Comprehensive CBSE Question Bank In Physics-XI (Term-II)

CBSE II Question Bank in Physics CLASS 11 Features Short Answer Type Questions Long Answer Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs Chapter Summary Very Short Answer Type Questions



Comprehensive CBSE Question Bank in Physics Term–II (For Class-XI)



Comprehensive CBSE Question Bank in Physics Term–II (For Class-XI) (According to the Latest CBSE Examination Pattern) By NARINDER KUMAR M.Sc. PES(I) Formerly, Senior Lecturer Department of Physics S.D. Govt. College, Ludhiana Punjab   laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi

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Contents Chapters Pages Unit VII: Properties of Bulk Matter 1. Mechanical Properties of Solids (NCERT Textbook Chapter–9).............................1–33 2. Mechanical Properties of Fluids (NCERT Textbook Chapter–10).......................34–121 3. Thermal Properties of Matter (NCERT Textbook Chapter–11)........................122–184 Unit VIII: Thermodynamics 4. Thermodynamics (NCERT Textbook Chapter–12).............................................185–212 Unit IX: Behaviour of Perfect Gases and Kinetic Theory of Gases 5. Kinetic Theory (NCERT Textbook Chapter–13).................................................213–261 Unit X: Oscillations and Waves 6. Oscillations (NCERT Textbook Chapter–14)......................................................262–327 7. Waves (NCERT Textbook Chapter–15)...............................................................328–391



Syllabus Class XI (Code N. 042) (2021-22) Term–II (Theory) Time: 2 Hours Max Marks: 35 Unit Properties of Bulk Matter Periods Marks Unit–VII Chapter–9: Mechanical Properties of Solids Chapter–10: Mechanical Properties of Fluid 22 Unit-VIII Chapter–11: Thermal Properties of Matter 23 Unit-IX Thermodynamics Unit-X Chapter–12: Thermodynamics 10 Behaviour of Perfect Gases and Kinetic Theory of Gases 08 Chapter–13: Kinetic Theory Oscillations and Waves 23 12 Chapter–14: Oscillations 63 35 Chapter 15: Waves Total Marks Syllabus assigned for Term II 22 Periods Unit VII: Properties of Bulk Matter Chapter–9: Mechanical Properties of Solids Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus. Chapter–10: Mechanical Properties of Fluids Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise. Chapter–11: Thermal Properties of Matter Heat, temperature, (recapitulation only) thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv-calorimetry; change of state-latent heat capacity.

Heat transfer-conduction, convection and radiation (recapitulation only), thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law, Greenhouse effect. Unit VIII: Thermodynamics 10 Periods Chapter–12: Thermodynamics Thermal equilibrium and definition of temperature (zeroth law of thermodynamics), heat, work and internal energy. First law of thermodynamics, isothermal and adiabatic processes. Second law of thermodynamics: reversible and irreversible processes. Unit IX: Behaviour of Perfect Gases and Kinetic Theory of Gases 08 Periods Chapter–13: Kinetic Theory Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases-assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number. Unit X: Oscillations and Waves 23 Periods Chapter–14: Oscillations Periodic motion - time period, frequency, displacement as a function of time, periodic functions. Simple harmonic motion (S.H.M) and its equation; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period. Free, forced and damped oscillations (qualitative ideas only), resonance. Chapter–15: Waves Wave motion: Transverse and longitudinal waves, speed of travelling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, Beats.

Chapter 6: Oscillations (NCERT Textbook Chapter-14) SUMMARY OF THE CHAPTER • A periodic motion repeats itself over and over again after a fixed interval of time. • An oscillatory motion repeats itself over and over again about its mean position and remains confined within two extreme positions on either side of the mean position. • The time period of an oscillatory motion refers to the time in which one oscillation is completed. • The number of vibrations that occur in a unit time is called the frequency of the periodic motion. The SI unit of frequency is hertz (Hz). • The displacement of a particle at any instant is the distance of the oscillating particle from its mean position at that instant. • A simple harmonic motion (S.H.M.) is written as 2 2 F(t) = a1 cos T t + b1 sin T t or F(t) = A cos  2 t   = A cos (t + )  T or F(t) = A sin  2 t   = A sin (t + )  T where A = a12  b12 is called amplitude,  2 t   is called phase, and  = 2 t  T T is called angular frequency. • The acceleration in S.H.M. is directly proportional to displacement and is always towards the mean position. Acceleration = – 2 × displacement • The velocity amplitude in S.H.M. is x and the acceleration amplitude of S.H.M. is x, where x is the displacement and  is the angular frequency. • The time period (T) of S.H.M. is independent of the amplitude of S.H.M. displacement 1 T = 2 acceleration = 2  acceleration per unit displacement • When a mass m is attached to some massless spring and pulled downwards, it will execute S.H.M. with time period given by, m T = 2 k where k is called force constant or spring factor. 262

Oscillations 263 • SI unit of K is N m–1 and its dimensional formula is [ML0L–2]. • A simple pendulum is a heavy point mass suspended by a weightless, inextensible and perfectly flexible string from a rigid support about which it can freely oscillate. l T = 2 g where l is the length of the string, from the point of suspension to the centre of gravity of the bob. • A second’s pendulum has a time period of two seconds. • When a particle of mass m executes S.H.M., then at a displacement x from the mean position, the particle has both kinetic energy and potential energy. Potential energy (at displacement x) = 1 m 2 x2 2 Kinetic energy (at displacement x) = 1 m 2 (A2 – x2) 2 Total energy = 1 m 2 x2 + 1 m 2 (A2 – x2) = 1 m 2 A2 = 1 kA2 ( k = m2) 2 2 2 2 • All oscillatory motions are periodic but all periodic motions are not oscillatory. • The oscillations produced by an oscillator of frequency equal to its natural frequency, such that no external periodic force acts on it, are called free oscillations. • For a spring, the magnitude of restoring force is given by F = kl, where l is the extension in the spring.  k = F   MLT 2   [ML0T 2 ] l [L] The force constant is a dimensional constant. • The oscillations produced by an oscillator under the effect of an external periodic force of frequency other than the natural frequency of the oscillator, are known as forced oscillations. • The phenomenon of setting an oscillator into oscillations of its natural frequency by bringing it near to another oscillating body (oscillator) of same frequency is known as resonance. • When the amplitude of oscillations constantly decreases with time, the oscillations are known as damped oscillations. Important Results 1. The least interval of time after which the periodic motion of a body repeats itself, is called time period (T). The number of periodic motions executed by a body per second is called frequency (). 1 = T 2 Angular frequency () = 2 or  = T

264 Physics—XI 2. Periodic functions are used to represent periodic motion. A function f(t) is said to be periodic, if f(t) = f(t + T) = f(t + 2T) where T is called the period of periodic function. sin  and cos  are the examples of periodic functions in the period equal to 2  radians. The infinite sets of periodic functions of period T may be expressed as, fn(t) = sin 2 nt and gn(t) = cos 2 nt , where n = 1, 2, 3, ... T T 3. The S.H.M. is the projection of a uniform circular motion on the diameter of a circle of reference. Consider a reference particle P moving with uniform speed along the circumference of a circle of radius a, with centre O. The foot (M) of the perpendicular executes S.H.M. along the diameter of the circle. y or y = a sin t a = sin  = sin t Fig. 6.1 If the projection is considered along XX, then we take, x = a cos t If the particle starts from Q, then 0 is initial phase, also called epoch of S.H.M. QOX = 0 y = a sin (t + 0) Also, x = a cos (t + 0) Fig. 6.2 4. In S.H.M., the direction of displacement (from mean position) is always away from the mean position whether the particle is moving from or coming towards the mean position.

