JEE-Physics Example Calculate the molar specific heat at constant volume. Given : specific heat of hydrogen at constant pressure is 6.85 cal mol–1 K–1 and density of hydrogen = 0.0899 g cm–3. One mole of gas = 2.016 g, J = 4.2 × 107 erg cal–1 and 1 atmosphere = 10 6 dyne cm–2. Solution Since the density of hydrogen is 0.0899 g cm–3 therefore volume occupied by 0.0899 g of hydrogen at NTP is 1000 cm3. So, volume of 1 mole (2.016 g) of gas, V 1000 2.016 cm3 0.0899 R PV 106 1000 2.016 CP – CV = J T J 0.0899 273 4.2 107 = 1.96 cal mol–1 K–1 C = C – 1.96 = (6.85 – 1.96) = 4.89 cal mol–1 K–1 VP Example The specific heat of argon at constant volume is 0.075 kcal/kg K. Calculate its atomic weight, [R = 2cal/mol K] Solution As argon is monoatomic, its molar specific heat at constant volume will be 33 2 CV R 3 cal/mol K, CV =Mw cV and cV = 0.075 cal/g K 2 2 So 3 = M × 0.075 M = 3 = 40 gram/mole w w 0.075 Effeciency of a cycle () : total Mechanical work done by the gas in the whole process Heat absorbed by the gas (only ve) area un der th e cy cle in P -V curve \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 = Heat injected into the system Example n moles of a diatomic gas has undergone a cyclic process ABC as shwon in figure. Temperature at a is T . Find 0 (i) Volume at C. (ii) Maximum temprature. (iii) Total heat given to gas. (iv) Is heat rejected by the gas, if yes how much heat is rejected. (v) Find out the efficiency. 66 E
JEE-Physics Solution 2 P0 P0 V = 2V (i) Since triangle OAV and OCV are simillar therefore V V0 0 0 (ii) Since process AB is isochoric hence PA PB T = 2T TA TB B 0 Since process BC is isobaric therefore TB TC T = 2T = 4 T VB VC C B 0 1 (iii) Since process is cyclic therefore Q = W = area under the cycle = P V . 2 00 (iv) Since U and W both are negative in process CA Q is negative in process CA and heat is rejected in process CA Q = W + U CA CA CA 15 = 2 [P0 + 2P0] V0 – 2 nR (Tc – Ta ) = 1 [P + 2P ] V – 5 nR 4P0 V0 P0 V0 0 00 2 nR nR 2 = –9P0V0 = Heat injected. (v) = efficiency of the cycle = work done by the gas P0 V0 / 2 × 100 h ea t in je cte d Q injected where Q = Q + Q = 5 nR (2T0 T0 ) 5 nR (2T0 ) 2P0 (2 V0 V0 ) 19 PV. inj AB BC 2 2 2 00 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 100 Therefore = % 19 CARNOT CYCLE Carnot devised an ideal engine which is based on a P Q1 P1 A T1 reversible cycle of four operations in succession : on standon source (i) Isothermal expansion, A B P2 on standD on sink B (ii) Adiabatic expansion, B C Q2 T2 (iii) Isothermal compression C D P4 C P3 V (iv) Adiabatic compression. D A V1 V4 V2 V3 E 67
JEE-Physics MAIN PARTS OF CARNOT'S ENGINE ARE • Source of heat : It is a hot body of very large heat capacity kept at a constant high temperature T . Its upper surface is perfectly 1 conducting so that working substance can take heat from it. • Mechanical arrangements and working substance : It is a hollow cylinder whose walls are perfectly non–conducting and its base is perfectly conducting fitted with non–conducting piston. This piston move without any friction. Ideal gas enclosed in cylinder as a working substance. • Heat sink : It is a cold body at low temperature T . It is a body of large heat capacity and its upper surface is highly 2 conducting so that working substance can reject heat to it. • Stand : It is made by perfectly insulating material. When cylinder placed on it working substance can expended or compressed adiabatic. • Working : A set of reversible processes through them working substance is taken back to initial condition to get maximum work from this type of ideal engine. Processes of Carnot's cycle can be denoted by an indicator diagram. • Isothermal expansion A B Initially the cylinder is taken to be in thermal equilibrium with the high temperature Q T , this is initial state of working substance denoted by point A (P , V , T ). After that source 1 111 the piston is allowed to move outward slowly. With the movement of the piston. The process is very slow so that it is isothermal. Heat from reservoir flows through the base of cylinder into the gas so temperature of the gas remains T . Gas expand and 1 receive heat Q1 from source and gets state B(P2, V2, T1) This heat input Q to the gas from path A to B is utilized for doing work W . 11 By path A to B the heat input to the gas = the work done against the external V2 PdV V2 µR T1 n V2 V1 V V1 V1 pressure. W1 Q1 dV µ R T1 • Adiabatic expansion B C Now cylinder is put in contact with a non–conducting stand and piston is allowed to move outward, because no heat can enter in or leave out so the expansion of gas is adiabatic. The temperature falls to T K and gas describes the adiabatic from B to 2 point C (P , V , T ). During this expansion more work is done (W ) at the expense of 332 2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 the internal energy. µR Q=0 Work done in adiabatic path BC is W2 1 T1 T2 • Isothermal compression C D Q2 Now the gas cylinder is placed in contact with sink at temperature T . The piston is 2 moved slowly inward so that heat produced during compression passes to the sink. The gas is isothermally compressed from C to point D (P , V , T ). The heat rejected 4 42 Q to the cold reservoir (sink) at T occurs over this path. Amount of work done on 22 gas W = amount of heat rejected to the sink 3 Q = W = RT2n V4 Q = µRT n V4 sink 2 3 V3 2 2 V3 E 68
JEE-Physics • Adiabatic compression D A The cylinder is removed from the sink and is put in contact with insulating stand now piston moves inward. Heat is not allowed to go out and it increases the internal energy of the system. Now work is done on the gas during adiabatic compression from state D to initial point A (P , V , T ). No heat exchanges occur over the adiabatic 111 µR T1 ) 1 (T2 path. Work done on the system W = 4 This cycle of operations is called a Carnot cycle. In first two steps work is done by engine W and W are positive Q=0 12 In last two steps work is done on gas W and W are negative 34 The work done in complete cycle W = the area of the closed part of the P–V cycle. W = W + W + W + W 1234 W µ R T1 n V2 µR T2 ) µRT2n V4 µR T1 ) µR T1 n V2 µRT2n V4 V1 1 (T1 V3 1 (T2 V1 V3 W µR T1n V2 µR T2 n V4 Efficiency of Carnot Engine, Q1 V1 V3 B to C and D to A are adiabatic paths V2 µR T1n V1 so T V ( – 1) = T V ( – 1) and T V ( – 1) = T V ( – 1) V2 V3 12 2 3 11 2 4 V1 V4 T1 T2 Q1 Q2 1 Q2 1 T2 Q1 Q2 T1 = Q1 Q1 T1 T1 T2 T1 T2 100% Q1 Q2 100% T1 Q1 The efficiency for the Carnot engine is the best that can be obtained for any heat engine. The efficiency of a Carnot engine is never 100% because it is 100% only if temperature of sink T = 0 which is impossible. 2 CARNOT THEOREM No irreversible engine (I) can have efficiency greater than Carnot reversible engine (R) working between same \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 hot and cold reservoirs. R I 1 T2 1 Q 2 T1 Q1 HEAT ENGINE source hot reservoir Heat engine is a device which converts heat into work. T1 Three parts of a heat engine: Q1 (i) Source of high temperature reservoir at temperature T 1 working W substance (ii) Sink of low temperature reservoir at temperature T 2 Q2 (iii) Working substance. sink cold reservoir In a cycle of heat engine the working substance extracts heat Q from 1 T2 source, does some work W and rejects remaining heat Q to the sink. 2 Efficiency of heat engine work done (W ) T1 T2 = Q1 Q2 heat taken from source (Q1 ) T1 Q1 E 69
JEE-Physics TYPES OF HEAT ENGINE • External combustion engine Steam engine is the example of external combustion engine. Its efficiency is 10 to 20% (small). • Internal combustion engine The examples are petrol engine and diesel engine. There are four strokes in internal combustion engine. (i) Intake stroke or Charging stroke (ii) Compression stroke (iii) Working (or power) stroke of Expansion stroke (iv) Exhaust stroke The useful work is done in third stroke called work stroke or power stroke. The efficiency of internal combustion engine is 40 to 60%. Example A carnot engine working between 400 K and 800 K has a work output of 1200 J per cycle. What is the amount of heat energy supplied to the engine from source per cycle? Solution W = 1200 J, T = 800 K, T = 400 K = 1 – T2 W 1 400 1200 1200 1 2 T1 Q1 800 0.5 = Q1 Q1 1200 Heat energy supplied by source Q1 0.5 = 2400 joule per cycle Example The temperatures T and T of the two heat reservoirs in an ideal carnot engine are 1500°C and 500°C 12 respectively. Which of the following : increasing T by 100°C or decreasing T by 100°C would result in a 12 greater improvement in the efficiency of the engine? Solution T = 1500°C = 1500 + 273 = 1773 K and T = 500°C = 500 + 273 = 773 K. 12 The efficiency of a carnot's engine 1 T2 T1 When the temperature of the source is increased by 100°C, keeping T unchanged, the new temperature of the 2 source is T´1 = 1500 + 100 = 1600°C = 1873 K. The efficiency becomes ´ 1 T2 1 773 0.59 T1 ' 1873 On the other hand, if the temperature of the sink is decreased by 100°C, keeping T unchanged, the new \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 1 temperature of the sink is T´ = 500 – 100 = 400°C = 673 K. The efficiency now becomes 2 ´´ 1 T´2 1 673 0.62 T1 1773 Since ´´ is greater than ´, decreasing the temperature of the sink by 100°C results in a greater efficiency than increasing the temperature of the source by 100°C. Example A heat engine operates between a cold reservoir at temperature T = 300 K and a hot reservoir at temperature 2 T . It takes 200 J of heat from the hot reservoir and delivers 120 J of heat to the cold reservoir in a cycle. What 1 could be the minimum temperature of hot reservoir? 70 E
JEE-Physics Solution Work done by the engine in a cycle is W = 200 – 120 = 80 J. = W 80 0.4 Q 200 From carnot's Theorem 0.4 1 T2 1 300 300 0.6 T1 300 T 500 T1 T1 T1 0.6 1 REFRIGER ATOR Hot reservoir T1 It is inverse of heat engine. It extracts heat (Q ) from a cold reservoir, Q1 = Q2 + W 2 W same external work W is done on it and rejects heat (Q ) to hot reservoir. 1 The coefficient of performance of a refrigerator. Heat extracted fro m cold reservo ir Q2 Q2 1 Q2 Work don e o n refrigerato r W Q1 Q2 Q1 1 Q2 Cold reservoir T2 For Carnot reversible refrigerator Q1 T1 Q2 1 1 = T2 Q2 T2 W T1 T2 Q1 1 T1 1 Q 2 T2 Example A carnot engine works as a refrigrator between 250 K and 300 K. If it receives 750 cal of heat from the reservoir at the lower temperature. Calculate the amount of heat rejected at the higher temperature. Solution T = 300 K T = 250 K Q = 750 Q = ? 1 22 1 Q1 T1 300 Q 2 T2 Q1 250 750 = 900 cal Example The temperature inside and outside of refrigerator are 260 K and 315 K respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done. Solution T = 260 K, T = 315 K ; W = 1 joule 21 Coefficient of performance of Carnot refrigerator Q2 T2 W T1 T2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Q2 260 260 Q2 260 = 6.19 J 1 315 260 42 42 Example A refrigerator takes heat from water at 0°C and transfer it to room at 27°C. If 100 kg of water is converted in ice at 0°C then calculate the work done. (Latent heat of ice is 3.4 × 105 J/kg) Solution Coefficient of performance (COP) = T2 273 273 T1 T2 300 273 27 W Q2 mL 100 3.4 105 100 3.4 105 27 = 3.36 × 107 J COP COP 273 / 27 273 E 71
