, U(x) = 1 k x 2 2 = 1 k A2 cos2(ωt + φ) (14.17) 2 Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period T/2, being zero at the mean position and maximum at the extreme displacements. 2020-21

OSCILLATIONS 351 It follows from Eqs. (14.15) and (14.17) that Observe that both kinetic energy and the total energy, E, of the system is, potential energy in SHM are seen to be always positive in Fig. 14.16. Kinetic energy can, of E =U+K course, be never negative, since it is proportional to the square of speed. Potential = 1 k A2 cos2(ωt + φ ) + 1 k A2 sin2(ωt + φ ) energy is positive by choice of the undermined 22 constant in potential energy. Both kinetic energy and potential energy peak twice during = 1 k A2 cos2 (ωt +φ) + sin2 (ωt + φ ) each period of SHM. For x = 0, the energy is 2 kinetic; at the extremes x = ±A, it is all potential energy. In the course of motion between these Using the familiar trigonometric identity, the limits, kinetic energy increases at the expense value of the expression in the brackets is unity. of potential energy or vice-versa. Thus, Example 14.7 A block whose mass is 1 kg E = 1 k A2 (14.18) is fastened to a spring. The spring has a 2 spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its The total mechanical energy of a harmonic equilibrium position at x = 0 on a frictionless oscillator is thus independent of time as expected surface from rest at t = 0. Calculate the for motion under any conservative force. The kinetic, potential and total energies of the time and displacement dependence of the block when it is 5 cm away from the mean potential and kinetic energies of a linear simple position. harmonic oscillator are shown in Fig. 14.16. Answer The block executes SHM, its angular frequency, as given by Eq. (14.14b), is ω= k m 50 N m– 1 = 1kg Fig. 14.16 Kinetic energy, potential energy and total = 7.07 rad s–1 energy as a function of time [shown in (a)] Its displacement at any time t is then given by, and displacement [shown in (b)] of a particle in SHM. The kinetic energy and potential x(t) = 0.1 cos (7.07t) energy both repeat after a period T/2. The total energy remains constant at all t or x. Therefore, when the particle is 5 cm away from the mean position, we have 0.05 = 0.1 cos (7.07t) Or cos (7.07t) = 0.5 and hence sin (7.07t) = 3 = 0.866 2 2020-21

352 PHYSICS Then, the velocity of the block at x = 5 cm is Fig. 14.17 A linear simple harmonic oscillator consisting of a block of mass m attached = 0.1 × 7.07 × 0.866 m s–1 to a spring. The block moves over a frictionless surface. The box, when pulled = 0.61 m s–1 or pushed and released, executes simple harmonic motion. Hence the K.E. of the block, spring is in equilibrium. The positions marked = 1 m v2 as –A and +A indicate the maximum 2 displacements to the left and the right of the mean position. We have already learnt that = ½[1kg × (0.6123 m s–1 )2 ] springs have special properties, which were first discovered by the English physicist Robert = 0.19 J Hooke. He had shown that such a system when deformed, is subject to a restoring force, the The P.E. of the block, magnitude of which is proportional to the deformation or the displacement and acts in = 1 k x2 opposite direction. This is known as Hooke’s 2 law (Chapter 9). It holds good for displacements small in comparison to the length of the spring. = ½(50 N m–1 × 0.05 m × 0.05 m) At any time t, if the displacement of the block = 0.0625 J from its mean position is x, the restoring force F The total energy of the block at x = 5 cm, acting on the block is, = K.E. + P.E. F (x) = –k x (14.19) = 0.25 J The constant of proportionality, k, is called the spring constant, its value is governed by the we also know that at maximum displacement, elastic properties of the spring. A stiff spring has K.E. is zero and hence the total energy of the large k and a soft spring has small k. Equation system is equal to the P.E. Therefore, the total (14.19) is same as the force law for SHM and energy of the system, therefore the system executes a simple harmonic motion. From Eq. (14.14) we have, = ½(50 N m–1 × 0.1 m × 0.1 m ) ω= k (14.20) = 0.25 J which is same as the sum of the two energies at m a displacement of 5 cm. This is in conformity with the principle of conservation of energy. and the period, T, of the oscillator is given by, 14.8 SOME SYSTEMS EXECUTING SIMPLE T = 2π m (14.21) HARMONIC MOTION k There are no physical examples of absolutely Stiff springs have high value of k (spring pure simple harmonic motion. In practice we come across systems that execute simple constant). A block of small mass m attached to harmonic motion approximately under certain conditions. In the subsequent part of this a stiff spring will have, according to Eq. (14.20), section, we discuss the motion executed by some such systems. large oscillation frequency, as expected 14.8.1 Oscillations due to a Spring physically. The simplest observable example of simple harmonic motion is the small oscillations of a block of mass m fixed to a spring, which in turn is fixed to a rigid wall as shown in Fig. 14.17. The block is placed on a frictionless horizontal surface. If the block is pulled on one side and is released, it then executes a to and fro motion about the mean position. Let x = 0, indicate the position of the centre of the block when the 2020-21