Oscillations 265 5. In S.H.M., the maximum displacement on either side of mean position is called amplitude of motion. 6. The velocity of a particle executing S.H.M., is given by, v =  a2  y2 7. The acceleration of a particle executing S.H.M. is given by, dv = – 2 y dt 8. The velocity amplitude of the particle executing S.H.M. is given by, v0 = a  Also, acceleration amplitude is given by, a0 = a2 9. The differential equation of a harmonic oscillator is given by, d2 y + w2 y = 0 dt2 displacement Time period, T = 2 acceleration 10. The force equation of a particle executing S.H.M. is given by, F = – ky, where k is called force constant or spring factor, k = m2 m or T = = 2 k 11. In case of a simple harmonic motion: (i) Potential energy, P.E.= 1 m w2y2 = 1 ky2 2 2 (ii) Kinetic energy, K.E. = 1 m 2 (a2 – y2) = 1 k(a2 – y2) 2 2 (iii) Total energy, E= P.E. + K.E. = 1 (m 2) a2 = 1 ka2 2 2 12. Figure below shows the variations of kinetic energy K, potential energy U and total energy E with displacement y. The graphs for K and U are parabolic while that for E is a straight line parallel to the displacement – axis. At y = 0, the energy is all kinetic and for y =  a the energy is all potential. Fig. 6.3

266 Physics—XI 13. Figure below shows the variations of energiesK, U and E of a harmonic oscillator with time t. Fig. 6.4 14. The spring constant of parallel combination of springs is, kp = k1 + k2 15. The spring constant of series combination of spring is, = VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. What is the time period of second’s pendulum? Ans. 2 second. 2. When will the motion of a simple pendulum be simple harmonic? Ans. The motion of a simple pendulum will be simple harmonic when the angular displacement  of the bob is small. 3. Two exactly similar simple pendula are vibrating with amplitudes 1 cm and 3 cm. What is the ratio of their energies of vibration? Ans. = =   =   4. Which of the following relationships between the acceleration and displacement of the particle involve SHM? a = 0.5 x, a = 400 x2, a = – 20 x, a = – 3 x2 Here, the letters have usual meanings. Ans. a = – 20 x. [Note that acceleration is directly proportional to displacement and is directed towards the mean position.] 5. How is angular frequency related to time period?  Ans.  = . 6. Is oscillation of a mass suspended by a spring simple harmonic? Ans. Yes. 7. A particle has maximum velocity in mean position and zero velocity at extreme position. Is it a sure test for simple harmonic motion? Ans. No.

Oscillations 267 8. Imagine a situation where the motion is not simple harmonic but the particle has maximum velocity in mean position and zero velocity at extreme position. Ans. Projection of a particle in non-uniform circular motion satisfies all the given conditions. 9. We know that, in SHM, the time period is given by   Does T depend upon displacement? Ans. No. T is independent of displacement. Note that the ‘displacement term’ is contained in acceleration also. 10. Name the trigonometric functions which are suitable for the analytical treatment of periodic motions. Ans. Sine or cosine functions or their linear combination. 11. Sometimes, when an automobile picks up speed, its body begins to rattle. Why? Ans. This is because of resonant vibrations. 12. Where is the potential energy of a simple harmonic oscillator maximum? Ans. At the extreme position. 13. Where is the kinetic energy of a simple harmonic oscillator maximum? Ans. At the mean position. 14. What is the total energy of a simple harmonic oscillator? Ans.  , where the letters have usual meanings. 15. The total energy of a particle executing SHM is E. What is the kinetic energy when the displacement is equal to one-half the amplitude? Ans.     .       16. Why is simple harmonic motion so named? Ans. This is because it is the simplest of all the harmonic motions. 17. Why a simple pendulum vibrating in air eventually stops? Ans. This is because its energy is dissipated in the form of heat. 18. A simple pendulum is transferred from Earth to Moon. Will it go faster or slower? Ans. Due to decrease in value of g, T shall increase. So, pendulum will vibrate slower. 19. A pendulum-controlled clock is transferred from Earth to Moon. What would be the effect on the clock? Ans. Due to decrease in the value of g, T shall increase. So, the clock shall slow down. 20. A spring-controlled wrist watch is taken from Earth to Moon. What shall be the effect on the watch? Ans. Since time period of a spring is independent of g therefore there will be no effect on the watch.

268 Physics—XI 21. A spring-mass system oscillating vertically has a time period T. What shall be the time period if oscillated horizontally? Ans. The time period shall remain T. 22. The time period of a body executing SHM is 0.05 s and the amplitude of vibration is 4 cm. What is the maximum velocity of the body? Ans. Maximum velocity = A = 4 × 10–2 ×  m s–1 = 1.6  m s–1. 23. The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change if the water begins to drain out of the hollow sphere? Ans. The time period will increase at first, then decrease until the sphere is empty. Finally, the period will be the same as when the sphere was full of water. 24. The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of aluminium? Ans. There will be no effect because the time period does not depend upon the nature of material of the bob. 25. A body of mass m, when hung on a spiral spring, stretches it by 20 cm. What is its period of oscillation when pulled down and released? Ans.    . 26. A body executes SHM with a period of second and an amplitude of 0.025 m. What is the maximum value of acceleration? Ans. Maximum acceleration =       = 0.4 m s–2. 27. A particle is in SHM of amplitude 2 cm. At the extreme position, the force is 4 N. What is the force at a point midway between mean and extreme positions? Ans. 2 N. 28. Can a simple pendulum be used in an artificial satellite? Ans. No. This is because there exists a state of weightlessness in a satellite. 29. When a particle oscillates simple harmonically, its potential energy varies periodically. If  is the frequency of oscillation of the particle, then what is the frequency of variation of potential energy? Ans. 2. 30. What fraction of the total energy is kinetic energy when the displacement is one-half of amplitude? Ans.        31. What fraction of the total energy is potential energy when the displacement is one- half of the amplitude?

Oscillations 269 Ans.     32. What happens to the time period of a simple pendulum if its length is doubled? Ans. The time period is increased by a factor of . 33. Two simple pendula of equal length cross each other at mean position. What is their phase difference? Ans.  radian i.e., 180°. 34. Can a simple pendulum vibrate at the centre of Earth? Ans. No. This is because of zero value of g at the centre of Earth. 35. Is restoring force necessary in SHM? Ans. Yes. 36. A body of mass 16 kg is oscillating on a spring of force constant 100 N m –1. Deduce the angular frequency. Ans. m = 16 kg, k = 100 N m–1,  = ?;  =   = 2.5 rad s–1 37. A spring has time period T. It is cut into n equal parts. What will be the time period of each part of the spring? Ans. T = 2 , T = 2 = 38. Spring of spring constant 1200 N m–1 is mounted on a smooth frictionless surface and attached to a block of mass 3 kg. Block is pulled 2 cm to the right and released. Find angular frequency of oscillation. Fig. 6.5Displacement (cm) Ans.  = k = 1200 rad s–1 = 20 rad s–1. m3 39. Fig. 6.6 shows the displacement-time graph of a simple harmonic oscillator. What is the amplitude, time period and initial phase of the oscillator? 3 2 1 0 1 234 5 6 7 8 –1 –2 –3 Time (s) Fig. 6.6 Ans. Amplitude is 2 cm and time period is 4 s. Graph begins from the origin. So, initial phase is zero.

270 Physics—XI 40. 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 rev/min., what is its maximum speed? Ans. A = m,  = 200 rev/min vmax. = A = × 200 m/min = 100 m/min. 41. When will the motion of a simple pendulum be simple harmonic? Ans. The motion of a simple pendulum will be simple harmonic when the bob of the pendulum is displaced from mean position so that sin . 42. What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator? Ans.  = . A  43. What is the ratio between the distance travelled by the oscillator in one time period and amplitude? Ans. = 4. P 44. In Fig. 6.7, what will be the sign of the velocity of  + the point P, which is the projection of the velocity O P A of the reference particle P. P is moving in a circle of radius R in anticlockwise direction. Fig. 6.7 Ans. The projection of the velocity of the reference particle is along negative x-axis. So, the sign of the velocity at P is negative. SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. For an oscillating simple pendulum, (i) what is the direction of acceleration of the bob at (a) the mean position, (b) the end points? (ii) is the tension in the string constant throughout the oscillation? If not, when is it (a) the least, (b) the greatest? Ans. (i) (a)Radial towards the point of suspension (b) tangential towards the mean position (ii) No. Tension is the (a) least at end-points, and (b) greatest at the mean position. 2. What is the phase difference between velocity and displacement in simple harmonic motion? Ans. We know that y = A sin t              Comparing, we find that velocity is ahead of displacement by a phase angle of  . 3. At what displacement, (i) the P.E. of a simple harmonic oscillator is maximum (ii) the K.E. is maximum? Ans. The potential energy of a simple harmonic oscillator is maximum at the extreme position. The kinetic energy is maximum at the mean position.