JEE-Physics SOME WORKED OUT EXAMPLES Example#1 In a 20m deep lake, the bottom is at a constant temperature of 4°C. The air temperature is constant at –10°C. The thermal conductivity of ice is 4 times that water. Neglecting the expansion of water on freezing, the maximum thickness of ice will be 20 200 (C) 20 m (D) 10 m (A) m (B) m 11 11 Solution Ans. (B) The rate of heat flow is the same through water and ice in the steady state so 10°C x 4K ice KA 4 0 4KA 0 10 x 200 m 0°C 20-x K water 20 x x 11 4°C Example#2 The figure shows two thin rods, one made of aluminum [ = 23 × 106 (C°)1] and the other of steel [ = 12 × 106 (C°)–1]. Each rod has the same length and the same initial temperature. They are attached at one end to two separate immovable walls. Temperature of both the rods is increased by the same amount, until the gap between the rods vanishes. Where do the rods meet when the gap vanishes? M idpo in t Aluminum Steel (A) The rods meet exactly at the midpoint. Ans. (B) (B) The rods meet to the right of the midpoint . Ans. (B) (C) The rods meet to the left of the midpoint. (D) Information insufficient E Solution As A steel so expansion in aluminum rod is greater. Example#3 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Certain perfect gas is found to obey PVn = constant during adiabatic process. The volume expansion coefficient at temperature T is 1n 1 n 1 (A) T (C) T (D) nT Solution (B) 1 n T PVn = constant & PV = RT V 1 V 1 T V 1 n T T 1 n volume expansion coefficient V 1 1 VT nT 72
JEE-Physics Example#4 Figures shows the expansion of a 2m long metal rod with temperature. The volume expansion coefficient of the metal is :– (mm) 2.5 2.0 1.5 1.0 0.5 0 5 10 15 20 25 T(K) (A) 3 × 10–4 K–1 (B) 1.5 × 10–4 K–1 (C) 3 × 10–5 K–1 (D) 1.5 × 10–5 K–1 Solution Ans. (B) 0.5 103 T 25 = 5 × 10–5 K–1 = 3 = 1.5 × 10–4 K–1 Example#5 The temperature of a body rises by 44°C when a certain amount of heat is given to it. The same heat when supplied to 22 g of ice at – 8°C, raises its temperature by 16°C. The water eqivalent of the body is [Given : s = 1 cal/g°C & L = 80 cal/g, s = 0.5 cal/g°C] water f ice (A) 25g (B) 50 g (C) 80 g (D) 100 g Solution Ans. (B) Supplied heat = (22) (0.5) (8) + (22) (80) + (22) (1) (16) = 88 + 1760 + 352 = 2200 cal 2 2 00 cal Heat capacityof the body = 44°C = 50 cal/°C Water equavalent of the body = Heat capacity of the body = 50 cal/°C 50g spcific heat capacity of water 1 cal/g°C Example#6 A fine steel wire of length 4m is fixed rigidly in a heavy brass frame as shown in figure. It is just taut at 20°C. The tensile stress developed in steel wire if whole system is heated to 120°C is :– \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 (Given = 1.8 × 10–5 °C–1, =1.2 × 10–5 °C–1,Y =2 × 1011 Nm–2,Y =1.7 × 107 Nm–2) brass steel steel brass (A) 1.02 × 104 Nm–2 (B) 1.2 × 108 Nm–2 (C) 1.2 × 106 Nm–2 (D) 6 × 108 Nm–2 Solution Ans. (B) S t r e s s = Y ( s t r a i n ) = Y ( b – s) T = ( 2 × 1 0 11 ) ( 0 . 6 × 1 0 –5) ( 1 0 0 ) = 1 . 2 × 1 0 8 N m – 2 S Example#7 540 g of ice at 0°C is mixed with 540 g of water at 80°C. The final temperature of the mixture is (Given latent heat of fusion of ice = 80 cal/g and specific heat capacity of water = 1 cal/g0C) (A) 0°C (B) 40°C (C) 80°C (D) less than 0°C E 73
JEE-Physics Solution Ans. (A) Heat taken by ice to melt at 0°C is Q = mL = 540 × 80 = 43200 cal 2 Hea t gi ven by water to cool up to 0°C i s Q = ms = 540 × 1× ( 80–0 ) = 432 00 c al 2 Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is 0°C. Example#8 A refrigerator converts 100 g of water at 25°C into ice at – 10°C in one hour and 50 minutes. The quantity of heat removed per minute is (specific heat of ice = 0.5 cal/g°C, latent heat of fusion = 80 cal/g) (A) 50 cal (B) 100 cal (C) 200 cal (D) 75 cal Solution Ans. (B) Heat removed in cooling water from 25°C to 0°C = 100 × 1 × 25 = 2500 cal Heat removed in converting water into ice at 0°C = 100 × 80 = 8000 cal Heat removed in cooling ice from 0° to –15°C = 100 × 0.5 × 10 = 500 cal Total heat removed in1 hr 50min = 2500 + 8000 + 500 = 11000 cal Heat removed per minute = 11000 =100 cal/min 110 Example#9 5n, n and 5n moles of a monoatomic, diatomic and non-linear polyatomic gases (which do not react chemically with each other) are mixed at room temperature. The equivalent degree of freedom for the mixture is- 25 48 52 50 (A) 7 (B) (C) (D) Solution 11 11 11 Ans. (D) feq f1n1 f2 n2 f3 n 3 (5n)(3) (n)(5) (5n)(6) 50 n1 n2 n3 5n n 5n 11 Example#10 Figure shows the adiabatic curve on log–log scale performed on a ideal gas. The gas must be :– logT7 6 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P655 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 log V (A) Monoatomic (B) Diatomic (C) A mixture of monoatomic and diatomic (D) A mixture of diatomic and polyatomic Solution Ans. (C) For adiabatic process TV–1 = constant E log T + (–1) log V = constant slope = – (–1 = – 5 3 10 2 5 7 7 35 For monoatomic gas ; For diatomic gas As 3 5 5 23 Hence, the gas must be a mixture of monoatomic & diatomic gas. 74
JEE-Physics Example#11 Figure demonstrates a polytropic process (i.e. PVn = constant) for an ideal gas. The work done by the gas will be in process AB is P 16P0 A P0 B V V0 2V0 15 14 (C) 8P V (D) Insufficient information (A) 2 P0 V0 (B) 3 P0 V0 00 Solution Ans. (B) For a polytropic process PVn = constant 16P V n = P (2V )n n=4 00 00 Work done = P1 V1 P2 V2 16P0 V0 P0 2V0 14 3 P0 V0 n 1 4 1 Example#12 A vessel contains 14 g (7 moles) of hydrogen and 96 g (3 moles) of oxygen at STP. Chemical reaction is induced by passing electric spark in the vessel till one of the gases is consumed. The temperature is brought back to it's starting value 273 K. The pressure in the vessel is- +– H2 O2 Spark (A) 0.1 atm (B) 0.2 atm (C) 0.3 atm (D) 0.4 atm Solution Ans. (A) When electric spark is passed, hydrogen reacts with oxygen to form water (H O). Each gram of hydrogen reacts 2 with eight grams of oxygen. Thus 96 g of oxygen will be totally consumed together with 12 g of hydrogen. 2 The gas left in thevessel will be 2g of hydrogen i.e. number of moles 1 . 2 Using PV = RT P \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65P22P21 P2 0.1atm P1 1 1 10 Vf(m3) Example#13 Suppose 0.5 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the final volume V versus Q. The temperature of the gas is (use n 9 = 2 and R= 25/3 J/mol-K) f 0.3 0.2 0.1 (A) 293 K (B) 360 K 0 500 1000 1500 (D) 412 K Q(J) E (C) 386 K 75
JEE-Physics Solution Ans. (B) Q = W = nRT n Vf ; T Q / Vi 1500 1500 360K Vi n R n Vf 0.5 25 / 3 n3 0.5 25 / 3 1 Example#14 The respective speeds of five molecules are 2,1.5,1.6,1.6 and 1.2 km/s. The most probable speed in km/s will be (A) 2 (B) 1.58 (C) 1.6 (D) 1.31 Solution Ans. (C) Since maximum number of molecules travel with speed 1.6 km/s so v = 1.6 km/s mp Example#15 When water is boiled at 2 atm pressure, the latent heat of vaporization is 2.2 × 106 J/kg and the boiling point is 120°C. At 2 atm pressure, 1 kg of water has volume of 10–3 m3 and 1 kg of steam has a volume of 0.824 m3. The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120°C is [1 atm pressure = 1.013 × 105 N/m2] (A) 2.033 J (B) 2.033 × 106 J (C) 0.167 × 106 J (D) 2.267 × 106 J Solution Ans. (B) Total heat given to correct convert water into steam at 120 °C is Q = mL = 1 × 2.2 × 106 = 2.2 × 106 J The work done by the system against the surrounding is PV = 2 × 1.013 × 105 (0.824 × 0.001) = 0.167 × 106 J U = Q – W = 2.033 × 106 J Example#16 Given T–P curve for three processes. If initial and final pressure are same for all processes then work done in process 1, 2 and 3 is W , W & W respectively. Correct order is 12 3 1 T2 3 P (A) W < W > W (B) W > W > W (C) W < W < W (D) W = W = W 123 123 123 123 Ans. (B) Solution graph-2, n =1 W = 0; graph-3, n <1 W < 0 Here T Pn where for : graph-1, n >1 W > 0 ; Example#17 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 The acceleration of a particle moving rectilinearly varies with displacement as a = – 4x. At x =4 m and t =0, particle is at rest. Select the INCORRECT alternative (A) The maximum speed of the particle is 8 m/s. (B) The distance travelled by the particle in first second is 20 m. (C) The velocity - acceleration graph of the particle is an ellipse. (D) The kinetic energy-displacement graph of the particle is a parabola. Solution Ans. (B) The equation shows a SHM : a 2 x 4 2 Time period = sec x A sin t From the given condition, A =4 & ; VPmax A 8m / s 2 Distance travelled in first second <2A or < 8 m KE = 1 m2 A2 x2 which represent a parabola. 2 76 E
JEE-Physics Example#18 Figure shows the variation of the internal energy U with density of one mole of an ideal monatomic gas for thermodynamic cycle ABCA. Here process AB is a part of rectangular hyperbola :- U(J) AC B kg m3 (A) process AB is isothermal & net work in cycle is done by gas. Ans. (B) (B) process AB is isobaric & net work in cycle is done by gas. (C) process AB is isobaric & net work in cycle is done on the gas. (D) process AB is adiabatic & net work in cycle is done by gas. Solution For the process AB : U = constant (hyperbola) U 3 RT (monoatomic ideal gas); RT = constant 22 Comparing it with ideal gas equation P 1 RT P is constant. M PC . Thus work done in cycle is positive P-V graph for the cycle is A B X Example#19 One mole of an ideal gas undergoes a process whose molar heat capacity is 4R and in which work done by gas CP for small change in temperature is given by the relation dW = 2RdT, then the ratio C V is (A) 7/5 (B) 5/3 (C) 3/2 (D) 2 Solution Ans. (C) For small change dQ = dU + dW nCdT = nC dT + 2nRdT C = C + 2R; 4R = C + 2R C = 2R V VV V RR 3 Also C = = 2R 2 – 2 = 1 = V 1 1 2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Example#20 An irregular rod of same uniform material as shown in figure is conducting heat at a steady rate. The tempera- ture gradient at various sections versus area of cross section graph will be Heat dT/dx dT/dx (A) (B) A A E 77