OSCILLATIONS 353 Example 14.8 A 5 kg collar is attached of pendulum. You can also make your own to a spring of spring constant 500 N m–1. It pendulum by tying a piece of stone to a long slides without friction over a horizontal rod. unstretchable thread, approximately 100 cm The collar is displaced from its equilibrium long. Suspend your pendulum from a suitable position by 10.0 cm and released. Calculate support so that it is free to oscillate. Displace (a) the period of oscillation, the stone to one side by a small distance and (b) the maximum speed and let it go. The stone executes a to and fro motion, (c) maximum acceleration of the collar. it is periodic with a period of about two seconds. Answer (a) The period of oscillation as given by We shall show that this periodic motion is Eq. (14.21) is, simple harmonic for small displacements from the mean position. Consider simple pendulum T = 2π m = 2π 5.0 kg — a small bob of mass m tied to an inextensible k 500 N m−1 massless string of length L. The other end of the string is fixed to a rigid support. The bob = (2π/10) s oscillates in a plane about the vertical line through the support. Fig. 14.18(a) shows this = 0.63 s system. Fig. 14.18(b) is a kind of ‘free-body’ diagram of the simple pendulum showing the (b) The velocity of the collar executing SHM is forces acting on the bob. given by, v(t ) = –Aω sin (ωt + φ) The maximum speed is given by, vm = Aω k = 0.1 × m = 0.1 × 500 N m –1 (a) 5 kg (b) = 1 m s–1 Fig. 14.18 (a) A bob oscillating about its mean and it occurs at x = 0 (c) The acceleration of the collar at the position. (b) The radial force T-mg cosθ displacement x (t ) from the equilibrium is provides centripetal force but no torque given by, about the support. The tangential force mg sinθ provides the restoring torque. a (t) = –ω2 x(t) k = – x(t) m Therefore, the maximum acceleration is, amax = ω2 A = 500 N m– 1 x 0.1 m 5 kg = 10 m s–2 and it occurs at the extremities. 14.8.2 The Simple Pendulum It is said that Galileo measured the periods of a swinging chandelier in a church by his pulse beats. He observed that the motion of the chandelier was periodic. The system is a kind 2020-21

354 PHYSICS Let θ be the angle made by the string with SHM - how small should the amplitude be? the vertical. When the bob is at the mean When you perform the experiment to position, θ = 0 determine the time period of a simple There are only two forces acting on the bob; pendulum, your teacher tells you to keep the tension T along the string and the vertical the amplitude small. But have you ever force due to gravity (=mg). The force mg can be resolved into the component mg cosθ along the asked how small is small? Should the string and mg sinθ perpendicular to it. Since amplitude to 50, 20, 10, or 0.50? Or could it the motion of the bob is along a circle of length be 100, 200, or 300? L and centre at the support point, the bob has a radial acceleration (ω2L) and also a tangental To appreciate this, it would be better to acceleration; the latter arises since motion along measure the time period for different the arc of the circle is not uniform. The radial amplitudes, up to large amplitudes. Of acceleration is provided by the net radial force course, for large oscillations, you will have T –mg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to to take care that the pendulum oscillates in a vertical plane. Let us denote the time period for small-amplitude oscillations as work with torque about the support since the T (0) and write the time period for amplitude radial force gives zero torque. Torque τ about the support is entirely provided by the tangental θ0 as T(θ0) = cT (0), where c is the multiplying factor. If you plot a graph of c versus θ0, component of force you will get values somewhat like this: τ = –L (mg sinθ ) (14.22) θ : 200 450 500 700 900 0 1.02 1.04 1.05 1.10 1.18 This is the restoring torque that tends to reduce c: angular displacement — hence the negative This means that the error in the time period is about 2% at an amplitude of 200, sign. By Newton’s law of rotational motion, 5% at an amplitude of 500, and 10% at an amplitude of 700 and 18% at an amplitude τ = Iα (14.23) of 900. where I is the moment of inertia of the system In the experiment, you will never be able about the support and α is the angular to measure T (0) because this means there are no oscillations. Even theoretically, acceleration. Thus, sin θ is exactly equal to θ only for θ = 0. There will be some inaccuracy for all other I α = –m g sin θ L (14.24) values of θ . The difference increases with increasing θ . Therefore we have to decide Or, how much error we can tolerate. No measurement is ever perfectly accurate. α = − m g L sin θ (14.25) You must also consider questions like I these: What is the accuracy of the stopwatch? What is your own accuracy in We can simplify Eq. (14.25) if we assume that starting and stopping the stopwatch? You the displacement θ is small. We know that sin θ will realise that the accuracy in your measurements at this level is never better can be expressed as, than 5% or 10%. Since the above table shows that the time period of the pendulum sin θ = θ − θ3 + θ 5 ± ... (14.26) increases hardly by 5% at an amplitude of 3! 5! 500 over its low amplitude value, you could very well keep the amplitude to be 50° in where θ is in radians. your experiments. Now if θ is small, sin θ can be approximated by θ and Eq. (14.25) can then be written as, α = − mgL θ (14.27) I In Table 14.1, we have listed the angle θ in degrees, its equivalent in radians, and the value 2020-21