Oscillations 271 4. The length of a second’s pendulum on the surface of Earth is 1 m. What will be the length of a second’s pendulum on the surface of Moon? Ans.   In both the cases, T is same.  l g On the Moon, the value of acceleration due to gravity is one-sixth of that on the surface of Earth. So, the length of second’s pendulum is m. 5. Will a pendulum clock gain or lose time when taken to the top of a mountain? Ans. The time period of a simple pendulum is given by T = 2 On top of a mountain, the value of ‘g’ will decrease. Therefore, the time period will increase. Hence the pendulum clock will lose time. 6. The maximum acceleration of a simple harmonic oscillator is a 0 and the maximum velocity is v0 . What is the displacement amplitude? Ans. Let x be the displacement amplitude and  be the angular frequency of the simple harmonic oscillator. Then, a0 = 2x and v0 = x  =x or x= . =  7. The bob of a vibrating simple pendulum is made of ice. How will the time period change when the ice starts melting? Ans. The time period is independent of the mass of bob. 8. A ball of radius r is made to oscillate in a bowl of radius R. What is the time period of oscillations of the ball? Ans. The distance from centre of ball to centre of bowl gives the equivalent length of simple pendulum. It is (R – r).  Time period = 2  . 9. A particle of mass m rests on a horizontal plane which is displaced up and down with simple harmonic motion. Show that the particle will remain in contact with the plane only if the downward acceleration at any instant is smaller than g. Ans. When the system has downward acceleration a, then normal reaction N = mg – ma = m(g – a). If a > g, then N will be negative and the particle shall not be in contact with the plane. 10. In the previous question, show that the amplitude should be less than .  Ans. 2A < g, A < . 

272 Physics—XI 11. What is the phase difference between acceleration and displacement in simple harmonic motion? Ans. We know that y = A sin t      a = y = – 2 A sin t = 2A sin (t + ) Comparing, we find that acceleration is ahead of displacement by a phase angle of . 12. At what distance from the mean position in SHM, the energy is half kinetic and half potential? Ans. Ek = m2 (A2 – x2). Here A is amplitude of motion. Ep = m2x2 As, Ek = Ep  m2(A2 – x2) = m2x2 or 2x2 = A2 or x = ± . 13. A glass window may be broken by a distant explosion. Is it correct? Ans. Yes. The sound waves can cause forced vibrations in glass due to difference between the frequency of sound waves and the natural frequency of glass. This can break the glass window. 14. The displacement in SHM is given by y = A sin (t  ) . Prove that y would remain the  same when t is increased by .  Ans.         = A sin [t + 2 + ] = A sin [(t + ) + 2]        = A sin (t + ) = y. 15. Define spring factor and give its dimensional formula. Ans. Spring factor is force per unit extension.[Spring factor] =  .  16. Alcohol in a U-tube is executing simple harmonic motion of time period T. Now, alcohol is replaced by water up to the same height in U-tube. What will be the effect on the time period? Ans. There will be no effect. This is because the time period of oscillation of a liquid column in a U-tube does not depend upon the density of the liquid. 17. A girl is swinging a swing in the sitting position. What shall be the effect on the frequency of oscillation if she stands up? Ans. The system may be roughly regarded as a simple pendulum. When the girl stands up, the ‘length of the simple pendulum’ is reduced. We know that   So, frequency of oscillation is increased. 18. A girl is swinging a swing. Another girl sits gently by her side. How shall the periodic time be affected?

Oscillations 273 Ans. As discussed in the previous question, the system may be regarded as a simple pendulum. The time period of the simple pendulum does not depend upon the mass of the bob. So, the periodic time shall remain unaffected. 19. The bob of a simple pendulum is positively charged. The pendulum is made to oscillate over a negatively charged plate. How shall the time period change as compared to the natural time period of the simple pendulum? Ans. Due to electric force of attraction between the bob and the plate, the effective value of acceleration due to gravity shall increase. Since T =  therefore T shall decrease. 20. What are two basic characteristics of an oscillating system? Ans. (i) The motion of an oscillating system remains confined within two extreme limits on either side of mean position. (ii) The motion of an oscillating system repeats itself over and over again about the mean position of the system. 21. Refer to Fig. 6.8. What is the spring constant of the system? Ans. The given arrangement is a parallel combination of three springs. 2  Spring constant of the system = 2k + k + k = 4k. 22. Two masses 1 kg and 3 kg are attached to opposite ends of a horizontal spring of spring constant 300 N m–1 (Fig. 6.9). What is the natural frequency of the system? 1 kg Fig. 6.9 3 kg Fig. 6.8 Ans. Reduced mass,  =   kg  =      Hz = 3.18 Hz 23. On an average, a human heart is found to beat 75 times in a min ute. Calculate its frequency and period. [NCERT Solved Example] Ans. The beat frequency of heart = 75/(1 min) = 75/(60 s) = 1.25 s–1 = 1.25 Hz Time period T = 1/(1.25 s–1) = 0.8 s 24. Obtain the equation of simple harmonic motion of a particle whose amplitude is 0.04 m and whose frequency is 50 Hz. The initial phase is π. 3 Ans. y = A sin (t + 0) But A = 0.04 m,  = 50 Hz,     y = 0.04     or y =         25. The shortest distance travelled by a particle executing SHM from mean position in 2 second is equal to 3 times its amplitude. Determine its time period. 2

274 Physics—XI Ans. t = 2 s,  , T =? Using y =A sin t, we get    or    or  =  or T = 12 second 26. A spring has a load of 0.50 kg attached at its end. A weight of 4.0 kg extends the spring by 16 cm. This 4.0 kg wt is removed and the block is allowed to execute harmonic vibration with the 0.5 kg load. Calculate its time period. Given : g = 9.8 m s–2. Ans. Since 4.0 kg wt. extends the spring by 16 cm, i.e., 0.16 m,  N m–1 = 245 N m–1  force constant k = T=    = 0.28 s 27. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is 5T . increased by m, the time period becomes 3 . Calculate the ratio Ans. T = 2 and = 2     So,    = 2   or M = M + m or = m or  28. Two particles A and B of equal masses are suspended from two massless springs of spring constants k1 and k2 respectively. If the maximum velocities during oscillations are equal, what is the ratio of the amplitudes of A and B? Ans. A11 = A22, A1 = A2 , So,  29. A 50 g mass hangs at the end of a spring. When 20 g more is added to the end of the spring, it stretches 7 cm more. (a) Calculate the spring constant. (b) If the 20 g is now removed, what will be the period of the motion? Ans. (a) Restoring force = 20 × 10–3 × 9.8 N Extension due to 20 g wt = 7 × 10–2 m Spring constant, k =     = 2.8 N m–1  (b) T =      = 0.84 s 30. When a mass is hung from the lower end of a spring of negligible mass, an extension of 10 cm is produced in the spring. The mass is set into vertical oscillations. What is the period of oscillations? Ans. Time period, T =  =   = 0.635 s