JEE-Physics dT/dx dT/dx (C) (D) AA Solution Ans. (B) dT dT H = KA is same in steady state condition, A = constant rectangular hyperbolic graph dx dx Example#21 8 3 A gas undergoes an adiabatic process in which pressure becomes 3 3 times and volume become 4 of initial volume. If initial absolute temperature was T, the final temperature is 32T 2T (C) T3/2 3T (A) 9 3 (B) 3 (D) Solution 2 Ans. (B) 8 3 3 4 For adiabatic process, P1V1 =P2V2 P0 V0 3 P0 V0 ..(i) 3 1 2T 4 3 Also, T1 V1 1 T2 V2 1 TV0 1 T2 V0 ...(ii) Solving (i) & (ii), T2 Example#22 Which of the following graph(s) shows the correct variation in intensity of heat radiations by black body and frequency at a fixed temperature– E E E E UV Visible Infra-red UV Visible Infra-red Infra-red Visible Ultra-voilet Infra-red Visible Ultra-voilet (A) 3500K 1500K 3500K 1500K Solution 2500K 2500K (C) (D) 2500K (B) 3500K 3500K 2500K 1500K 1500K Ans. (AC) 1 According to Wien's law m T mT. As the temperature of body increases, frequency corresponding to maximu m energy in rad i ati on ( ) inc r eases. \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 m Also area under the curve Ed E d T4 Example#23 The temperature drop through a two layer furnace wall is 900°C. Each layer is of equal area of cross–section. Which of the following action(s) will result in lowering the temperature of the interface? 78 E
JEE-Physics (A) By increasing the thermal conductivity of outer layer. Ans. (AD) (B) By increasing the thermal conductivity of inner layer. (C) By increasing thickness of outer layer. (D) By increasing thickness of inner layer. Solution Rate of heat flow KiA 1000 K0A 100 100 1 900 1 K0i i 0 K i0 Now, we can see that can be decreased by increasing thermal conductivity of outer layer (K ) and thickness of o inner layer ( ). i Example#24 5 kg of steam at 100°C is mixed with 10 kg of ice at 0°C. Choose correct alternative/s (Given s = 1 cal/g°C, L = 80 cal/g, L = 540 cal/g) water FV (A) Equilibrium temperature of mixture is 160°C (B) Equilibrium temperature of mixture is 100°C 1 (C) At equilibrium, mixture contains 13 kg of water 3 2 (D) At equilibrium, mixture contains 1 kg of steam 3 Solution Ans. (B,C,D) Required heat Available heat 10 kg ice (0°C) 5 kg steam (100°C) 800 kcal 2700 cal 10 g water (0°C) 5 g water (100°C) 1000 kcal 10 g water (100°C) So available heat is more than required heat therefore final temperature will be 100°C. Mass of heat condensed = 800 1000 10 kg. Total mass of water = 10 10 40 13 1 kg 540 3 3 3 3 Total mass of steam = 5 10 5 1 2 kg 33 3 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Example#25 Water contained in a jar at room temperature (20°C) is intended to be cooled by method-I or method-II given below : Method -I : By placing ice cubes and allowing it to float. Method-II : By wrapping ice cubes in a wire mesh and allowing it to sink. Choose best method(s) to cool the water. (A) Method-I from 20°C to 4°C (B) Method-I from 4°C to 0°C (C) Method-II from 20°C to 4°C (D) Method-II from 4°C to 0°C Solution Ans. (AD) Initially (above 4°C), a decrease in temperature, increases the density of water and consequently it descends, replacing the relatively warm water. Convention currents set up in this way demands the location of ice to be on the water surface. Below 4°C, a decrease in temperature decreases the water density and as a result it ascends up displacing the relatively warm water. To setup convention currents in this way, the position of ice cubes should be at the bottom. E 79
JEE-Physics Example#26 n moles of an ideal triatomic linear gas undergoes a process in which the temperature changes with volume as T = k1V2 where k1 is a constant. Choose correct alternative(s). 5 (A) At normal temperature Cv = 2 R (B) At any temperature Cp – Cv = R (C) At normal temperature molar heat capacity C=3R (D) At any temperature molar heat capacity C=3R Solution Ans. (ABC) At normal temperature C = f R = 5 R ; At any temperature C – C = f 1 f R = R v 2 2 pv 2 2 from process T = k V2 & ideal gas equation PV = nRT we have PV–1 = constant x =–1 1 R RR 5R C CV CV CV At normal temperature C = R + = 3R 1x 1 1 2 2 2 Example#27 An ideal gas expands in such a way that PV2 = constant throughout the process. Select correct alternative (A) This expansion is not possible without heating (B) This expansion is not possible without cooling (C) Internal energy remains constant in this expansion (D) Internal energy increases in this expansion Solution Ans. (B) 1 PV = nRT & PV2 = constant V gas can expand only if it cools T As tempeature decreases during expansion so internal energy will decrease. Example#28 Which of the following processes must violate the first law of thermodynamics (Q = W + Eint)? (A) W > 0, Q < 0 and Eint > 0 (B) W > 0, Q < 0 and Eint < 0 (C) W < 0, Q > 0 and Eint < 0 (D) W > 0, Q > 0 and Eint < 0 Solution Ans. (AC) For (A) : W > 0 & Eint > 0 Q > 0 For (B) : W > 0 & Eint < 0 Q > 0 or Q < 0 For (C) : W < 0 & Eint < 0 Q < 0 For (D) : W > 0 & Eint < 0 Q > 0 or Q < 0 Example#29 In a process on a closed system of ideal gas, the initial pressure and volume is equal to the final pressure and volume (A) initial internal energy must be equal to the final internal energy \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 (B) the work done on the system is zero (C) the work done by the system is zero (D) the initial temperature must be equal to final temperature Solution Ans. (AD) Here n = constant so if P = P and V = V then T = T 21 21 21 Also work done by the system may be zero or may not be zero. Example#30 d In Newton's law of cooling, dt = – k(–0), the constant k is proportional to (A) A, surface area of the body (B) S, specific heat of the body 1 (D) e, emissivity of the body (C) , m being mass of the body m 80 E
JEE-Physics Solution Ans. (AC) dQ 4 04 dt eA 3 e A 03 dQ 4 04 dt eA 3 e A 03 Example#31 V The graph below shows V-P curve for three processes. 3 21 Choose the correct statement(s) (A) Work done is maximum in process 1. (B) Temperature must increase in process 2 & 3. P (C) Heat must be supplied in process 1. (D) If final volume of gas in process 1, 2 and 3 are same then temperature must be same. Solution Ans. (AC) Area under P-V curve and volume axis represent work. Internal energy in 1 increases (expansion at constant pressure) In process 2 and 3 internal energy may decrease depending on the process. Since final pressure of 1, 2 & 3 are different, final temperature must also be different. Example#32 P2 During an experiment, an ideal gas is found to obey a condition = constant [density of the gas]. The gas is initially at temperature T, pressure P and density . The gas expands such that density changes to 2 (A) The pressure of the gas changes to 2P (B) The temperature of the gas changes to 2T. (C) The graph of the above process on the P-T diagram is parabola. (D) The graph of the above process on the P-T diagram is hyperbola. Solution Ans. (B,D) PV = nRT P T .....(i) T 1 ...(ii) T 1 ...(iii) Example#33 to 35 At 20°C a liquid is filled upto 10 cm height in a container of glass of length 20 cm and cross-sectional area 100 cm2 Scale is marked on the surface of container. This scale gives correct reading at 20°C. Given = 5 × 1 0 –5 k –1, = 1 × 1 0 –5 ° C –1 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65L g 20cm 10cm 3 3 . The volume of liquid at 40°C is :– A0 = 100cm2 (A) 1002 cc (B) 1001 cc (C) 1003 cc (D) 1000.5 cc (D) 10.1 cm 3 4 . The actual height of liquid at 40°C is- (A) 10.01 cm (B) 10.006 cm (C) 10.6 cm E 81
JEE-Physics 3 5 . The reading of scale at 40°C is- (A) 10.01 cm (B) 10.004 cm (C) 10.006cm (D) 10.04 cm Solution 33. Ans. (B) V = V (1+LT) = (10) (100) [ 1 + 5 × 10–5 × 20] = 1000 (1 + 0.001) = 1001 cm3 = 1001 cc 0 34. Ans. (B) Cross sectional area of vessel at 40°C A = A (1 + 2 gT) = 100 (1 + 2 × 10–5 × 20) = 100.04 cm2 0 Actual volume of liquid 1001 Actual height of liquid = cross sec tional area of vessel = 100.04 = (1001) (100 + 0.04)–1 1001 1 0.04 1 (1001) 1 0.04 1 (1001 – 0.4) = 10.006 cm 100 100 100 100 100 35. Ans. (B) TV = SR (1 + T) = (TV) (1 – T) gg SR = (TV) (1 + g T)–1 = (10.006) (1 – 10–5 × 20) = 10.006 – 0.002 = 10.004 c Example#36 to 38 One mole of a monoatomic ideal gas occupies two chambers of a cylinder partitioned by means of a movable piston. The walls of the cylinder as well as the piston are thermal insulators. Initially equal amounts of gas fill both the chambers at (P , V , T ). A coil is burnt in the left chamber which absorbs heat and expands, pushing the 0 00 piston to the right. The gas on the right chamber is compressed until to pressure becomes 32 P . 0 P0,V0,T0 P0,V0,T0 3 6 . The final volume of left chamber is (A) V0 15 7 9 8 (B) 8 V0 (C) 8 V0 (D) 8 V0 3 7 . The work done on the gas in the right chamber is 17 (D) 2 P0 V0 9 9 13 (A) 2 P0 V0 (B) – 2 P0 V0 (C) 2 P0 V0 131 (D) 4 RT0 3 8 . The change in internal energy of the gas in the left chamber is \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 E 186 177 59 (A) 4 RT0 (B) 4 RT0 (C) 2 RT0 Solution 36. Ans. (B) Since the compression on the right is adiabatic so P V = P V 00 RR P0 V05 / 3 32P0 VR5 / 3 VR V0 V = V + 7 15 V 8 L0 V = 0 8 0 8 37. Ans. (A) Work done on the gas= PR VR P0 V0 4P0 V0 P0 V0 9 1 5 / 3 1 2 P0 V0 82
JEE-Physics 38. Ans. (B) For mechanical equilibrium P = P = 32 P LR 0 So P V = (32P ) 1 5 V0 = 60P V = 60nRT = nRT T = 60T 8 L0 LL 0 00 0L The change in the internal energy of the gas in the left chamber 13 177 U nC V T 2 2 R 59 T0 4 RT0 Example#39 to 41 One mole of an ideal monoatomic gas undergoes a cyclic process as shown in figure. Temperature at point 1 = 300 K and process 2-3 is isothermal. P 2 3P0 P0 1 3 V0 3V0 v 3 9 . Net work done by gas in complete cycle is (A) 9n3 12 P0 V0 (B) 9n3 4 P0 V0 (C) 9n3 4 P0 V0 (D) 9n3 8 P0 V0 4 0 . Heat capacity of process 1 2 is 5R (D)2R (C) R 3R (A) (B) 2 2 2 4 1 . The efficiency of cycle is 9n3 4 9n3 4 9n3 4 9n3 12 (A) 9n3 12 (B) 9n3 12 (C) 9n3 16 (D) 9n3 16 Solution 39. Ans. (C) 1 W W12 W23 W31 2 4P0 2V0 nRTn 3 P0 8 V0 4P0 V0 9P0 V0n3 9n3 4 P0 V0 40. Ans. (D) Q n 3R .T W 3R 4P0 V0 3R P0 V0 3R R nT 2 2 nT 2 C n 2T0 2R \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 nT 2 2 Q123 n 2R T12 W23 n 2R 8 T0 nR 9T0 n 3 nRT0 16 9n3 41. Ans. (C) W 9n3 4 Q 1 23 9n3 16 Example#42 to 44 A body cools in a surrounding of constant temperature 30°C. Its heat capacity is 2J/°C. Initial temperature of the body is 40°C. Assume Newton’s law of cooling is valid. The body cools to 36°C in 10 minutes. 4 2 . In further 10 minutes it will cool from 36°C to : (A) 34.8°C (B) 32.1°C (C) 32.8°C (D) 33.6°C E 83