OSCILLATIONS 355 of the function sin θ . From this table it can be = 9.8(m s–2 ) × 4(s2 ) seen that for θ as large as 20 degrees, sin θ is 4π 2 nearly the same as θ expressed in radians. =1m Table 14.1 sin θ as a function of angle θ 14.9 DAMPED SIMPLE HARMONIC MOTION (degrees) (radians) sin We know that the motion of a simple pendulum, Equation (14.27) is mathematically, identical swinging in air, dies out eventually. Why does it happen ? This is because the air drag and the to Eq. (14.11) except that the variable is angular friction at the support oppose the motion of the pendulum and dissipate its energy gradually. displacement. Hence we have proved that for The pendulum is said to execute damped small θ, the motion of the bob is simple harmonic. oscillations. In dampled oscillations, the energy From Eqs. (14.27) and (14.11), of the system is dissipated continuously; but, for small damping, the oscillations remain ω = mgL approximately periodic. The dissipating forces I are generally the frictional forces. To understand the effect of such external forces on the motion and of an oscillator, let us consider a system as shown in Fig. 14.19. Here a block of mass m connected to an elastic spring of spring constant k oscillates vertically. If the block is pushed down a little and released, its angular frequency of oscillation is ω = k m , as seen in Eq. (14.20). T = 2π I (14.28) However, in practice, the surrounding medium mgL (air) will exert a damping force on the motion of the block and the mechanical energy of the Now since the string of the simple pendulum block-spring system will decrease. The energy is massless, the moment of inertia I is simply loss will appear as heat of the surrounding mL2. Eq. (14.28) then gives the well-known medium (and the block also) [Fig. 14.19]. formula for time period of a simple pendulum. T = 2π L (14.29) g Example 14.9 What is the length of a simple pendulum, which ticks seconds ? Answer From Eq. (14.29), the time period of a simple pendulum is given by, T = 2π L g From this relation one gets, L = gT 2 4π 2 The time period of a simple pendulum, which Fig. 14.19 The viscous surrounding medium exerts ticks seconds, is 2 s. Therefore, for g = 9.8 m s–2 a damping force on an oscillating spring, and T = 2 s, L is eventually bringing it to rest. 2020-21

356 PHYSICS The damping force depends on the nature of ω '= k − b2 (14.34) the surrounding medium. If the block is m 4m2 immersed in a liquid, the magnitude of damping will be much greater and the dissipation of In this function, the cosine function has a energy much faster. The damping force is generally proportional to velocity of the bob. period 2π/ω′ but the function x(t) is not strictly [Remember Stokes’ Law, Eq. (10.19)] and acts opposite to the direction of velocity. If the periodic because of the factor e–b t/2m which damping force is denoted by Fd, we have decreases continuously with time. However, if the decrease is small in one time period T, the motion Fd = –b v (14.30) represented by Eq. (14.33) is approximately periodic. where the positive constant b depends on characteristics of the medium (viscosity, for The solution, Eq. (14.33), can be graphically example) and the size and shape of the block, represented as shown in Fig. 14.20. We can etc. Eq. (14.30) is usually valid only for small regard it as a cosine function whose amplitude, velocity. which is Ae–b ,t/2m gradually decreases with time. When the mass m is attached to the spring (hung vertically as shown in Fig. 14.19) and released, the spring will elongate a little and the mass will settle at some height. This position, shown by O in Fig 14.19, is the equilibrium position of the mass. If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is FS = –kx, where x is the displacement* of the mass from its equilibrium position. Thus, the total force acting on the mass at any time t, is F = –kx –bv. If a(t) is the acceleration of mass at time t, Fig. 14.20 A damped oscillator is approximately periodic with decreasing amplitude of then by Newton’s Law of Motion applied along oscillation. With greater damping, oscillations die out faster. the direction of motion, we have m a(t) = –k x(t) – b v(t) (14.31) Here we have dropped the vector notation because we are discussing one-dimensional Now the mechanical energy of the undamped oscillator is 1/2 kA2. For a damped oscillator, motion. the amplitude is not constant but depends on time. For small damping, we may use the same Using the first and second derivatives of x (t) expression but regard the amplitude as A e–bt/2m. for v (t) and a (t), respectively, we have m d2x + b dx + k x = 0 (14.32) dt 2 dt E(t ) = 1 k A2 e–b t/m The solution of Eq. (14.32) describes the 2 (14.35) motion of the block under the influence of a damping force which is proportional to velocity. Equation (14.35) shows that the total energy The solution is found to be of the form of the system decreases exponentially with time. x(t) = A e–b t/2m cos (ω′t + φ ) Note that small damping means that the (14.33) where A is the amplitude and ω ′ is the angular dimensionless ratio  b  is much less than 1. frequency of the damped oscillator given by, km * Under gravity, the block will be at a certain equilibrium position O on the spring; x here represents the displacement from that position. 2020-21