Oscillations 275 31. What is the length of a simple pendulum, which ticks seconds? [NCERT Solved Example] Ans. T = 2 , L =  Now, T = 2 s and g = 9.8 m s–2 L =  m = 0.99 m  1 m  32. The length of a simple pendulum executing simple harmonic motion is increased by 21%. What is the percentage increase in the time period of the pendulum? Ans. T = 2 , T = 2 So,  = = 1.1    × 100 = (1.1 – 1) × 100 = 10    33. The bob of a simple pendulum executes SHM in water with a period t. The period of oscillation of the bob in air is t0. What is the relation between t and t0? Given : density of the bob is kg m–3. Neglect frictional force of water. Ans.  = kg m–3,  = 1000 kg m–3 g =     g =    or g =        t0 = 2 , t = 2  = 2 × 2 or t = 2t0 34. Two pendulums whose lengths differ by 22 cm oscillate at the same place so that one of them makes 30 oscillations and the other 36 oscillations during the same time. Find the lengths of the pendulum. Ans. 30 × 2 = 36 × 2 or =  =  Also, L1 – L2 = 22 cm, –1= or = –1 or = or L2 = 50 cm L1 = 22 + L2 = (22 + 50) cm = 72 cm 35. A light pointer fixed to one prong of a tuning fork touches a vertical plate. The fork is set into vibration and simultaneously the plate is allowed to fall freely. When the plate falls through 10 cm, eight complete oscillations are counted. What is the frequency of the tuning fork? 11 8 Ans. Using S = ut + 2 at2, t = 7 s and  = 1 / 7 Hz = 56 Hz 36. The displacements of two particles executing SHM are given by : x1 = 5 cos 10t and x2 = 4 sin (10t + ). What is the phase difference between the velocities of these particles? Ans. v1 = 50 (– sin 10t) or v1 = 50 cos   10t  v2 = 40 cos (10t + )  2  Phase difference between velocities = (10t + ) –    10t  =–   2  2

276 Physics—XI 37. A block of mass m1 is fastened to a spring of spring constant k and a block of mass m2 is placed against it (Fig. 6.10). What is the compression of the spring at the equilibrium position? m2 m1 k  Ans. (m1 + m2) g sin  = kx or x = Fig. 6.10  .  38. The potential energy of a particle in motion along x-axis is given by U = 4(1 – cos 2x) joule, where x is in metre. Determine the time period of small oscillations, if the mass of the particle is 1 kg. Ans. F=– U =–  sin 2x  2x, when x is small. x x [4(1 – cos 2x)] = – 8 sin 2x  – kx = – 16x, k = 16 Nm–1 T = 2     s. 39. A silver atom in a solid oscillates in SHM in a certain direction with a frequency of 1012 s–1. What is the force constant of the bonds connecting one atom with the other? Molecular weight of silver = 108, Avogadro’s number = 6 × 1026 (kg mol)–1. Ans. m =  kg = 1.8 × 10–25 kg = or 2 =    or k = 42m2 = 4(3.14)2 × 1.8 × 10–25 × (1012)2 N m–1 = 7.1 Nm–1. 40. Three blocks A, B and C weighing 700 g, 500 g and 400 g respectively are suspended at the end of a spring. When A is removed, system moves with a period of 3 s. When B is also removed, with what period the system will oscillate? Ans. 3 = 2 , T = 2 So,   Rigid support k Spring T = 2 s. Mass, m 41. Here are three sets of values for the spring constant, Vane damping constant and mass for the dampled oscillator of Fig. 6.11. Rank the sets according to the time for Damping, b the mechanical energy to decrease to one-fourth of its initial value, greatest first. Fig. 6.11 Set1 2k0 b0 m0 Set 2 k0 6b0 4m0 Set 3 k0 3b0 m0 Ans. 1, 2, 3 The value of k is not important. The ratio m matters. b

Oscillations 277 42. Which of the following relationships between the force F on a particle and the particle’s position x implies simple harmonic oscillation? (a) F = – 5x (b) F = – 400x2 (c) F = 10x (d) F = 3x2. Ans. In simple harmonic motion, force is directly proportional to displacement and is directed towards the mean position. These conditions of SHM are satisfied only by F = – 5x mentioned in (a). So, (a) is the right choice. 43. A particle of mass m is executing oscillations about the origin on the X-axis. Its potential energy is V = kx3, where k is a positive constant. If the amplitude of oscillation is r, prove that the time period of the motion is inversely proportional to the square root of the amplitude of motion. Ans. V = kx3, F = – = – 3kx2 ; Also, F = – m2x,  m2x = – 3kx2 or  = or T = 2 . 44. Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other 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. Ans. (b) and (c). 45. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7 x (b) a = – 200x2 (c) a = – 10x (d) a = 100 x3. Ans. Only (c) i.e., a = –10x represents SHM. This is because acceleration is proportional and opposite to displacement (x). 46. 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). Ans. gm = 1.7 m s–2, ge = 9.8 m s–2, Tm =?, Te = 3.5 s Te = 2 and Tm = 2 Now,  or Tm = Te or Tm = 3.5 s = 8.4 s.

278 Physics—XI 47. Displacement versus time curve for a particle Displacement executing SHM is shown in Fig. 6.12. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum. Ans. (i) (A), (C), (E), (G) CE G (ii) (B), (D), (F), (H) AB DF H Time (s) Velocity is zero when particle is at the extreme Fig. 6.12 position. Velocity is maximum when particle is at the mean position. 48. Two identical springs of spring constant K are attached to a block of mass m and to fixed supports as shown in Fig. 6.13. When the K K mass is displaced from equilibrium position by a distance x towards right, find the restoring force. Fig. 6.13 Ans. The spring towards right hand side of mass gets compressed by a distance x. The spring towards left hand side of mass gets stretched by distance x. Restoring force = – Kx – Kx = – 2Kx. The –ve sign indicates that force is opposite to direction of x. The restoring force acts towards left. 49. What are the two basic characteristics of a simple harmonic motion? Ans. (a) Acceleration is directly proportional to displacement. (b) Acceleration is directed opposite to displacement. 50. Show that for a particle executing SHM, velocity and displacement have a phase difference of /2. Ans. y = A sin t, v =A cos t = A sin       1 = t, 2 = t +  So, 2 – 1 =  51. The length of a second’s pendulum on the surface of Earth is 1 m. What will be the length of a second’s pendulum on the Moon? Ans. 2 =      Inextensible string So, 6l= l or l=  m = 0.17 m Pulley M 52. Find the time period of mass M when displaced from its equilibrium position and then released for the system Fig. 6.14 shown in Fig. 6.14.

Oscillations 279 Ans. If mass M moves down by h, then the spring extends by 2h (because each side expands by h). The tension along the string and spring is same. In equilibrium Mg = 2(k × 2h) where k is the spring constant. On pulling the mass down by x, F = Mg – 2k(2h + 2x) = – 4kx So, T = 2 M 4k 53. Show that the motion of a particle represented by y = sin t – cos t is simple harmonic with a period of 2/. Ans. y =     =      =           This is the equation of SHM of angular frequency . =    54. Find the displacement of a simple harmonic oscillator at which its P.E. is half of the maximum energy of the oscillator. Ans.     or y2 = or y =  55. A mass of 2 kg is attached to the spring of spring constant 50 N m–1. The block is pulled to a distance of 5 cm from its equilibrium position at x = 0 on a horizontal frictionless surface from rest at t = 0. Write the expression for its displacement at any time t. Ans. =  = 5 rad s–1, A = 5 × 10–2 m Using x = A sin t, x = 5 × 10–2 sin 5t SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. A particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8 × 10–3 J. Obtain the equation of motion of this particle if the initial phase of oscillation is 45°. Ans. y =   ...(1) A = 0.1 m,      Now,  = 8 × 10–3 2 =  or  = 4 s–1  From equation (1), y =    .  