JEE-Physics 4 3 . The temperature of the body in °C denoted by the variation of versus time t is best denoted as 40°C 40°C (A) 30°C (B) 30°C (0,0) t (0,0) t 40°C 40°C (C) (D) 30°C (0,0) t (0,0) t 4 4 . When the body temperature has reached 36°C, it is heated again so that it reaches to 40°C in 10 minutes. Assume that the rate of loss of heat at 38°C is the average rate of loss for the given time. The total heat required from a heater by the body is : (A) 7.2 J (B) 0.728 J (C) 16 J (D) 32 J Solution 42. Ans. (D) 40 36 = k(38 – 30) and 36 x k 36 x 30 x 33.6 10 10 2 43. Ans. (D) d kA dt ms (T – T0) Magnitude of slope will decrease with time. 44. Ans. (C) 40 36 k (38 – 30) k 4 8 1 10 10 20 d when the block is at 38°C and room temperature is at 30°C the rate of heat loss ms × = ms k (38 – 30) dt 1 Total heat loss in 10 minute Q = ms k (38 – 30) × 10 = 2 × 20 × 8 × 10 = 8 J Now heat agained by the object in the said 10 minutes. Q = ms = 2 × 4 = 8 J Total heat required = 8 + 8 = 16 J Example#44 Column II (Answers) \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Column I (Questions) (P) 2 0 (A) The temperature of an iron piece is increased from 20° to 70°. What is change in its temperature on the Fahrenheit scale (in °F)? (B) At what temperature (in °C) do the Celsius and Fahrenheit readings (Q) 4 0 have the same numerical value? (C) 100 g ice at 0°C is converted into steam at 100 °C. (R) – 40 Find total heat required (in kcal) (S) 7 2 (T) 9 0 (L = 80 cal/g, s = 1 cal/g°C,L = 540 cal/g) fw v (D) A ball is dropped on a floor from a height of 5 m. After the collision it rises upto a height of 3m. Assume that 50% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in temperature (in milli centigrade) of the ball in the collision. (s = 500 J/K, g = 10 m/s2) ball 84 E
JEE-Physics Solution Ans.(A) (S); (B) (R); (C) (T); (D) (P) (A) C0 F 32 F 9 C 9 50 90F 100 180 5 5 (B) x 0 x 32 x 40 100 180 (C) Total heat required = mL + ms + mL = 72000 cal = 72 kcal f v (D) m s 50 mg h1 h2 1 C 20 103C 100 50 Example#46 7 An ideal gas whose adiabatic exponent equals to is expanded according to the law P=2V. The initial 5 25 volume of the gas is equal to V0= 1 unit. As a result of expansion the volume increases 4 times. (Take R = 3 units) Column - I Column - II (A) Work done by the gas (P) 25 units (B) Increment in internal energy of the gas (Q) 45 units (C) Heat supplied to the gas (R) 75 units (D) Molar heat capacity of the gas in the process (S) 15 units (T) 55 units Solution Ans. (A) (S); (B) (R); (C) (Q); (D) (P) 4 V0 V 2 4 V0 15 V02 = 15 units V0 2VdV W PdV V0 From PV = nRT, 2V2 = nRT 2 V22 V12 nR T nRT = 30V 2 0 U nC V T nR 30 V02 30 12 30 5 = 75 units T 1 2 7 1 1 5 Q = W + U = 15 + 30 = 45 units Molar heat capacity : C CV R 5R 1 R 5 R R 3R 3 25 = 25 units 1x 2 2 2 3 1 Example#47 The figure given below show different process for a given amount for an ideal gas. W is work done by the \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 system and Q is heat absorbed by the system. PP P PP (i) (ii) (iii) (iv) (v) adiabatic V V 1/V V V Column-I Column-II (A) Q > 0 (P) In figure (i) (B) W < 0 (Q) In figure (ii) (C) Q < 0 (R) In figure (iii) (D) W > 0 (S) In figure (iv) (T) In figure (v) E 85
JEE-Physics Solution Ans. (A) (PRT) ; (B) (S) ; (C) (S) ; (D) (PQRT) 1 figure (i) P V PV = constant Isothermal (T = constant), so U = 0 1 V is decreasing; So V is increasing hence, W > 0 Q = U +W = W > 0 Example#48 In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C . To do this steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam is required per hour. (Specific heat of steam = specific heat of water = 1 cal/g°C, Latent heat of vaporisation = 540 cal/g) Solution Ans. 1 Suppose m kg steam is required per hour Heat is released by steam in following three steps (i) When 150°C steam Q1 100°C steam Q = mc = m × 1 (15 0 –100) = 5 0 m cal 1 steam (ii) When 100°C steam Q2 100°C water Q = mL = m× 540= 540 m cal 2v (iii) When 100 °C water Q2 90°C water Q = mc W = m × 1 × (10 0 –90 ) = 10 m c al 3 Hence total heat given by the steam Q = Q + Q + Q = 600 m cal ....(i) 123 Heat taken by 10 kg waterQ' = mcW = 10 × 103 × 1 × (80–20) = 600 × 103 cal Hence Q = Q' 600 m = 600 × 103 m = 103 gm = 1 kg Example#49 The specific heat of a metal at low temperatures varies according to S = (4/5)T3 where T is the absolute temperature. Find the heat energy needed to raise unit mass of the metal from T = 1 K to T = 2 K. Solution Ans. 3 Q mSdT mT4 Q 15 3 5 m5 Example#50 A rod has variable co-efficient of linear expansion x . If length of the rod is 1m. Determine increase in 5000 length of the rod in (cm) on increasing temperature of the rod by 100°C. x \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Solution Ans. 1 Increase in length of dx = 0T = x dx 1 00 x dx 5000 50 Total thermal expansion = 1 x x2 1 1 1cm dx m 0 50 100 0 100 Example#51 A clock pendulum made of invar has a period of 2 s at 20°C. If the clock is used in a climate where average temperature is 40°C, what correction (in seconds) may be necessary at the end of 10 days to the time given by clock ? (invar = 7 × 10–7 °C–1, 1 day = 8.64 × 104 s) 86 E
JEE-Physics Solution Ans. 6 T 2 T 1 1 T 10 8.64 104 7 10 7 20 6 s g T 2 T 2 2 2 Example#52 Two different rods A and B are kept as shown in figure. Temp.(°C) 100° 70°C 35°C 70° 100°C 35° AB 0 30 Distance 100 (cm) The variation of temperature of different cross sections is plotted in a graph shown in figure. Find the ratio of thermal conductivities of B to A. Solution Ans. 2 Q Q K A A 100 70 KB A 70 35 KA KB KB 2 t t 2 KA A B 30 70 Example#53 A body cools from 50°C to 49.9°C in 5 sec. It cools from 40°C to 39.9°C in t sec. Assuming Newtons law of cooling to be valid and temperature of surrounds at 30°C, value of t/5 will be? Solution Ans. 2 From 1 2 k 1 2 t 2 0 We have 0.1 k 39.5 and 0.1 k 19.5 t 39.5 2 t 10 t 2 5 2 t 2 5 19.5 5 Example#54 A cylinder of cross-section area A has two pistons of negligible mass separated by distances loaded with spring of negligible mass. An ideal gas at temperature T is in the cylinder where the springs are relaxed. When the gas 1 is heated by some means its temperature becomes T and the springs get compressed by each. If P is 2 20 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 2P0 A , then find the ratio of T and T . \\\\\\\\\\\\\\\\\\\\\\\\ 21 \\\\\\\\\\\\\\\\\\\\\\\\ atmospheric pressure and spring constant k = Solution Ans. 4 P1 V1 P2 V2 where V = A and V = 2A and P = P and P = kx P0 2P0 T2 P2 V2 4 T1 T2 T1 P1 V1 12 10 2A E 87
JEE-Physics Example#55 A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of xP. Find the value of x. Solution Ans. 2 PV = nRT, V is constant Example#56 P B One mole of a gas is taken from state A to state B as shown in figure. 2P0 2V0 V Work done by the gas is × 10 J. Find the value of . P0 T1 Ans. 7 25 A (Given : T1=320 K, R= 3 ) Solution V0 1 3 Work done = P0 2P0 2V0 V0 2 P0 V0 2 and P V = R × 320 00 3 3 25 So work done R 320 320 4000J 4 103 J = 4 + 3 = 7 22 3 Example#57 container P0 A container having base area A . Contains mercury upto a height . At its bottom a 00 A0 Hg thin tube of length 4 and cross-section area A (A<<A ) having lower end closed is 00 A attached. Initially the length of mercury in tube is 3 . In remaining part 2 mole of a gas gas 0 at temperature T is closed as shown in figure. Determine the work done (in joule) by gas if all mercury is displaced from tube by heating slowly the gas in the rear end of the tube b y mea ns of a he ate r. (Given : dens it y of me rc u r y = , a tmospher ic p re ss u re P = 20g, 0 0 C of gas = 3/2 R, A= (3/)m2, = (1/9) m, all units in S.I.) V Solution Ans. 5 If x is length of mercury in tube then pressure of gas 0 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 = + g0 + gx = 3g0 gx 3g0 gx Adx 2 P' P + W 1 3 . 5 g A 0 5 0 3 0 Example#58 The relation between internal energy U, pressure P and volume V of a gas in an adiabatic process is: U=a+bPV where a = b = 3. Calculate the greatest integer of the ratio of specific heats []. Solution Ans. 1 For an adiabatic process, dQ=dU+PdV =0 d[a+bPV] + PdV=0 bPdV + bV dP + PdV =0 (b+1)PdV+bV dP=0 (b 1) dV b dP 0 V P (b+1)logV + b logP= constant; Vb+1Pb =constant b 1 = constant b 1 4 1.33 b 3 PV b 88 E
JEE-Physics Example#59 One mole of an ideal monatomic gas undergoes the process P = T1/2, where is constant. If molar heat capacity of the gas is R when R = gas constant then find the value of . Solution Ans.2 R 3R R P T1/2 and PV nRT PV 1 constant C C V 1 1 2R 2 2 Example#60 One mole of a monoatomic gas is enclosed in a cylinder and occupies a volume of 4 liter at a pressure 100 N/ m2. It is subjected to a process T =V2, where is a positive constant, V is volume of the gas and T is Kelvin temperature. Find the work-done by gas (in joule) in increasing the volume of gas to six times initial volume. Solution Ans. 7 nRT 6 V0 nR 2 2 V 2 nR V0 W 6 V0 PdV where P nRV W VdV V0 W hR 35 V02 P0 V0 35 P0 2 2 nRV0 Example#61 Two taps A and B supply water at temperatures 10º and 50° C respectively. Tap A alone fills the tank in 1 hour and tap B alone fills the tank in 3 hour. If we open both the taps together in the empty tank, if the final temperature of the water in the completely filled tank is found to be 5 (in °C). Find the value of . Neglect loss of heat to surrounding and heat capacity of the tank. Solution Ans. 4 m (T–10) s = m S 50 T T 20 3 Example#62 Two identical metal plates are welded end to end as shown in figure-(i). 20 cal of heat flows through it in 4 minutes. if the plates are welded as shown in figure-(ii), find the time (in minutes) taken by the same amount of heat to flow through the plates. T1 T2 T1 T2 Fig. (i) Fig. (ii) Solution Ans. 1 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-5\\Thermal physics\\Eng\\Theory.P65 Rate of heat flow Q kA T1 T2 t t A E 89
JEE-Physics WAVE THEORY, SOUND WAVES & DOPPLER'S EFFECTS INTRODUCTION OF WAVES What is wave motion ? • When a particle moves through space, it carries KE with itself. Wherever the particle goes, the energy goes with it. (One way of transport energy from one place to another place) • There is another way (wave motion) to transport energy from one part of space to other without any bulk motion of material together with it. Sound is transmitted in air in this manner. E x . You (Kota) want to communicate your friend (Delhi) Write a letter Use telephone 1st option involves the concept of particle & the second choice involves the concept of wave. Ex . When you say \"Namaste\" to your friend no material particle is ejected from your lips to fall on your friends ear. Basically you create some disturbance in the part of the air close to your lips. Energy is transferred to these air particles either by pushing them ahead or pulling them back. The density of the air in this part temporarily increases or decreases. These disturbed particles exert force on the next layer of air, transferring the disturbance to that layer. In this way, the disturbance proceeds in air and finally the air near the ear of the listener gets disturbed. Note :- In the above example air itself does not move. A wave is a disturbance that propagates in space, transports energy and momentum from one point to another without the transport of matter. Few examples of waves : The ripples on a pond (water waves), the sound we hear, visible light, radio and TV signals etc. CLASSIFICATION OF WAVES Wave classification According to Necessity Propagation Dimension Vibration of medium of energy of particle \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (i) Elastic or mechanical wave (i) Progressive (i) One dimensional (i) Transverse (ii) Electro magnetic wave (ii) Sationary (ii) Two dimensional (ii) Longitudinal (E.M. wave) or non-mech. (iii) Three dimentional 1 . Based on medium necessity :- A wave may or may not require a medium for its propagation. The waves which do not require medium for their propagation are called non-mechanical, e.g. light, heat (infrared), radio waves etc. On the other hand the waves which require medium for their propagation are called mechanical waves. In the propagation of mechanical waves elasticity and density of the medium play an important role therefore mechanical waves are also known as elastic waves. Example : Sound waves in water, seismic waves in earth's crust. 2 . Based on energy propagation :- Waves can be divided into two parts on the basis of energy propagation (i) Progressive wave (ii) Stationary waves. The progressive wave propagates with constant velocity in a medium. In stationary waves particles of the medium vibrate with different amplitude but energy does not propagate. E1