OSCILLATIONS 357 Of course, as expected, if we put b = 0, all Or ½ = exp (–bt1/2/m) equations of a damped oscillator in this section reduce to the corresponding equations of an ln (1/2) = –(bt1/2/m) undamped oscillator. Or t1/2 = 0.693 × 200 g Example 14.10 For the damped oscillator 40 g s–1 shown in Fig. 14.19, the mass m of the block is 200 g, k = 90 N m–1 and the damping = 3.46 s constant b is 40 g s–1. Calculate (a) the This is just half of the decay period for period of oscillation, (b) time taken for its amplitude. This is not surprising, because, amplitude of vibrations to drop to half of according to Eqs. (14.33) and (14.35), energy its initial value, and (c) the time taken for depends on the square of the amplitude. Notice its mechanical energy to drop to half its that there is a factor of 2 in the exponents of initial value. the two exponentials. Answer (a) We see that km = 90×0.2 = 18 kg N 14.10 FORCED OSCILLATIONS AND RESONANCE m–1 = kg2 s–2; therefore km = 4.243 kg s–1, and b = 0.04 kg s–1. Therefore, b is much less than When a system (such as a simple pendulum or km . Hence, the time period T from Eq. (14.34) a block attached to a spring) is displaced from is given by its equilibrium position and released, it oscillates T = 2π m with its natural frequency ω, and the oscillations k are called free oscillations. All free oscillations eventually die out because of the ever present damping forces. However, an external agency can maintain these oscillations. These are called forced or driven oscillations. We consider the = 2π 0.2 kg case when the external force is itself periodic, 90 N m –1 with a frequency ω called the driven frequency. d The most important fact of forced periodic = 0.3 s oscillations is that the system oscillates not with (b) Now, from Eq. (14.33), the time, T1/2, for the its natural frequency ω, but at the frequency ω amplitude to drop to half of its initial value is d given by, of the external agency; the free oscillations die out due to damping. The most familiar example of forced oscillation is when a child in a garden swing periodically presses his feet against the ln(1/2) ground (or someone else periodically gives the b/2m T1/2 = child a push) to maintain the oscillations. Suppose an external force F(t) of amplitude = 0.693 × 2 × 200 s F0 that varies periodically with time is applied 40 to a damped oscillator. Such a force can be = 6.93 s represented as, (c) For calculating the time, t1/2, for its F(t ) = Fo cos ω t (14.36) mechanical energy to drop to half its initial value d we make use of Eq. (14.35). From this equation we have, The motion of a particle under the combined action of a linear restoring force, damping force E (t1/2)/E (0) = exp (–bt1/2/m) and a time dependent driving force represented by Eq. (14.36) is given by, m a(t ) = –k x(t ) – bv(t ) + Fo cos ω t (14.37a) d Substituting d2x/dt2 for acceleration in Eq. (14.37a) and rearranging it, we get 2020-21

358 PHYSICS m d2 x + b dx + kx = Fo cos ωd t (14.37b) b=50g/s (least dt 2 dt damping) This is the equation of an oscillator of mass b=70g/s m on which a periodic force of (angular) b=140g/s ωd is frequency with applied. The oscillator, initially, oscillates its natural frequency ω. When we apply the external periodic force, the oscillations with the natural frequency die out, and then the body oscillates with the (angular) frequency of the external periodic force. Its displacement, after the natural oscillations die out, is given by x(t) = A cos (ωdt + φ ) (14.38) where t is the time measured from the moment Fig. 14.21 The displacement amplitude of a forced when we apply the periodic force. oscillator as a function of the angular The amplitude A is a function of the forced frequency of the driving force. The frequency ωd and the natural frequency ω. ω /ω Analysis shows that it is given by amplitude is the greatest at d =1, the resonance condition. The three curves Fο correspond to different extents of damping { ( ) }A = m2 ω2 − ω 2 2 + ω 2 2 1/2 (14.39a) present in the system. The curves 1 and d d b 3 correspond to minimum and maximum damping in the system. and tan φ = – vο (14.39b) If we go on changing the driving frequency, ωd xο the amplitude tends to infinity when it equals where m is the mass of the particle and v0 and the natural frequency. But this is the ideal case x0 are the velocity and the displacement of the particle at time t = 0, which is the moment when of zero damping, a case which never arises in a real system as the damping is never perfectly we apply the periodic force. Equation (14.39) zero. You must have experienced in a swing that shows that the amplitude of the forced oscillator when the timing of your push exactly matches depends on the (angular) frequency of the with the time period of the swing, your swing driving force. We can see a different behaviour gets the maximum amplitude. This amplitude of itshceloossecitlloaωto.rWwehceonnωsdidiserfathr efrsoemtwωoacnadsewsh. en is large, but not infinity, because there is always it some damping in your swing. This will become (a) Small Damping, Driving Frequency far clear in the (b). from Natural Frequency : In this case, ωd b will (b) Driving Frequency Close to Natural be much smaller than m(ω2 –ω2d), and we can Frequency : If ω is very close to ω, m (ω2 – ω2 ) d d neglect that term. Then Eq. (14.39) reduces to would be much less than ω b, for any reasonable d F value of b, then Eq. (14.39) reduces to ο ( )A = (14.40) mω2 − 2 A = Fο (14.41) ω d ωd b Fig. 14.21 shows the dependence of the This makes it clear that the maximum displacement amplitude of an oscillator on the possible amplitude for a given driving frequency angular frequency of the driving force for is governed by the driving frequency and the different amounts of damping present in the damping, and is never infinity. The phenomenon system. It may be noted that in all cases the of increase in amplitude when the driving force amplitude is the greatest when ω /ω = 1. The is close to the natural frequency of the oscillator d curves in this figure show that smaller the is called resonance. damping, the taller and narrower is the In our daily life, we encounter phenomena resonance peak. which involve resonance. Your experience with 2020-21