280 Physics—XI 2. A vibrating simple pendulum of time period T is placed in a lift which is accelerating upwards. What will be the effect on the time period? F + F Ans. When the lift accelerates upwards, mg – mg = ma R(= mg) + F mg = mg + ma or g = g + a So, there is an effective increase in the value of acceleration due to gravity. a We know that T =  . Clearly, T decreases. mg Fig. 6.15 3. A spring of force constant k is cut into two pieces, such that one piece is double the length of the other. What is the force constant of the longer piece of the spring? Ans. Let the length of original piece be l. Then, lengths of shorter piece and longer piece will be l/3 and 2l/3 respectively. As F = kl or kl = constant or k  Force constant of the longer piece, k1  = = or k1 = k. 4. A particle is performing SHM along x-axis with amplitude 4.0 cm and time period 1.2 s. What is the minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm? 2 or t= T sin–1 x Ans. x = A sin t = A sin T t 2  A  At x = 2, t =  sin–1   =    = 0.1 s  At x = 4, t =  sin–1      = = 0.3 s  Required time is (0.3 – 0.1) s = 0.2 s. d 5. A body is dropped in a hole drilled across a diameter of the Earth. This body shall execute SHM. What is the time period of SHM? Ans. At depth d below the surface of Earth, gd = g    y   = g   or gd = y   F = mgd = y Force constant, k = T = 2 = 2 = 2 . Fig. 6.16

Oscillations 281 6. Refer to the system shown in Fig. 6.17. What is the force constant 1 of the system? Ans. The given arrangement is a series combination of three springs of spring constants k1, k2 and (k3 + k4) respectively.   2      k=   . 34  7. A simple pendulum consisting of an inextensible length l and mass m is oscillating in a stationary lift. The lift then accelerates upwards Fig. 6.17 with a constant acceleration of 4.5 m s–2. Write expression for the time period of pendulum in two cases. Does the time period increase, decrease or remain the same, when lift is accelerating upwards? Ans. Case I : When the lift is stationary, Time period, T1 = 2 = 2 Case II : When the lift accelerates upwards Time period, T2 = 2 = 2 = 2   Clearly, T2 < T1. Therefore time period of pendulum decreases, when the lift accelerates upwards. 8. In damped oscillations, the amplitude of oscillations is reduced to one-fourth of its initial value 10 cm at the end of 50 oscillations. What will be its amplitude when the oscillator completes 150 oscillations? Ans. Using A =          = A0   = 10   cm = 0.156 cm. A = A0 = A0     9. Two simple harmonic motions are represented by the equations y1 = 0.1 sin       and y2 = 0.1 cos t. Calculate the initial phase of the velocity of particle 1 with respect to the velocity of particle 2? Ans. v1 = (y1) = 0.1 cos      × 100= 10 cos          v2 = (y2) = 0.1 (– sin t) ×  = – 0.1  sin t = 0.1 (– sin t) = 0.1 cos          Required phase difference = – = –

282 Physics—XI 10. Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case. (a) sin t – cos t (b)sin2 t [NCERT Solved Example] Ans. sin t – cos t = sin t – sin   t  = 2 cos  sin  t   = 2 sin  t    2   4   4   4  2 This function represents a simple harmonic motion having a period T =  and a phase angle    or  7  .  4   4  (b) sin2 t = 1 – 1 cos 2 t 2 2  The function is periodic having a period T =  . It also represents a harmonic motion with the point of equilibrium occurring at 1 instead of zero. 2 11. Given : y = sin2 t. Show that it does not represent simple harmonic motion. What is the period of the given function? Ans. = 2 sin t cos t ×  =  sin 2t acceleration = =  (cos 2t) 2 = 22 cos 2t Since the acceleration is not proportional to displacement therefore the given function does not represent simple harmonic motion. Now, y = sin2 t = (1 – cos 2t) The given function is clearly a periodic function of angular frequency 2. Time period, T =  =   12. A particle of mass 1 kg is moving with SHM. Its greatest velocity is 20 m s–1 and its amplitude is 10 m. Find the period and the force of attraction towards the centre when the particle is at its greatest distance. Ans. Maximum velocity, A = 20 m s–1 Amplitude, A = 10 m Angular velocity,  =   Now,  = 2 or T =  second = 3.143 s Force of attraction towards centre = Restoring force = m2A = 1 × 2 × 2 × 10 N = 40 N 13. A body executes SHM under the influence of one force with a time period of 0.8 second. It has a time period of 0.6 second under the action of another force. Calculate the time period when both the forces act in the same direction simultaneously.

Oscillations 283 Ans. We know that, in SHM,  or T =    Now,  ...(1) and  ...(2) Dividing (1) by (2),   Again,  ...(3)  Dividing (3) by (1), we get   or  = 0.48 s  14. A sphere is hung with a wire. A 30° rotation of the sphere about the wire generates a restoring torque of 4.6 N m. If the moment of inertia of the sphere about the wire is 0.082 kg m2, deduce the frequency of angular oscillations. Ans. Angular displacement,      [...  rad = 180°] Torque, C = 4.6 N m where C is the restoring torque per unit angular displacement.  C=   Moment of inertia, I = 0.082 kg m2 Frequency,  =  = 1.65 Hz    15. Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time, will its kinetic energy be 75% of the total energy? Ans. y = A sin t, v = A cos t Kinetic energy, K = mv2 = mA22 cos2 t Now, × mA22 = mA22 cos2 t or cos2 t = or cos t =  t=  or 1 or t = or t= = s= 6s

284 Physics—XI 16. A body executing linear SHM has a velocity of 0.03 m s–1 when its displacement is 0.04 m and a velocity of 0.04 m s–1 when its displacement is 0.03 m. (a) Find the amplitude and period of the oscillation. (b) If the mass of the body is 50 × 10 –3 kg, calculate the total energy of oscillation. Ans. (a) v =   or v2 =    Now, 0.032 =   and     On simplification, A = 0.05 m and  = 1 rad s–1 Time period,    T=     (b) Energy =          = 6.25 × 10–5 J 17. A mass M attached to a spring oscillates every 2 second. If the mass is increased by 2 kg, the period increases by one second. Find the initial mass M, assuming that Hooke’s law is obeyed. Ans. Period of oscillation, T =  where M is the suspended mass and k is the spring constant of the spring. In the first case, 2=  In the second case, 3=  Squaring and dividing, = or M = 1.6 kg 18. A pendulum clock gives correct time. What is the error in time per day if the length increases by 0.05%? Ans. Let T be the correct time period and T' the time period when the length is increased by 0.05%. Then T=          Dividing, =             Applying Binomial theorem and neglecting squares and higher powers, we get =      = 0.00025 or    Loss of time per second = 0.00025 s Loss of time per day = 0.00025 × 24 × 60 × 60 s = 21.6 s

Oscillations 285 19. A cylindrical wooden block of cross-section 15 cm2 and mass 230 g is floated over water with an extra weight 50 g attached to its bottom. The cylinder floats vertically. From the state of equilibrium, it is slightly depressed and released. If the specific gravity of wood is 0.30 and g = 9.8 m s–2, deduce the frequency of oscillation of the block. Ans. Cross-sectional area, A = 15 cm2 = 15 × 10–4 m2 Total mass, m = (230 + 50) g = 280 g = 0.28 kg Density of wood,  = 0.30 g cm–3 = 300 kg m–3 Density of water,  = 103 kg m–3 If the cylinder be depressed through a distance y, then restoring force, F = weight of extra water displaced = (A × y)g = 15 × 10–4 × y× 103 × 9.8 N m–1 spring factor, k = = 15 × 10–4 × 103 × 9.8 N m–1 inertia factor, m = 0.28 kg Frequency, =    Hz = 1.15 Hz  20. Is y = a sin t cos t a SHM? Explain your answer. 1 Ans. y = a sin t cos t y = 2 a (2 sin t cos t) 1 dy 1 or y = 2 a sin 2t dt = 2 a × 2 cos 2t d2y =– 1 a × (2)2 sin 2t d2y = – 42  1 a sin 2t  dt2 2 dt2  2  d2y = – 42y acceleration  – displacement. dt2 Thus the given equation represents SHM. 21. In SHM of frequency , what is the frequency of oscillation of kinetic energy? Ans. In SHM, y = A sin t = A sin 2t v = = (A sin 2t) = 2A cos 2t Kinetic energy, K = mv2 = m(2A)2 cos2 2t Maximum kinetic energy, K0 = m(2A)2  K = K0 cos2 2t or K = [1 + cos 2 (2t)] or K = [1 + cos 2(2)t] So, the frequency of oscillation of kinetic energy is 2.