JEE-Physics 3 . Based on direction of propagation :- Waves can be one, two or three dimensional according to the number of dimensions in which they propagate energy. Waves moving along strings are one-dimensional. Surface waves or ripples on water are two dimensional , while sound or light waves from a point source are three dimensional. 4. Based on the motion of particles of medium : Direction of Direction of Disturbance Propagation Direction of Waves are of two types on the basis of Disturbance Direction of Longitudinal wave motion of particles of the medium. Propagation (i) Longitudinal waves Transverse wave (ii) Transverse waves In the transverse wave the direction associated with the disturbance (i.e. motion of particles of the medium) is at right angle to the direction of propagation of wave while in the longitudinal wave the direction of disturbance is along the direction of propagation. TR ANSVERSE WAVE MOTION Mechanical transverse waves produce in such type of medium which have shearing property, so they are known as shear wave or S-wave Note :- Shearing is the property of a body by which it changes its shape on application of force. Mechanical transverse waves are generated only in solids & surface of liquid. Crest In this individual particles of the medium execute Trough SHM about their mean position in direction r Particle Crest Normal level to the direction of propagation of wave motion. Trough \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 A crest is a portion of the medium, which is raised temporarily above the normal position of rest of particles of the medium , when a transverse wave passes. A trough is a portion of the medium, which is depressed temporarily below the normal position of rest of particles of the medium , when a transverse wave passes. LONGITUDINAL WAVE MOTION In this type of waves, oscillatory motion of the medium particles produces regions of compression (high pressure) and rarefaction (low pressure) which propagated in space with time (see figure). RC R C R C Particle Wave Note : The regions of high particle density are called compressions and regions of low particle density are called rarefactions. The propagation of sound waves in air is visualized as the propagation of pressure or density fluctuations. The pressure fluctuations are of the order of 1 Pa, whereas atmospheric pressure is 105 Pa. 2E
JEE-Physics Mechanical Waves in Different Media • A mechanical wave will be transverse or longitudinal depends on the nature of medium and mode of excitation. • In strings mechanical waves are always transverse when string is under a tension. In gases and liquids mechanical waves are always longitudinal e.g. sound waves in air or water. This is because fluids cannot sustain shear. • In solids, mechanical waves (may be sound) can be either transverse or longitudinal depending on the mode of excitation. The speed of the two waves in the same solid are different. (Longitudinal waves travels faster than transverse waves). e.g., if we struck a rod at an angle as shown in fig. (A) the waves in the rod will be transverse while if the rod is struck at the side as shown in fig. (B) or is rubbed with a cloth the waves in the rod will be longitudinal. In case of vibrating tuning fork waves in the prongs are transverse while in the stem are longitudinal. CCC C R C RC T T Transverse-wave Longitudinal-wave (B) (A) Further more in case of seismic waves produced by Earthquakes both S (shear) and P (pressure) waves are produced simultaneously which travel through the rock in the crust at different speeds [v 5 km/s while v 9 km/s] S–waves are transverse while P–waves longitudinal. S P Some waves in nature are neither transverse nor longitudinal but a combination of the two. These waves are called 'ripple' and waves on the surface of a liquid are of this type. In these waves particles of the medium vibrate up and down and back and forth simultaneously describing ellipses in a vertical plane. v A B Ripple \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 CHAR ACTERISTICS OF WAVE MOTION Some of the important characteristics of wave motion are as follows : • In a wave motion, the disturbance travels through the medium due to repeated periodic oscillations of the particles of the medium about their mean positions. • The energy is transferred from place to another without any actual transfer of the particles of the medium. • Each particle receives disturbance a little later than its preceding particle i.e., there is a regular phase difference between one particle and the next. • The velocity with which a wave travels is different from the velocity of the particles with which they vibrate about their mean positions. • The wave velocity remains constant in a given medium while the particle velocity changes continuously during its vibration about the mean position. It is maximum at the mean position and zero at the extreme position. • For the propagation of a mechanical wave, the medium must possess the properties of inertia, elasticity and minimum friction amongst its particles. SOME IMPORTANT TERMS CONNECTED WITH WAVE MOTION • Wavelength () [length of one wave] Distance travelled by the wave during the time, any one particle of the medium completes one vibration about its mean position. We may also define wavelength as the distance between any two nearest particles of the medium, vibrating in the same phase. • Frequency (n) :Number of vibrations (Number of complete wavelengths) complete by a particle in one second. E3
JEE-Physics • Time period (T) : Time taken by wave to travel a distance equal to one wavelength. • Amplitude (A) : Maximum displacement of vibrating particle from its equilibrium position. • Angular frequency () : It is defined as = 2 2n • Phase : Phase is a quantity which contains all information related to any vibrating particle in a wave. For equation y = A sin (t – kx); (t – kx) = phase. 2 • Angular wave number (k) : It is defined as k = • Wave number ( ) : It is defined as = 1 k = number of waves in a unit length of the wave pattern. 2 • Particle velocity, wave velocity and particle's acceleration : In plane progressive harmonic wave particles of the medium oscillate simple harmonically about their mean position. Therefore, all the formulae what we have read in SHM apply to the particles here also. For example, maximum particle velocity is ± A at mean position and it is zero at extreme positions etc. Similarly maximum particle acceleration is ±2A at extreme positions and zero at mean position. However the wave velocity is different from the particle velocity. This depends on certain characteristics of the medium. Unlike the particle velocity which oscillates simple harmonically (between + A and – A) the wave velocity is constant for given characteristics of the medium. • Particle velocity in wave motion : The individual particles which make up the medium do not travel through the medium with the waves. They simply oscillate about their equilibrium positions. The instantaneous velocity of an oscillating particle of the medium, through which a wave is travelling, is known as \"Particle velocity\". y Particle velocity (v) y t wave (vP) x • Wave velocity : The velocity with which the disturbance, or planes of equal phase (wave front), travel through the medium is called wave (or phase) velocity. • Relation between particle velocity and wave velocity : y \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Wave equation :- y = A sin (t - kx), Particle velocity v= t = A cos (t - kx). Wave velocity = v = = = , y = - Ak cos (t–kx) =- A k cos (t - kx) = - 1 y y 1 y P T 2 k x vp t x v P t y Note : x represent the slope of the string (wave) at the point x. Particle velocity at a given position and time is equal to negative of the product of wave velocity with slope of the wave at that point at that instant. • Differential equation of harmonic progressive waves : 2 y 2 y 2 y 1 2 y E t2 = –A2 sin (t – kx) x2 = – Ak2 sin (t–kx) x2 = vP2 t2 4
JEE-Physics • Particle velocity (v ) and acceleration (a ) in a sinusoidal wave : pp The acceleration of the particle is the second particle is the second partial derivative of y (x, t) with respect to t, aP 2 y(x, t) 2 A sin(kx t) 2 y(x, t) t2 i.e., the acceleration of the particle equals –2 times its displacement, which is the result we obtained for SHM. Thus, a = –2 (displacement) P • Relation between Phase difference, Path difference & Time difference y , T, 2 A A Phase () 3 2 5 0 3 T 22 2 Wave length () 0 3 5 3 42 4 42 Time-period (T) 0 T T 3T 5T 3T 42 42 4 FHG KJI T 2 T Path difference = 2 Phase difference Example A progressive wave of frequency 500 Hz is travelling with a velocity of 360 m/s. How far apart are two points 60o out of phase. Solution v 360 We know that for a wave v = f So = f = 500 = 0.72 m 0.72 Phase difference = 60o = (/180) x 60 = (/3) rad, so path difference x= ()= x 3 =0.12 m 2 2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 THE GENER AL EQUATION OF WAVE MOTION Some physical quantity (say y) is made to oscillate at one place and these oscillations of y propagate to other places. The y may be, (i) displacement of particles from their mean position in case of transverse wave in a rope or longitudinal sound wave in a gas. (ii) pressure difference (dP) or density difference (d) in case of sound wave or (iii) electric and magnetic fields in case of electromagnetic waves. The oscillations of y may or may not be simple harmonic in nature. Consider one-dimensional wave travelling along x-axis. In this case y is a function of x and t. i.e. y = f(x, t) But only those function of x & t, represent a wave motion which satisfy the differential equation. 2 y v2 2 y ...(i) t2 x 2 The general solution of this equation is of the form y (x, t) = f (ax ± bt) ...(ii) Thus, any function of x and t and which satisfies equation (i) or which can be written as equation (ii) represents a wave. The only condition is that it should be finite everywhere and at all times, Further, if these conditions are satisfied, then speed of wave (v) is given by v coefficient of t b coefficient of x a E5