OSCILLATIONS 359 swings is a good example of resonance. You motion is gradually damped and not sustained. might have realised that the skill in swinging to Their frequencies of oscillation gradually greater heights lies in the synchronisation of change, and ultimately, they oscillate with the the rhythm of pushing against the ground with frequency of pendulum 1, i.e., the frequency of the natural frequency of the swing. the driving force but with different amplitudes. They oscillate with small amplitudes. The To illustrate this point further, let us response of pendulum 4 is in contrast to this consider a set of five simple pendulums of set of pendulums. It oscillates with the same assorted lengths suspended from a common rope frequency as that of pendulum 1 and its as shown in Fig. 14.22. The pendulums 1 and 4 amplitude gradually picks up and becomes very have the same lengths and the others have large. A resonance-like response is seen. different lengths. Now, let us set pendulum 1 This happens because in this the condition for into motion. The energy from this pendulum gets resonance is satisfied, i.e. the natural frequency transferred to other pendulums through the of the system coincides with that of the connecting rope and they start oscillating. The driving force. driving force is provided through the connecting rope. The frequency of this force is the frequency We have so far considered oscillating systems with which pendulum 1 oscillates. If we observe which have just one natural frequency. In the response of pendulums 2, 3 and 5, they first general, a system may have several natural start oscillating with their natural frequencies frequencies. You will see examples of such of oscillations and different amplitudes, but this systems (vibrating strings, air columns, etc.) in the next chapter. Any mechanical structure, like Fig. 14.22 Five simple pendulums of different a building, a bridge, or an aircraft may have lengths suspended from a common several possible natural frequencies. An support. external periodic force or disturbance will set the system in forced oscillation. If, accidentally, the forced frequency ωd happens to be close to one of the natural frequencies of the system, the amplitude of oscillation will shoot up (resonance), resulting in possible damage. This is why, soldiers go out of step while crossing a bridge. For the same reason, an earthquake will not cause uniform damage to all buildings in an affected area, even if they are built with the same strength and material. The natural frequencies of a building depend on its height, other size parameters, and the nature of building material. The one with its natural frequency close to the frequency of seismic wave is likely to be damaged more. SUMMARY 1. The motion that repeats itself is called periodic motion. 2. The period T is the time required for one complete oscillation, or cycle. It is related to the frequency ν by, T = 1 ν 2020-21

360 PHYSICS The frequency ν of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz : 1 hertz = 1 Hz = 1 oscillation per second = 1s–1 3. In simple harmonic motion (SHM), the displacement x (t) of a particle from its equilibrium position is given by, x (t) = A cos (ωt + φ ) (displacement), in which A is the amplitude of the displacement, the quantity (ωt + φ ) is the phase of the motion, and φ is the phase constant. The angular frequency ω is related to the period and frequency of the motion by, ω = 2π = 2πν (angular frequency). T 4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs. 5. The particle velocity and acceleration during SHM as functions of time are given by, v (t) = –ωA sin (ωt + φ ) (velocity), a (t) = –ω2A cos (ωt + φ ) = –ω2x (t) (acceleration), Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude vm=ω A and acceleration amplitude am =ω 2A, respectively. 6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion. 7. A particle executing simple harmonic motion has, at any time, kinetic energy K = ½ mv2 and potential energy U = ½ kx2. If no friction is present the mechanical energy of the system, E = K + U always remains constant even though K and U change with time. 8. A particle of mass m oscillating under the influence of Hooke’s law restoring force given by F = – k x exhibits simple harmonic motion with ω= k (angular frequency) m T = 2π m (period) k Such a system is also called a linear oscillator. 9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by, T = 2π L g 10. The mechanical energy in a real oscillating system decreases during oscillations because external forces, such as drag, inhibit the oscillations and transfer mechanical energy to thermal energy. The real oscillator and its motion are then said to be damped. If the 2020-21