286 Physics—XI 22. In a HCl molecule, we may treat Cl to be of infinite mass and H alone oscillating. If the oscillation of HCl molecule shows frequency 9 × 1013 s–1, deduce the force constant. Given : Avogadro’s number = 6 × 1026 per kg mole. Ans. 1 kg of H has 6 × 1026 atoms.  m=   =       or k = 422m  = 4(3.14)2 (9 × 1013)2 ×   = 5.32 × 102 N m–1 23. A particle at the end of a spring executes simple harmonic motion with a period t 1 while the corresponding period for another spring is 2t . What is the period of oscillation when the two springs are joined in series? Ans. t1 = 2 , t2 = 2 T = 2 = 2  = 2   T2 = 42    T2 = 42 + 42 = t12 + t22    24. Two masses m1 and m2 are suspended together by a massless spring of spring constant k as shown in Fig. 6.18. When the masses are in equilibrium, m1 is removed without disturbing the system. Calculate m2 the amplitude of oscillation and the angular frequency of m2 . m1 Ans. Let l1 and l2 be the extensions produced by masses m1 and m2 respectively. Fig. 6.18 Then, (m1 + m2) g = k (l1 + l2) ...(1) Fig. 6.19 After m1 is removed, m2 g = kl2 ...(2) Equations (1) and (2) gives m1 g = kl1 or l1 = Now, angular frequency of m2,     . 25. One end of a long metallic wire of length l is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. Find the time period of oscillations of mass m. The cross-sectional area and Young’s modulus of the wire are given to be A and Y respectively. Fl Fl F Ans. Y = or y1 = AY Ay1 F = ky2 or y2 = k y = y1 + y2 = F l + 1 =  l k  AY  F or F = AYk y,  AY k   AYk  lk  AY  AYk m = 2 m(lk  AY) . K = lk  AY and T = 2 K AYk

Oscillations 287 26. Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread l = 20 cm in length, if it is located in a liquid whose density is n = 3.0 times less than that of the ball. The resistance of the liquid is to be neglected. Ans. Let, V = volume of bob  = density of material of the bob g = effective value of acceleration due to gravity Now, effective weight = weight of bob – upthrust of liquid  or g =    or  Vg = Vg – V      Again, T =  or T =   or     Substituting values, we get T =      = 1.1 s  27. You are provided with a light spring, a metre scale and a known mass. How will you find the time period of oscillation of a mass without the use of clock? Ans. Suspend the spring from a rigid support. Attach the given mass to the free end of the spring. Measure the extension l with the help of the given metre scale. We know that the time period of the spring is given by  Also, mg = kl or    Since both l and g are known therefore T can be B calculated. m 28. A small block of mass m is attached to three identical springs A, B and C, each of spring constant k. If the 45° C block is pushed slightly against the spring A and A released, then what will be the period of oscillations? Ans. Let x be the distance through which the block is pushed against spring A. Extension of spring C is x cos 45° or . Extension of spring B is x sin 45° or . Fig. 6.20 Force on block due to all the springs      = kx +     = kx + k x = 2 k x Acceleration = T = 2 = 2 .

288 Physics—XI 29. A block of mass m is attached to three identical springs, C 90° B each of force constant k, as shown in Fig. 6.21. If the block 135° is pushed slightly against spring A and released, find the time period of oscillation. A Ans. Total restoring force on block, F = FA + FB cos 45° + FC cos 45° = – ky – 2 ky cos 45° = – ky – 2ky cos2 45° = – 2 ky  K = 2k T = 2 m = 2 m K 2k . Fig. 6.21 30. Springs of spring constants k, 2k, 4k, 8k, ...... are connected in series. A mass m kg is attached to the lower end of the last spring. The system is allowed to vibrate. Determine the time period of oscillation. Ans. If K be the equivalent spring constant of the given combination, then                =  or K = T = 2 = 2 .      31. In Fig. 6.22, mass m is released when the spring is unstretched. Neglecting the inertia and friction of the pulley, what is the amplitude of the resulting oscillation? Ans. Suppose the mass falls through a distance h before stopping. Then,   In its upward motion, the mass will rise through distance h. k m  Amplitude of oscillation =  . 32. A particle undergoing simple harmonic oscillation of period T Fig. 6.22 is at –xm at time t = 0. Is it at – xm , at + xm , at 0, between – xm and 0, or between 0 and + xm when (a) t = 2.00 T (b) t = 3.50 T and ) (c) t = 5.25 T? Ans. (a) At t = 2.00 T, the particle is back at the initial position. So, x = – xm (b) At t = 3.50 T, the particle is half vibration away from the initial position. So, x = + xm (c) At t = 5.25 T, the particle is one-fourth vibration away from the initial position. So, x = 0. 33. In Fig. 6.23, the block has a kinetic energy of 3 J and the spring has an elastic potential energy of 2 J when the block is at x = + 2.0 cm. (a) What is the kinetic energy at x = 0? What are the elastic potential energies at (b) x = – 2.0 cm and (c) x = – xm?

Oscillations 289 Frictionless x – =0 Fig. 6.23 Ans. (a) At x = 0, the system is at its mean position. So, total energy is kinetic. Thus, kinetic energy at x = 0 is equal to total energy i.e., (3 + 2) J or 5 J. (b) It follows from symmetry considerations that the elastic potential energy at x = – 2 cm will be the same as elastic potential energy at x = + 2 cm. So, elastic potential energy at x = – 2 cm is 2 J. (c) At x = – xm, the total energy is potential. So, potential energy at x = – xm is (2 + 3) J i.e., 5 J. 34. In damped oscillations, the amplitude after 50 oscillations is 0.8 a0 , where a0 is the initial amplitude. Determine the amplitude after 150 oscillations.  bt Ans. at = a0 e 2m , at = a0e–t 0.8a0 = a0e– (50T) or e(50 T) = 5 or 50T = loge 5 4 4 150 T = loge a0 ; 3 × 50 T = loge a0 ; at at 3 loge 5 = log a0 a0 =  5 3 = 125 , at = 64 a0. 4 at at  4  64 125 35. When a force F1 acts on a particle, frequency is 6 Hz. When a force F2 acts, frequency is 8 Hz. What is the frequency when both the forces act simultaneously in same direction? Ans. Let m be the mass of the particle and A be its amplitude of oscillations. Therefore, F1 = mA12 = mA(21)2 = mA 4212 and F2 = mA22 = mA(22)2 = mA 4222 When both forces act simultaneously in same direction, then Net force, F = F1 + F2 mA 422 = mA 4212 + mA 42 22 or 2 = 12 + 22 or  =    or  =  Hz = 10 Hz. 36. A solid sphere of radius R is floating in a liquid of density  with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing simple harmonic motion. Find the frequency of these oscillations. Ans. If the sphere is displaced through a small distance x vertically inside the liquid, then the effective restoring force on it is R2 xg. [Here R2 is area of cross-section of sphere]

290 Physics—XI  –m = R2 xg =– x         Therefore, motion is SHM. Now, 2 = or  =  Frequency =  . 37. A block executes SHM about mean position y0 with an amplitude A and angular frequency  . At t = 0, the block is at the mean position and moving upwards. At a certain height y from the mean position, the block gets detached from the spring (when the block gets detached assume that spring compresses immediately such that it does not interfere with subsequent motion of the block). Find y so that height h attained by the block be maximum. [2A > g]. Ans. When the block detaches from the spring at a distance y above the mean position, its velocity v =   Total height attained by the block after detachment, h = +y or h =  (A2 – y2) + y For hmax., =0  y= .  38. A point mass m is suspended at the end of a wire of negligible mass, length l and cross-section area A. If E is the Young’s modulus of the material of the wire, obtain an expression for the time period of oscillation for the simple harmonic motion along the vertical line. Ans. E = or E =    or F = AE Restoring force = – L Spring constant, k = Time period, T = 2 = 2 . 39. A particle moving in straight line has velocity v given by v2 =  –  y 2 where  and  are constants and y is the distance of the particle from a fixed point in the line. Determine the time period and amplitude. Ans. v2 =  – y2 dv dy Differentiating w.r.t. time t, 2v dt = – 2y dt = – 2yv or dv = – y or Acceleration = – y dt   2 = ,  =   Clearly,  Now, T=  When v = 0, y = A.  0 =  – A2 or A =  . 