JEE-Physics Example Which of the following functions represent a travelling wave ? (a) (x – vt)2 (b) n(x + vt) (c) e (xvt)2 1 (d) x vt Solution Although all the four functions are written in the form f (ax + bt), only (c) among the four functions is finite everywhere at all times. Hence only (c) represents a travelling wave. Equation of a Plane Progressive Wave If, on the propagation of wave in a medium, the particles of the medium perform simple harmonic motion then the wave is called a 'simple harmonic progressive wave'. Suppose, a simple harmonic progressive wave is propagating in a medium along the positive direction of the x-axis (from left to right). In fig. (a) are shown the equilibrium positions of the particles 1, 2, 3 ....... 12345 6 7 8 9 (a) (b) 2 3 9 x y 4 1 a 5 x6 8 7 Direction of wave When the wave propagates, these particles oscillate about their equilibrium positions. In Fig. (b) are shown the instantaneous positions of these particles at a particular instant. The curve joining these positions represents the wave. Let the time be counted from the instant when the particle 1 situated at the origin starts oscillating. If y be the displacement of this particle after t seconds, then y = a sin t...(i) where a is the amplitude of oscillation and = 2 n, where n is the frequency. As the wave reaches the particles beyond the particle 1, the particles start oscillating. If the speed of the wave be v, then it will reach particle 6, distant x from the particle 1, in x/v sec. Therefore, the particle 6 will start oscillating x/v sec after the particle 1. It means that the displacement of the particle 6 at a time t will be the same as that of the particle 1 at a time x/v sec earlier i.e. at time t – (x/v). The displacement of particle 1 at time t – (x/v) can be the particle 6, distant x from the origin (particle 1), at time t is given by y = a sin t x But = 2 n, y = a sin (t – kx) k ...(ii) v v y = a sin 2 t x y = a sin 2 t x ...(iv) T Also k = ...(iii) T This is the equation of a simple harmonic wave travelling along +x direction. If the wave is travelling along the \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 –x direction then inside the brackets in the above equations, instead of minus sign there will be plus sign. For example, equation (iv) will be of the following form : y =a sin 2 t x . If be the phase difference between T the above wave travelling along the +x direction and an other wave, then the equation of that wave will be y = a sin 2 t x T Example The equation of a wave is, y ( x, t) 0.05 sin (1 0 x 40t) m 2 4 Find : (a) The wavelength, the frequency and the wave velocity (b) The particle velocity and acceleration at x=0.5 m and t=0.05 s. 6E
JEE-Physics Solution The equation may be rewritten as, y ( x, t) 0.05 sin 5 x 2 0 t m (a) 4 (b) Comparing this with equation of plane progressive harmonic wave, y(x, t) A sin(kx t ) we have, wave number k 2 5rad / m = 0.4m The angular frequency is, 2f 20 rad /s f = 10Hz The wave velocity is, v f 4 m s 1 iny x direction 5 The particle velocity k (20)(0.05) 2 4 and acceleration are, v = cos =2.22m/s p t a = 2 y (2 0 )2 (0 .0 5 ) sin 5 =140 m/s2 p t2 2 4 INTENSITY OF WAVE The amount of energy flowing per unit area and per unit time is called the intensity of wave. It is represented by I. Its units are J/m2s or watt/metre2. I = 22f2A2v i.e. I A2 and I A2. =v t If P is the power of an isotropic point source, then intensity at a distance r is given by, P1 I 4r2 or I r2 (for a point source) If P is the power of a line source, then intensity at a distance r is given by, I P or I 1 (for a line source) As, I A2 2 r r Therefore, A 1 (for a point source) and A 1 (for a line source) rr SUPERPOSITION PRINCIPLE Two or more waves can propagate in the same medium without affecting the motion of one another. If several waves propagate in a medium simultaneously, then the resultant displacement of any particle of the medium at any instant is equal to the vector sum of the displacements produced by individual wave. The phenomenon of intermixing of two or more waves to produce a new wave is called Superposition of waves. Therefore according to superposition principle. \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 The resultant displacement of a particle at any point of the medium, at any instant of time is the vector sum of the displacements caused to the particle by the individual waves. If y1, y2 , y3 , ... are the displacement of particle at a particular time due to individual waves, then the resultant displacement is given by y y1 y2 y3 ... Principle of superposition holds for all types of waves, i.e., mechanical as well as electromagnetic waves. But this principle is not applicable to the waves of very large amplitude. Due to superposition of waves the following phenomenon can be seen • Interference : Superposition of two waves having equal frequency and nearly equal amplitude. • Beats : Superposition of two waves of nearly equal frequency in same direction. E7
JEE-Physics • Stationary waves : Superposition of equal wave from opposite direction. • Lissajous' figure : Superposition of perpendicular waves. Superposition of waves Interference Beats Stationary waves Lissaju's figures Longitudinal Transverse Application Application Organ pipe Resonance tube Stretched string Sonometer INTERFERENCE OF WAVES : When two waves of equal frequency and nearly equal amplitude travelling in same direction having same state of polarisation in medium superimpose, then intensity is different at different points. At some points intensity is large, whereas at other points it is nearly zero. Consider two waves y = A sin (t - kx) and y = A sin (t – kx + ) By principle of superposition 1 1 2 2 where y = y + y = A sin (t – kx + ) 1 2 A2 A 2 A 2 2A1A2 cos and tan A2 sin 1 2 A1 A2 cos As intensity I A2 so I I1 I2 2 I1I2 cos • Constructive interference (maximum intensity) : Phase difference = 2n or path difference = n where n = 0, 1, 2, 3, ... 2 A = A + A and Imax I1 I2 2 I1I2 = I1 I2 max 1 2 • Destructive interference (minimum intensity) : where n = 0, 1, 2, 3, ... Phase difference = (2n+1), or path difference = (2n–1) 2 2 A = A – A and Imin I1 I2 2 I1I2 = I1 I2 min 1 2 GOLDEN KEY POINTS • Maximum and minimum intensities in any interference wave form. IMax I1 I2 2 A 1 A2 2 IMin = I1 I2 A 1 A2 = • Average intensity of interference wave form % &< I >or I = Imax Imin = I + I \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 av 2 12 if A = A = A and I = I = I then I = 4I, I = 0 and I =2I 12 12 max min AV • Degree of interference Pattern (f) : Degree of hearing (Sound Wave) or Imax Imin Degree of visibility (Light Wave) f = Imax Imin × 100 In condition of perfect interference degree of interference pattern is maximum f = 1 or 100% max • Condition of maximum contrast in interference wave form a =a and I = I then I = 4I and I = 0 12 12 max min For perfect destructive interference we have a maximum contrast in interference wave form. 8E
JEE-Physics VELOCITY OF TR ANSVERSE WAVE A= R2 T r2 d W Mass of per unit length m = , m = r2d, where d = Density of matter m,d Velocity of transverse wave in any wire v = T or T T r2 =A density m r2 d = Ad • If m is constant then, v T it is called tension law. 1 • If tension is constant then v it is called law of mass m 1 • If T is constant & take wire of different radius for same material then v it is called law of radius r • If T is constant & take wire of same radius for different material. Then v 1 law of density d REFLECTION FROM RIGID END When the pulse reaches the right end which is clamped at the wall, the element at the right end exerts a force on the clamp and the clamp exerts equal and opposite force on the element. The element at the right end is thus acted upon by the force from the clamp. As this end remains fixed, the two forces are opposite to each other. The force from the left part of the string transmits the forward wave pulse and hence, the force exerted by the clamp sends a return pulse on the string whose shape is similar to a return pulse but is inverted. The original pulse tries to pull the element at the fixed end up and the return pulse sent by the clamp tries to pull it down, so the resultant displacement is zero. Thus, the wave is reflected from the fixed end and the reflected wave is inverted with respect to the original wave. The shape of the string at any time, while the pulse is being reflected, can be found by adding an inverted image pulse to the incident pulse. Equation of wave propagating in +ve x-axis Incident wave y = a sin (t – kx) 1 Reflected wave y = a sin (t + kx + ) 2 or y = – a sin (t + kx) 2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 REFLECTION FROM FREE END The right end of the string is attached to a light frictionless ring which can freely move on a vertical rod. A wave pulse is sent on the string from left. When the wave reaches the right end, the element at this end is acted on by the force from the left to go up. However, there is no corresponding restoring force from the right as the rod does not exert a vertical force on the ring. As a result, the right end is displaced in upward direction more than the height of the pulse i.e., it overshoots the normal maximum displacement. The lack of restoring force from right can be equivalent described in the following way. An extra force acts from right which sends a wave from right to left with its shape identical to the original one. The element at the end is acted upon by both the incident and the reflected wave and the displacements add. Thus, a wave is reflected by the free end without inversion. Incident wave y = a sin (t – kx) Reflected wave y = a sin (t + kx) 1 2 STATIONARY WAVES * Definition : The wave propagating in such a medium will be reflected at the boundary and produce a wave of the same kind travelling in the opposite direction. The superposition of the two waves will give rise to a stationary wave. Formation of stationary wave is possible only and only in bounded medium. E9