OSCILLATIONS 361 damping force is given by Fd = –bv, where v is the velocity of the oscillator and b is a damping constant, then the displacement of the oscillator is given by, x (t) = A e–bt/2m cos (ω′t + φ ) where ω′, the angular frequency of the damped oscillator, is given by ω′ = k − b2 m 4m 2 If the damping constant is small then ω′ ≈ ω, where ω is the angular frequency of the undamped oscillator. The mechanical energy E of the damped oscillator is given by E(t) = 1 kA2e −bt /m 2 11. If an external force with angular frequency ωd acts on an oscillating system with natural angular frequency ω, the system oscillates with angular frequency ωd. The amplitude of oscillations is the greatest when ω = ω d a condition called resonance. POINTS TO PONDER 1. The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer. 2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = – k x is simple harmonic. 3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: –mω2r. In the latter case, the phases of motion, in two perpendicular directions (x and y) must differ by π/2. Thus, for example, a particle subject to a force –mω2r with initial position (0, A) and velocity (ωA, 0) will move uniformly in a circle of radius A. 4. For linear simple harmonic motion with a given ω, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase. 2020-21

362 PHYSICS 5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity. 6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes. 7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law). 8. The motion of a simple pendulum is simple harmonic for small angular displacement. 9. For motion of a particle to be simple harmonic, its displacement x must be expressible in either of the following forms : x = A cos ωt + B sin ωt x = A cos (ωt + α ), x = B sin (ωt + β ) The three forms are completely equivalent (any one can be expressed in terms of any other two forms). Thus, damped simple harmonic motion [Eq. (14.31)] is not strictly simple harmonic. It is approximately so only for time intervals much less than 2m/b where b is the damping constant. 10. In forced oscillations, the steady state motion of the particle (after the forced oscillations die out) is simple harmonic motion whose frequency is the frequency of the driving frequency ωd, not the natural frequency ω of the particle. 11. In the ideal case of zero damping, the amplitude of simple harmonic motion at resonance is infinite. Since all real systems have some damping, however small, this situation is never observed. 12. Under forced oscillation, the phase of harmonic motion of the particle differs from the phase of the driving force. Exercises 14.1 Which of the following examples represent periodic motion? 14.2 (a) A swimmer completing one (return) trip from one bank of a river to the other 14.3 and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position. Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? 2020-21

OSCILLATIONS 363 14.4 Fig. 14.23 14.5 14.6 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt – cos ωt (b) sin3 ωt (c) 3 cos (π/4 – 2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (–ω2t2) (f) 1 + ωt + ω2t2 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = –200x2 (c) a = –10x (d) a = 100x3 2020-21

364 PHYSICS 14.7 The motion of a particle executing simple harmonic motion is described by the displacement function, 14.8 14.9 x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ? A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Fig. 14.24 Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. 14.10 In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? 14.11 Figures 14.25 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 14.25 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case. 14.12 Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, 2020-21

OSCILLATIONS 365 and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s). (a) x = –2 sin (3t + π/3) (b) x = cos (π/6 – t) (c) x = 3 sin (2πt + π/4) (d) x = 2 cos πt 14.13 Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F. Fig. 14.26 14.14 (a) What is the maximum extension of the spring in the two cases ? 14.15 (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the 14.16 period of oscillation in each case ? The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ? The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2) Answer the following questions : (a) Time period of a particle in SHM depends on the force constant k and mass m of the particle: T = 2π m . A simple pendulum executes SHM approximately. Why then is k the time period of a pendulum independent of the mass of the pendulum? (b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π l Think of a qualitative argument to . g appreciate this result. (c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ? (d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ? 14.17 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ? 2020-21

366 PHYSICS 14.18 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period T = 2π hρ ρ1g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid). 14.19 One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion. Additional Exercises 14.20 An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.27). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.27]. Fig.14.27 14.21 You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg. 14.22 Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period. 14.23 A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = –α θ , where J is the restoring couple and θ the angle of twist). 14.24 A body describes simple harmonic motion with an amplitude of 5 cm and a period of 14.25 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm. A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x and pushed towards 0 xoths=eciaclleacntoitosrne(sωwtii+nthθt)eaarnmvedslonocofitttyehevth0paaattrttahimmeeeitnet ir=tsia0ωl.,vDxee0lotaecnritmdy ivins0.en[tHehgienatatim:veSp.tl]aitrutdweitohf the resulting the equation 2020-21