Oscillations 291 40. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) The rotation of the 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 lowermost point. (d) General vibrations of a polyatomic molecule about its equilibrium position. Ans. (b) and (c) SHM (a) and (d) represent periodic but not SHM Note : A polyatomic molecule has a number of natural frequencies ; So, in general, its vibration is a superposition of SHM’s of a number of different frequencies. This superposition is periodic but not SHM. 41. Figure 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)? Fig. 6.24 Ans. (b) and (d) are periodic, each with a period = 2 s ; (a) and (c) are not periodic. [Note in (c), repetition of merely one position is not enough for motion to be periodic ; the entire motion during one period must be repeated successively]. 42. 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 ( is any positive constant) : (a) sin t – cos t (b) sin3 t (c) 3 cos     (d) cos t + cos 3t + cos 5t   (f) 1 + t + 2t2 (e) exo (– 2t2) Ans. (a) Simple harmonic,   (b) periodic,   but not simple harmonic (c) simple  harmonic, T =   (d) Periodic,   but not simple harmonic (e) non-periodic (f)     non-periodic (physically not acceptable as the function  as t ).

292 Physics—XI 43. 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 Ans. (a) 0, +, + (b) 0, –, – (c) –,0,0 (d) –, –, – (e) +, +, + (f) –, –, – (here 0 stands for the magnitude of the quantity). 44. 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? Ans. M = 50 kg, y = 20 cm = 0.2 m, T = 0.60 s, mg =? F = ky or Mg = ky or k =  N m–1 = 2450 N m–1 = Now, T = 2 or T2 = 42 or m =  or m =    kg = 22.3 kg or mg = 22.3 × 9.8 N = 218.5 N = 22.3 kgf. 45. A spring having spring constant 1200 N m–1 is mounted on a horizontal table as shown in figure. 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. 6.25 Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Ans. k = 1200 N m–1, m = 3 kg, A = 2.0 cm = 2 × 10–2 m (i)  =   s–1 = 3.2 s–1.  (ii) maximum acceleration = 2A = A =   m s–2 = 8 m s–2. (iii) maximum speed = A = A = 2 × 10–2 m s–1 = 0.4 m s–1. 46. In question 45, let us take the position of the mass when the spring is unstretched at x = 0, and the direction from left to right as the positive direction of x-axis. Give x

Oscillations 293 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? Ans. A = 2 cm,  = = s–1 = 20 s–1 (a) Since time is measured from mean position,  x = A sin t = 2 sin 20t (b) At the maximum stretched position, the body is at the extreme right position. The initial phase is  .  x = A sin      = A cos t = 2 cos 20t   (c) At the maximum compressed position, the body is at the extreme left position. The initial phase is  .  x = A sin     = – A cos t = – 2 cos 20t   Note. The functions neither differ in amplitude nor in frequency. The functions differ only in initial phase. 47. Figs. 6.26 (a) & (b) 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. Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P in each case. T=2s P 2m T=4s ( = 0) 3 cm P( = 0) (a) (b) Fig. 6.26 Ans. (a) The particle begins its motion from the mean position and along negative direction of x- axis.  x = – A sin t or      Here x is in cm.

294 Physics—XI (b) The particle begins its motion from the extreme position, along positive direction of x-axis.   x = – A cos t or x = – 2 cos t or x = – 2 cos t Here x is in metre. 48. 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? Ans. In the given problem, T =  where ge is the effective value of acceleration due to gravity. While the acceleration due to gravity g acts in the downward direction, the radial acceleration acts in the horizontal direction. So, the effective value ge of acceleration due to gravity is given by ge =  or T =  .     T =       49. 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  is defined by the relation J = – , where J is the restoring couple and  the angle of twist). Ans. T =   or   or =  or      or =  or  =   = 1.97 N m rad–1. 50. A body of mass m is situated in a potential field U(x) = U0 (1 – cos x) when U0 and  are constants. Find the time period of small oscillations. Ans. F = – [U0 – U0 cos x] = – U0  sin x But sin x  x F = – U02x F x The –ve sign shows that F is directed towards mean position. So, the body will execute SHM. Spring constant = U02 Inertia factor = m T =  

Oscillations 295 51. Consider a pair of identical pendulums, which oscillates with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums? Ans. 1 = 0 sin (t + 1) 2 = 0 sin (t + 2) For first pendulum, 1 = 2° = 0 2 = 2sin (t + 1) or t + 1 = 90° For second pendulum, 2 = – 1° – 1 =2 sin (t + 2) or sin (t + 2) = – = sin (180° + 30°) = sin 210° t + 2 = 210° (t + 2) – (t + 1) = 210° – 90°  2 – 1 = 120° 52. A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand. (a) What is the amplitude of oscillation? (b) Find the frequency of oscillation? Ans. (a) The lowest position and position of hand are the extreme positions for the oscillation. Amplitude of oscillation = = 2 cm. (b)  =  where l is the extension in spring  =  = 3.52 s–1 LONG ANSWER TYPE QUESTIONS (5 Marks) 1. Explain what is meant by a periodic and an oscillatory motion. Distinguish between them. Ans. Periodic Motion. Periodic motion is that motion which repeats itself at regular intervals of time. The interval of time is called the time period of periodic motion. Examples. (i) Suppose an insect climbs up a ramp and falls down. It comes back to the initial point and repeats the process identically. If we draw a graph of its height above the ground versus time, it would look something like Fig. 6.27. () T Fig. 6.27. Periodic motion Fig. 6.28. Periodic motion (ii) If a child climbs up a step, comes down, and repeats the process, its height above the ground would look like that in Fig. 6.28.

296 Physics—XI (iii) When you play the game of bouncing a ball off the ground, between your palm and the ground, its height versus time graph would look like the one in Fig. 6.29. Note that both the curved parts in Fig. 6.29 are sections of a parabola given by the 1 Newton’s equation of motion, h = ut + 2 gt2 for downward motion, and h = ut – 1 2 gt2 for upward motion, with different values of u in each case. Fig. 6.29. Periodic motion (iv) Motion of planets around the Sun. (v) Rotation of Earth about its polar axis. (vi) Motion of Moon around the Earth. (vii) Motion of Halley’s comet around the Sun. (viii) Motion of the pendulum of a wall clock. (ix) Motion of the balance wheel of a watch. (x) Motion of the hands of a clock. Oscillatiory Motion If a body moves back and forth repeatedly about a mean position, it is said to possess oscillatory or vibratory motion. Examples. (i) Motion of the pendulum of a wall clock. (ii) Vibrations of the wire of a ‘sitar’. (iii) Vibrations of the drum of a ‘tabla’. (iv) Oscillations of a mass suspended from a spring. (v) Motion of liquid in a U-tube when the liquid is once compressed in one limb and then left to itself. (vi) A weighted test tube floating in a liquid executes oscillatory motion when pressed down and released. Difference between periodic motion and oscillatory motion. An oscillatory motion is always periodic. A periodic motion may or may not be oscillatory. So, oscillatory motion is merely a special case of periodic motion. As an example, the motion of the planets around the Sun is periodic but not oscillatory. 2. Define simple harmonic motion. Write its dimensional formula. Ans. Simple Harmonic Motion is a motion which is necessarily periodic and oscillatory about a fixed mean position. A particle executing such a motion is always in stable equilibrium about its mean position. So, if a particle is disturbed slightly from its mean position, it tends to return to its mean position. The force which tends to bring the particle back to the mean position is called the restoring force. The greater the displacement of the particle from the mean position, greater is the restoring force. Further, the restoring force is always directed towards the mean position. Thus, simple harmonic motion is defined as such an oscillatory motion about a fixed point (mean position) in which the restoring force is always proportional to the displacement from that point and is always directed towards that point. If a particle suffers a small displacementx from its mean position, then the magnitude of restoring force F is given by F = – kx ...(1)