JEE-Physics ANALYTICAL METHOD FOR STATIONARY WAVES • From rigid end : We know equation for progressive wave in positive x-direction y = a sin (t – kx) 1 After reflection from rigid end y = a sin (t + kx + ) = – a sin (t + kx) 2 By principle of super position. y = y + y = a sin (t – kx) – a sin (t + kx) = – 2a sin kx cos t 1 2 This is equation of stationary wave reflected from rigid end Amplitude =2a sin kx Velocity of particle v = dy =2a sin kx sin t pa dt dy stress dp dy Strain dx = – 2ak cos kx cos t Elasticity E = strain = dy Change in pressure dp = E dx dx x = 0, 2 , ......... • Node A = 0, V = 0, strain max. C h a ng e i n p r e s sur e max pa 3 A max, -V max. strain = 0 Ch ang e i n p ress ure = 0 • Antinode x = , .......... pa 44 • From f ree end : we know equation for progressive wave in positive x-direction y = a sin (t – kx) 1 After reflection from free end y = a sin (t + kx) 2 By Principle of superposition y = y + y = a sin (t – kx) + a sin (t + kx) = 2 a sin t cos kx 1 2 A m p l i t u d e = 2a cos kx, Velocity of particle = v = dy = 2a cos t cos kx Pa dt Strain dy dy dx = – 2ak sin t sin kx Change in pressure dp = E dx A Max, V = dy max. Strain = 0, dp = 0 • Antinode : x = 0, , ......... pa dt 2 3 5 dy = 0, strain max , dp max • Node : x = , , ............ A = 0, V = 44 4 pa dt PROPERTIES OF STATIONARY WAVES The stationary waves are formed due to the superposition of two identical simple harmonic waves travelling in opposite direction with the same speed. Important characteristics of stationary waves are:– \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (i) Stationary waves are produced in the bounded medium and the boundaries of bounded medium may be rigid or free. (ii) In stationary waves nodes and antinodes are formed alternately. Nodes are the points which are always in rest having maximum strain. Antinodes are the points where the particles vibrate with maximum amplitude having minimum strain. (iii) All the particles except at the nodes vibrate simple harmonically with the same period. (iv) The distance between any two successive nodes or antinodes is /2. (v) The amplitude of vibration gradually increases from zero to maximum value from node to antinode. (vi) All the particles in one particular segment vibrate in the same phase, but the particle of two adjacent segments differ in phase by 180° (vii) All points of the medium pass through their mean position simultaneously twice in each period. (viii) Velocity of the particles while crossing mean position varies from maximum at antinodes to zero at nodes. (ix) In a stationary wave the medium is splited into segments and each segment is vibrating up and down as a whole. 10 E
JEE-Physics (x) In longitudinal stationary waves, condensation (compression) and refraction do not travel forward as in progressive waves but they appear and disappear alternately at the same place. (xi) These waves do not transfer energy in the medium. Transmission of energy is not possible in a stationary wave. TR ANSMISSION OF WAVES We may have a situation in which the boundary is intermediate between these two extreme cases, that is, one in which the boundary is neither rigid nor free. In this case, part of the incident energy is transmitted and part is reflected. For instance, suppose a light string is attached to a heavier string as in (figure). When a pulse travelling on the light reaches the knot, same part of it is reflected and inverted and same part of it is transmitted to the heavier string. Incident pulse Incident pulse (a) (a) Transmitted pulse Transmitted Reflected pulse pulse Reflected pulse (b) (b) As one would expect, the reflected pulse has a smaller amplitude than the incident pulse, since part of the incident energy is transferred to the pulse in the heavier string. The inversion in the reflected wave is similar to the behaviour of a pulse meeting a rigid boundary, when it is totally reflected. When a pulse travelling on a heavy string strikes the boundary of a lighter string, as in (figure), again part is reflected and part is transmitted. However, in this case the reflected pulse is not inverted. In either case, the relative height of the reflected and transmitted pulses depend on the relative densities of the two string. In the previous section, we found that the speed of a wave on a string increases as the density of the string decreases. That is, a pulse travels more slowly on a heavy string than on a light string, if both are under the same tension. The following general rules apply to reflected waves. When a wave pulse travels from medium A to medium B and vA > vB (that is, when B is denser than A), the pulse will be inverted upon reflection. When a wave pulse travels from medium A to medium B and v < v (A is denser than B), it will not be inverted upon reflection. AB GOLDEN KEY POINTS Phenomenon of reflection and transmission of waves obeys the laws of reflection and refraction. The frequency of these wave remains constant i.e. does not change. i = r = t = From rarer to denser medium y = a sin (t – k x) y = –a sin (t + k x) y = a sin (t – k x) i i 1 r i 1 t t 2 From denser to rarer medium y = a sin (t – k x) y = a sin (t + k x) y = a sin (t – k x) ii 1 ri 1 tt 2 STATIONARY WAVE ARE OF TWO TYPES : (i) Transverse st. wave (stretched string) (ii) Longitudinal st. wave (organ pipes) \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 (i) Transverse Stationar y wave (Fixed at Both ends) Fundamental Harmonic N N = = 2 f = v A 2 2 Second Harmonic N N 2 = = f = 2v A 2 2 A Third Harmonic N A N 3 = = 2 f = 3v A A 2 3 2 E 11
JEE-Physics p = = 2 f = pv NN 2 p 2 p th harmonic AAAAAA • Law of length : For a given string, under a given tension, the fundamental frequency of vibration is inversely proportional to the length of the string, i.e, n 1 (T and m are constant) • Law of tension : The fundamental frequency of vibration of stretched string is directly proportional to the square root of the tension in the string, provided that length and mass per unit length of the string are kept constant. n T ( and m are constant) • Law of mass : The fundamental frequency of vibration of a stretched string is inversely proportional to the square root of its mass per unit length provided that length of the string and tension in the string are kept constant, i.e., n 1 ( and T are constant) m • Melde's experiment : In Melde's experiment, one end of a flexible piece of thread is tied to the end of a tuning fork. The other end passed over a smooth pulley carries a pan which can be loaded. There are two arrangements to vibrate the tied fork with thread. Transverse arrangement : Case 1. In a vibrating string of fixed length, the product of number of loops and square root of tension are constant or p T = constant. T = Mg \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 M Case 2. When the tuning fork is set vibrating as shown in fig. then the prong vibrates at right angles to the thread. As a result the thread is set into motion. The frequency of vibration of the thread (string) is equal to the frequency of the tuning fork. If length and tension are properly adjusted then, standing waves are formed in the string. (This happens when frequency of one of the normal modes of the string matched with the frequency of the tuning fork). Then, if p loops are formed in the thread, then the frequency of the pT tuning fork is given by n = 2 m Case 3. If the tuning fork is turned through a right angle, so that the prong vibrates along the length of the thread, then the string performs only a half oscillation for each complete vibrations of the prong. This is because the thread only makes node at the midpoint when the prong moves towards the pulley i.e. only once in a vibration. Longitudinal arrangement : The thread performs sustained oscillations when the natural frequency of the given length of the thread under tension is half that of the fork. T = Mg M 2p T E Thus if p loops are formed in the thread, then the frequency of the tuning fork is n = 2 m 12
JEE-Physics SONOMETER : Sonometer consists of a hollow rectangular box of light wood. One end of the experimental wire is fastened to one end of the box. The wire passes over a frictionless pulley P at the other end of the box. The wire is stretched by a tension T. B1 B2 A P The box serves the purpose of increasing the loudness of the sound produced by the vibrating wire. If the length the wire between the two bridges is , then the frequency of vibration is n = l T 2 m To test the tension of a tuning fork and string, a small paper rider is placed on the string. When a vibrating tuning fork is placed on the box, and if the length between the bridges is properly adjusted, then when the two frequencies are exactly equal, the string quickly picks up the vibrations of the fork and the rider is thrown off the wire. COMPARISON OF PROGRESSIVE AND STATIONARY WAVES Progressive waves Stationary waves 1. These waves travels in a medium These waves do not travel and remain confined with definite velocity. between two boundaries in the medium. 2. These waves transmit energy in the medium. These waves do not transmit energy in the medium. 3. The phase of vibration varies The phase of all the particles in between two nodes continuously from particle to particle. is always same. But particles of two Adjacent nodes differ in phase by 180° 4. No particle of medium is Particles at nodes are permanently at rest. Permanently at rest. 5. All particles of the medium vibrate The amplitude of vibration changes from particle and amplitude of vibration is same. to particle. The amplitude is zero for all at nodes and maximum at antinodes. 6 . All the particles do not attain the All the particles attain the maximum maximum displacement position displacement simultaneously. \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 SPEED OF LONGITUDINAL (SOUND) WAVES Newton Formula v = E (Use for every medium) medium Where E = Elasticity coefficient of medium & = Density of medium • For solid medi um v = Y solid Where E = Y = Young's modulas • For liquid Medium v = B Where E = B, where B = volume elasticity coefficient of liquid liquid E 13
JEE-Physics • For gas medium The formula for velocity of sound in air was first obtained by Newton. He assumed that sound propagates through air and temperature remains constant. (i.e. the process is isothermal) so Isothermal Elasticity = P vair = (P / P = 1.01 x 105 N/m2 kg/m3 so v = 1.01 105 At NTP for air and = 1.3 air = 279 m/s 1.3 However, the experimental value of sound in air is 332 m/s which is much higher than given by Newton's formula. • Laplace Correction In order to remove the discrepancy between theoretical and experimental values of velocity of sound, Laplace modified Newton's formula assuming that propagation of sound in air is adiabatic process, i.e. P Adiabatic Elasticity = p so that v i.e. v = 1.41 x 279 = 331.3 m/s [as air = 1.41] Which is in good agreement with the experimental value (332 m/s). This in turn establishes that sound propa- gates adiabatically through gases. The velocity of sound in air at NTP is 332 m/s which is much lesser than that of light and radio–waves (= 3 x 108 m/s). This implies that – (a) If we set our watch by the sound of a distant siren it will be slow. (b) If we record the time in a race by hearing sound from starting point it will be lesser than actual. (c) In a cloud–lightening, though light and sound are produced simultaneously but as c > v, light proceeds thunder. An in case of gases – v = P PV as mass M or v = RT [as PV = RT] or v = RT s = mass volume V s s Mw M mass M as where M w = Molecular weight Mw Mw And from kinetic-theory of gases v = (3RT / M w ) vs rms So v rms = 3 EFFECT OF VARIOUS QUANTITIES (1) Effect of temperature v2 T2 1 T v1 = T1 = For a gas & M is constant v t 273 v = v 1 t 2 W 273 t0 273 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 By applying Binomial theorem. (i) For any gas medium v = v 1 t (ii) For air : v = v + 0.61 t m/sec. (v = 332 m/sec. ) t0 546 t0 0 (2) Effect of Relative Humidity With increase in humidity, density decreases so in the light of v = P / ) We conclude that with rise in humidity velocity of sound increases. This is why sound travels faster in humid air (rainy season) than in dry air (summer) at same temperature. Due to this in rainy season the sound of factories siren and whistle of train can be heard more than summer. 14 E