CHAPTER FIFTEEN WAVES 15.1 Introduction 15.1 INTRODUCTION 15.2 Transverse and longitudinal waves In the previous Chapter, we studied the motion of objects 15.3 Displacement relation in a oscillating in isolation. What happens in a system, which is progressive wave a collection of such objects? A material medium provides 15.4 The speed of a travelling such an example. Here, elastic forces bind the constituents wave to each other and, therefore, the motion of one affects that of 15.5 The principle of the other. If you drop a little pebble in a pond of still water, superposition of waves the water surface gets disturbed. The disturbance does not 15.6 Reflection of waves remain confined to one place, but propagates outward along 15.7 Beats a circle. If you continue dropping pebbles in the pond, you 15.8 Doppler effect see circles rapidly moving outward from the point where the water surface is disturbed. It gives a feeling as if the water is Summary moving outward from the point of disturbance. If you put Points to ponder some cork pieces on the disturbed surface, it is seen that Exercises the cork pieces move up and down but do not move away Additional exercises from the centre of disturbance. This shows that the water mass does not flow outward with the circles, but rather a moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this Chapter, we will study such waves. Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of sig- nals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For exam- ple, sound waves may be first converted into an electric cur- rent signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a 2020-21

368 PHYSICS satellite. Detection of the original signal will usu- We shall illustrate this connection through ally involve these steps in reverse order. simple examples. Not all waves require a medium for their Consider a collection of springs connected to propagation. We know that light waves can one another as shown in Fig. 15.1. If the spring travel through vacuum. The light emitted by at one end is pulled suddenly and released, the stars, which are hundreds of light years away, disturbance travels to the other end. What has reaches us through inter-stellar space, which is practically a vacuum. Fig. 15.1 A collection of springs connected to each other. The end A is pulled suddenly The most familiar type of waves such as waves generating a disturbance, which then on a string, water waves, sound waves, seismic propagates to the other end. waves, etc. is the so-called mechanical waves. These waves require a medium for propagation, happened? The first spring is disturbed from its they cannot propagate through vacuum. They equilibrium length. Since the second spring is involve oscillations of constituent particles and connected to the first, it is also stretched or depend on the elastic properties of the medium. compressed, and so on. The disturbance moves The electromagnetic waves that you will learn from one end to the other; but each spring only in Class XII are a different type of wave. executes small oscillations about its equilibrium Electromagnetic waves do not necessarily require position. As a practical example of this situation, a medium - they can travel through vacuum. consider a stationary train at a railway station. Light, radiowaves, X-rays, are all electromagnetic Different bogies of the train are coupled to each waves. In vacuum, all electromagnetic waves other through a spring coupling. When an have the same speed c, whose value is : engine is attached at one end, it gives a push to the bogie next to it; this push is transmitted from c = 299, 792, 458 ms–1. (15.1) one bogie to another without the entire train being bodily displaced. A third kind of wave is the so-called Matter waves. They are associated with constituents of Now let us consider the propagation of sound matter : electrons, protons, neutrons, atoms and waves in air. As the wave passes through air, it molecules. They arise in quantum mechanical compresses or expands a small region of air. This description of nature that you will learn in your causes a change in the density of that region, later studies. Though conceptually more abstract say δρ, this change induces a change in pressure, than mechanical or electro-magnetic waves, they δp, in that region. Pressure is force per unit area, have already found applications in several so there is a restoring force proportional to devices basic to modern technology; matter the disturbance, just like in a spring. In this waves associated with electrons are employed case, the quantity similar to extension or compression of the spring is the change in in electron microscopes. density. If a region is compressed, the molecules in that region are packed together, and they tend In this chapter we will study mechanical to move out to the adjoining region, thereby waves, which require a material medium for increasing the density or creating compression their propagation. in the adjoining region. Consequently, the air in the first region undergoes rarefaction. If a The aesthetic influence of waves on art and region is comparatively rarefied the surrounding literature is seen from very early times; yet the air will rush in making the rarefaction move to first scientific analysis of wave motion dates back the adjoining region. Thus, the compression or to the seventeenth century. Some of the famous rarefaction moves from one region to another, scientists associated with the physics of wave making the propagation of a disturbance motion are Christiaan Huygens (1629-1695), possible in air. Robert Hooke and Isaac Newton. The understanding of physics of waves followed the physics of oscillations of masses tied to springs and physics of the simple pendulum. Waves in elastic media are intimately connected with harmonic oscillations. (Stretched strings, coiled springs, air, etc., are examples of elastic media). 2020-21