Oscillations 297 where k is known as the force constant. Its SI unit is N m–1. Its dimensional formula is [ML0 T – 2]. The negative sign in equation (1) indicates that the restoring force is directed towards the mean position. 3. Which of the following functions of time represent (a) periodic and (b) non-periodic motion? Give the period for each case of periodic motion [ is any positive constant]. (i) sin t + cos t (ii) sin t + cos 2t + sin 4t (iii) e–t (iv) log (t). [NCERT Solved Example] Ans. (i) sin t + cos t is a periodic function. It can also be written as sin (t + /4). Now              The periodic time of the function is 2/. (ii) This is an example of a periodic motion. It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its value, sin t has a period T = 2/ ; cos 2t has a period / =T/2 ; and sin 4t has a period 2/=T/4 . The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is T. Thus, the sum is a periodic function with a period 2/. (iii) The function e–t is non-periodic, it decreases monotonically with increasing time and tends to zero as t  and thus, never repeats its value. (iv) The function log (t) increases monotonically with time t. It, therefore, never repeats its value and is a non-periodic function. It may be noted that as t , log (t) diverges to . It, therefore, cannot represent any kind of physical displacement. 4. A harmonic oscillation is represented by x = 0.34 cos (3000t + 0.74) where x and t are in mm and s respectively. Deduce (i) amplitude, (ii) frequency, (iii) angular frequency, (iv) period and (v) initial phase. Ans. x = 0.34 cos (3000 t + 0.74) Comparing with general equation x= A cos (t + 0), we get A = 0.34 mm,  = 3000 rad s–1 and  0 = 0.74 rad (i) amplitude = 0.34 mm = 0.034 cm   = 477.27 Hz (ii) frequency,     (iii) angular frequency,  = 3000 rad s–1 (iv) period,     = 0.0021 s (v) initial phase, 0 = 0.74 rad 5. Show how simple harmonic motion can be realised in practice, as the projection of a uniform circular motion? Ans. Consider a particle moving with uniform speed along the circumference of a circle of radius ‘A’. This circle is known as the reference circle while the particle is known as the reference particle or generating point.

298 Physics—XI Let P be the position of the reference particle at any time. Y At that time, N is the projection of the reference particle P on the diameter XOX (Fig. 6.30). When the reference particle moves from X to Y, the projection N moves from O NX X to O. When the reference particle moves from Y to X, X the projection N moves from O to X. Similarly, when the reference particle moves from X to X via Y, the projection N moves from X to X. Thus as the reference particle Y completes one revolution, the projection N completes one vibration about the mean positionO. Fig. 6.30. SHM is projection The motion of N along XOX is simple harmonic motion. of uniform circular motion The motion of the projection of the reference particle along any other diameter of the circle of reference will also be simple harmonic. This leads to the following geometric definition of simple harmonic motion. Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle of reference. 6. Fig. 6.31 depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the f igures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. [NCERT Solved Example] Fig. 6.31 Ans. (a) At t = 0, OP makes an angle of 45° = /4 rad with the (positive direction of) x- axis. After time t, it covers an angle  in the anticlockwise sense, and makes an angle of    with the x-axis. The projection of OP on the x-axis at time t is given by,         For T = 4 s,      which is a SHM of amplitude A, period 4 s, and an    initial phase = rad. (b) In this case at t = 0, OP makes an angle of    rad with the x-axis. After a time t, it covers an angle of  in the clockwise sense and makes an angle of      

Oscillations 299 with the x-axis. The projection of OP on the x-axis at time t is given by              For T= 30 s,      Writing this as      and comparing with Eq. x = xm cos (t + ), we find   that this represents a SHM of amplitude B, period 30 s, and an initial phase of   rad. 7. A body oscillates with SHM according to the equation (in SI units), x = 5.0 m cos [(2  rad s–1) t + /4]. At t = 1.5 s, calculate the (a) displacement, (b) speed and (c) acceleration of the body. [NCERT Solved Example] Ans. The angular frequency  of the body = 2 rad/s and its time period T = 1 s. At t = 1.5 s (a) Displacement = (5.0 m) cos [(2 rad s–1) × 1.5 s + /4] = (5.0 m) cos [(3 + /4) rad] = – 5.0 × 0.707 m = – 3.535 m (b) Using v = – A sin (t + 0), Speed of body = – (5.0 m) (2 s–1) sin       = – (5.0 m) (2 s–1) sin          = 10 × 0.707 m s–1 = 22.22 m s–1 (c) Acceleration of the body = – (2 s–1)2 × displacement P = – (2 s–1)2 × (–3.535 m) = 139.7 m s–2 5 cm 8. A person normally weighing 60 kg stands on a platform which O oscillates up and down harmonically at a frequency of 2.0 s –1 and an 5 cm amplitude of 5.0 cm. If a machine on a platform gives the person’s weight against time, deduce the maximum and minimum reading it Q will show. Given : g = 10 m s–2. Fig. 6.32 Ans. Let P and Q be the extreme positions between which the platform vibrates. Let O be the mean position (Fig. 6.32). Then, OP = OQ = amplitude, A = 5 cm = 0.05 m Frequency,  = 2.00 per second Angular frequency,  = 2 = 4 rad s–1 At P or Q, the acceleration is maximum i.e., A2. At these positions, restoring force = mA2 = 60 × 0.05 × (4)2 N = 474.1 N = 47.41 kg wt.        

300 Physics—XI At P, both the weight and the restoring force are directed towards the mean position. So, the effective weight is maximum at P. It is given by W1 = (60 + 47.41) kg wt. = 107.41 kg wt. At Q, the weight and the restoring force are opposed to each other. So, the effective weight is minimum and is given by W2 = (60 – 47.41) kg wt. = 12.59 kg wt. 9. A body of mass 10–2 kg has a velocity of 6 × 10–2 m s–1 after one second of its starting from mean position. If time period is 6 second, determine the kinetic energy, potential energy, total energy, maximum kinetic energy, maximum potential energy. Ans. m = 10–2 kg, v = 6 × 10–2 m s–1, t = 1 s, T = 6 s W e k n ow th at v = A  cos t  6 × 10–2 =      or           1.8 × 10–5 J  Kinetic energy = Total energy =          = 7.2 × 10–5 J     Potential energy = Total energy – Kinetic energy = (7.2 × 10–5 – 1.8 × 10–5) J = 5.4 × 10–5 J Maximum kinetic energy = Maximum potential energy = Total energy = 7.2 × 10–5 J 10. A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position. [NCERT Solved Example] Ans. The block executes SHM. Its angular frequency is =     Its displacement at any time t is then given by, x (t) = 0.1 cos (7.07t) 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 or sin (7.07t) =  The velocity of the block at x = 5 cm = – 0.1 × 7.07 × 0.866 m s–1 = – 0.6123 m s–1 Kinetic energy of the block =    = 0.1874 J Potential energy of the block =    = 0.0625 J The total energy of the block at x = 5 cm = K.E. + P.E. = 0.2499 J

Oscillations 301 We also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system =   = 0.25 J which is same as the sum of the two energies at a displacement of 5 cm. It establishes that in SHM the total energy is constant. 11. Obtain an equation for the frequency of Oscillation of spring of mass ‘m’ and force constant ‘k’ on a vertical spring. Ans. Consider a light and elastic spring of length L suspended vertically from a rigid support [Fig. 6.33 (a)]. The spring is unstretched, i.e., in the relaxed position. Let a small mass m be suspended from the lower end of the spring such that the spring suffers an extension l in its natural length [Fig. 6.33 (b)]. Let F be the restoring force developed in the spring. Applying Hooke’s law, we get F=–kl ...(1) where k is a spring constant of the Fig. 6.33. Mass oscillating on a vertical spring spring. The minus sign in equation (1) indicates that while the extension l is directed downwards, the restoring force F is directed upwards. Let the mass be now pulled down through small distancey (< l) as shown in Fig. 6.33 (c). Now, the total extension is (l + y). Let (F + F) be the corresponding restoring force. Then, F + F = – k (l + y) ...(2) Subtracting (1) from (2), we get F = – k y This additional restoring force is responsible for producing acceleration in the suspended mass when it is released. When the mass is released, it begins to oscillate up and down. At extension y, it has acceleration . Now,   [Newton’s second law of motion] Put   where  is a positive constant of the given system because both k and m are fixed.   So, the acceleration is directly proportional to displacement and is directed towards the mean position. Hence the motion of the suspended mass is simple harmonic.