WAVES 369 In solids, similar arguments can be made. In Fig. 15.3 A harmonic (sinusoidal) wave travelling a crystalline solid, atoms or group of atoms are along a stretched string is an example of a arranged in a periodic lattice. In these, each transverse wave. An element of the string atom or group of atoms is in equilibrium, due to in the region of the wave oscillates about forces from the surrounding atoms. Displacing its equilibrium position perpendicular to the one atom, keeping the others fixed, leads to direction of wave propagation. restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, position as the pulse or wave passes through with springs between pairs of them. them. The oscillations are normal to the direction of wave motion along the string, so this In the subsequent sections of this chapter is an example of transverse wave. we are going to discuss various characteristic properties of waves. We can look at a wave in two ways. We can fix an instant of time and picture the wave in space. 15.2 TRANSVERSE AND LONGITUDINAL This will give us the shape of the wave as a WAVES whole in space at a given instant. Another way is to fix a location i.e. fix our attention on a We have seen that motion of mechanical waves particular element of string and see its involves oscillations of constituents of the oscillatory motion in time. medium. If the constituents of the medium oscillate perpendicular to the direction of wave Fig. 15.4 describes the situation for propagation, we call the wave a transverse wave. longitudinal waves in the most familiar example If they oscillate along the direction of wave of the propagation of sound waves. A long pipe propagation, we call the wave a longitudinal filled with air has a piston at one end. A single wave. sudden push forward and pull back of the piston will generate a pulse of condensations (higher Fig.15.2 shows the propagation of a single density) and rarefactions (lower density) in the pulse along a string, resulting from a single up medium (air). If the push-pull of the piston is and down jerk. If the string is very long compared continuous and periodic (sinusoidal), a Fig. 15.2 When a pulse travels along the length of a stretched string (x-direction), the elements of the string oscillate up and down (y- direction) to the size of the pulse, the pulse will damp out Fig. 15.4 Longitudinal waves (sound) generated in a before it reaches the other end and reflection pipe filled with air by moving the piston up from that end may be ignored. Fig. 15.3 shows a and down. A volume element of air oscillates similar situation, but this time the external in the direction parallel to the direction of agent gives a continuous periodic sinusoidal up wave propagation. and down jerk to one end of the string. The resulting disturbance on the string is then a sinusoidal wave. In either case the elements of the string oscillate about their equilibrium mean 2020-21

370 PHYSICS sinusoidal wave will be generated propagating Example 15.1 Given below are some in air along the length of the pipe. This is clearly examples of wave motion. State in each case an example of longitudinal waves. if the wave motion is transverse, longitudinal or a combination of both: The waves considered above, transverse or (a) Motion of a kink in a longitudinal spring longitudinal, are travelling or progressive waves since they travel from one part of the medium produced by displacing one end of the to another. The material medium as a whole spring sideways. does not move, as already noted. A stream, for (b) Waves produced in a cylinder example, constitutes motion of water as a whole. containing a liquid by moving its piston In a water wave, it is the disturbance that moves, back and forth. not water as a whole. Likewise a wind (motion (c) Waves produced by a motorboat sailing of air as a whole) should not be confused with a in water. sound wave which is a propagation of (d) Ultrasonic waves in air produced by a disturbance (in pressure density) in air, without the motion of air medium as a whole. vibrating quartz crystal. In transverse waves, the particle motion is Answer normal to the direction of propagation of the (a) Transverse and longitudinal wave. Therefore, as the wave propagates, each (b) Longitudinal element of the medium undergoes a shearing (c) Transverse and longitudinal strain. Transverse waves can, therefore, be (d) Longitudinal propagated only in those media, which can sustain shearing stress, such as solids and not 15.3 DISPLACEMENT RELATION IN in fluids. Fluids, as well as, solids can sustain A PROGRESSIVE WAVE compressive strain; therefore, longitudinal waves can be propagated in all elastic media. For mathematical description of a travelling For example, in medium like steel, both wave, we need a function of both position x and transverse and longitudinal waves can time t. Such a function at every instant should propagate, while air can sustain only give the shape of the wave at that instant. Also, longitudinal waves. The waves on the surface at every given location, it should describe the of water are of two kinds: capillary waves and motion of the constituent of the medium at that gravity waves. The former are ripples of fairly location. If we wish to describe a sinusoidal short wavelength—not more than a few travelling wave (such as the one shown in Fig. centimetre—and the restoring force that 15.3) the corresponding function must also be produces them is the surface tension of water. sinusoidal. For convenience, we shall take the Gravity waves have wavelengths typically wave to be transverse so that if the position of ranging from several metres to several hundred the constituents of the medium is denoted by x, meters. The restoring force that produces these the displacement from the equilibrium position waves is the pull of gravity, which tends to keep may be denoted by y. A sinusoidal travelling the water surface at its lowest level. The wave is then described by: oscillations of the particles in these waves are not confined to the surface only, but extend with y(x,t) = asin(kx − ωt + φ) (15.2) diminishing amplitude to the very bottom. The particle motion in water waves involves a The term φ in the argument of sine function complicated motion—they not only move up and means equivalently that we are considering a down but also back and forth. The waves in an linear combination of sine and cosine functions: ocean are the combination of both longitudinal and transverse waves. y(x,t ) = A sin(kx − ωt ) + B cos(kx − ωt ) (15.3) It is found that, generally, transverse and From Equations (15.2) and (15.3), longitudinal waves travel with different speed in the same medium. a= A2 + B2 and φ = tan −1  B  A To understand why Equation (15.2) represents a sinusoidal travelling wave, take a fixed instant, say t = t0. Then, the argument of the sine function in Equation (15.2) is simply 2020-21