JEE-Physics (3) Effect of Pressure E P RT As velocity of sound v = = = M So pressure has no effect on velocity of sound in a gas as long as temperature remain constant. This is why in going up in the atmosphere, though both pressure and density decreases, velocity of sound remains constant as long as temperature remains constant. Further more it has also been established that all other factors such as amplitude, frequency, phase, loudness pitch, quality etc. has partially no effect on velocity of sound. Velocity of sound in air is measured by resonance tube or Hebb's method while in gases by Quinke's tube. Kundt's tube is used to determine velocity of sound in any medium solid, liquid or gas. (4) Effect of Motion of Air If air is blowing then the speed of sound changes. If the actual speed of sound is v and the speed of air is w, then the speed of sound in the direction in which air is blowing will be (v+ w), and in the opposite direction it will be (v – w). (5) Effect of Frequency There is no effect of frequency on the speed of sound. Sound waves of different frequencies travel with the same speed in air although their wavelength in air are different. If the speed of sound were dependent on the frequency, then we could not have enjoyed orchestra. Example A piezo electric quartz plate of thickness 0.005 m is vibrating in resonant conditions. Calculate its fundamental frequency if for quartz Y = 8 1010 N/m2 and = 2.65 x 103 kg/m3 Solution Y 8 1010 We known that for longitudinal waves in solids v = , So v = 2.65 103 = 5.5 103 m/s Further more for fundamental mode of plate – (/2) = L So = 2 x 5 x 10–3 = 10–2 m But as v = f, i.e., f = (v/) so f = [5.5 x 103/10–2] = 5.5 x 105 Hz = 550 kHz Example Determine the change in volume of 6 liters of alcohol if the pressure is decreased from 200 cm of Hg to 75 cm. [velocity of sound in alcohol is 1280 m/s, density of alcohol = 0.81 gm/cc, density of Hg = 13.6 gm/cc and g = 9.81 m/s2] Solution For propagation of sound in liquid v = B / i.e., B = v2 \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 P P V (P ) But by definition B = – V So –V = v2, i.e. V = V v 2 V Here P = H g – H g = (75 – 200) 13.6 981 = –1.667 106 dynes/cm2 21 6 103 1.667 106 So V = 0.81 1.280 105 2 = 0.75 cc Example Speed of sound in air is 332 m/s at NTP. What will the speed of sound in hydrogen at NTP if the density (a) of hydrogen at NTP is (1/16) that of air. (b) Calculate the ratio of the speed of sound in neon to that in water vapour at any temperature. [Molecular weight of neon = 2.02 10–2 kg/mol and for water vapours = 1.8 10–2 kg/mol] E 15
JEE-Physics Solution E P RT The velocity of sound in air is given by v = = = M (a) vH PH air = air [as P = P ] In terms of density and pressure v air = H Pair air H H v air 16 v = air H = 332 = 1328 m/s H 1 ( b ) In terms of temperature and molecular weight v Ne Ne MW [as T = T ] v W = M Ne W NW Now as neon is mono atomic ( = 5/3) while water vapours poly atomic ( = 4/3) so b gv Ne5 / 3 1.8 10 2 5 1.8 b gv W = 4 / 3 2.02 10 2 = 4 2.02 = 1.055 VIBR ATION IN ORGAN PIPES When two longitudinal waves of same frequency and amplitude travel in a medium in opposite directions then by superposition, standing waves are produced. These waves are produced in air columns in cylindrical tube of uniform diameter. These sound producing tubes are called organ pipes. 1. Vibration of air column in closed organ pipe : Fundamental Ist Overtone IInd Overtone The tube which is closed at one end and open at the other end is AA A called close organ pipe. On blowing air at the open end, a wave travels towards closed end from where it is reflected towards open N end. As the wave reaches open end, it is reflected again. So two N A 53 longitudinal waves travel in opposite directions to superpose and 1 32 4 produce stationary waves. At the closed end there is a node since 4 4 particles does not have freedom to vibrate whereas at open end A N there is an antinode because particles have greatest freedom to A vibrate. N N N (i) (ii) (iii) Hence on blowing air at the open end, the column vibrates forming antinode at free end and node at closed end. If is length of pipe and be the wavelength and v be the velocity of sound in organ pipe then Case (i) =4 vv \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 4 n1 4 fundamental frequency. Case (ii) 3 4 v 3v 4 3 n2 4 First overtone / IIIrd Harmonic Case (iii) 5 4 n3 v 5v Second overtone / Vth Harmonic 4 5 4 When closed organ pipe vibrate in mth overtone then (2m 1) 4 So 4 n (2m 1) v (2m 1) 4 Hence frequency of overtones is given by n : n : n .......... = 1 : 3 : 5 .... 1 2 3 16 E
JEE-Physics 2 . Vi br at i on o f a i r c olu mns i n op en or ga n p i p e : Fundamental Ist Overtone II nd Overtone The tube which is open at both ends is called an open organ pipe. AA A On blowing air at the open end, a wave travel towards the other N N N 3 3 end after reflection from open end waves travel in opposite direction 2 to superpose and produce stationary wave. Now the pipe is open 1 A at both ends by which an antinode is formed at open end. Hence on 2A 2 2 2N blowing air at the open end antinodes are formed at each end and N A nodes in the middle. If is length of the pipe and be the wavelength N A and v is velocity of sound in organ pipe. AA (i) (ii) (iii) Case (i) 2 n1 v v Fundamental frequency. 2 2 Case (ii) 2 2 n2 v 2v First overtone frequency. 2 2 2 Case (iii) 3 2 v 3v 2 n 3 2 Second overtone frequency. 3 Hence frequency of overtones are given by the relation n : n : n .......... = 1 : 2 : 3 .... 1 2 3 When open organ pipe vibrate in mth overtone then (m 1) so 4 n (m 1) v 4 m 1 2 GOLDEN KEY POINTS • A rod clamped at one end or a string fixed at one end is similar to vibration of closed end organ pipe. • A rod clamped in the middle is similar to the vibration of open end organ pipe. • If an open pipe is half submerged in water, it becomes a closed organ pipe of length half that of open pipe i.e. frequency remains same. • Due to finite motion of air molecular in organ pipes reflection takes place not exactly at open end but some what above it so in an organ pipe antinode is not formed exactly at free–end but above it at a distance e = 0.6r (called end correction or Rayleigh's correction) with r being the radius of pipe. So for closed organ pipe L L + 0.6r while for open L L + 2 0.6r (as both ends are open) vv so that – f = while f = C 4(L 0.6r) 0 2(L 1.2r) This is why for a given v and L narrower the pipe higher will the frequency or pitch and shriller will be the sound. \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 • For an organ pipe (closed or open) if v = constant. f (1/L) So with decrease in length of vibrating air column, i.e., wavelength ( L), frequency or pitch will increase and vice–versa. This is why the pitch increases gradually as an empty vessel fills slowly. • v • For an organ pipe if f = constant. v or v L , f = = constant i.e. the frequency of an organ pipe will remain unchanged if the ratio of speed of sound in to its wave length remains constant. E As for a given length of organ pipe = constant. f v So ( a ) With rise in temperature as velocity will increase (v T the pitch will increase. (Change in length with temperature is not considered unless stated) ( b ) With change in gas in the pipe as v will change and so f will change (v / M ) ( c ) With increase in moisture as v will increase and so the pitch will also. 17
JEE-Physics Example For a certain organ pipe, three successive resonant frequencies are observed at 425, 595 and 765 Hz respec- tively. Taking the speed for sound in air to be 340 m/s (a) Explain whether the pipe is closed at one end or open at both ends (b) determine the fundamental frequency and length of the pipe. Solution ( a ) The given frequencies are in the ratio 425 : 595 : 765, i.e., 5 : 7 : 9 And as clearly these are odd integers so the given pipe is closed pipe. ( b ) From part (b) it is clear that the frequency of 5th harmonic (which is third overtone) is 425 Hz so 425 = 5fc fC = 85 Hz Further as v v 340 Example fC = 4L , L = 4fC = 4 85 = 1 m AB is a cylinder of length 1 m fitted with a thin flexible diaphram C at middle and two other thin flexible diaphram A and B at the ends. The portions AC and BC contain hydrogen and oxygen gases respectively. The diaphragms A and B are set into vibrations of the same frequency. What is the minimum frequency of these vibrations for which diaphram C is a node ? Under the condition of the experiment the velocity of sound in hydrogen is 1100 m/s and oxygen 300 m/s. Solution As diaphragm C is a node, A and B will be antinode (as in a organ pipe either both ends are antinode or one end node and the other antinode), i.e., each part will behave as closed end organ pipe so that vH 1100 v0 300 AC B f = 4L H = 4 0.5 = 550 Hz And f= 4L0 = 4 0.5 =150Hz 0 H H2 O2 As the two fundamental frequencies are different, the system will vibrate with a n H f0 150 3 common frequency f such that f = nHfH = n0f0 i.e. = = = n0 fH 550 11 i.e., the third harmonic of hydrogen and 11th harmonic of oxygen or 6th harmonic of hydrogen and 22nd harmonic of oxygen will have same frequency. So the minimum common frequency f = 3 550 or 11 150 = 1650 Hz APPAR ATUS FOR DETERMINING SPEED OF SOUND 1 . Quinck's Tube : A C T2 T1 B It consists of two U shaped metal tubes. Sound waves with the help of \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 tuning fork are produced at A which travel through B & C and comes D out at D where a sensitive flame is present. Now the two waves coming through different path interfere and flame flares up. But if they are not in phase, destructive interference occurs and flame remains undisturbed. Suppose destructive interference occurs at D for some position of C. If now the tube C is moved so that interference condition is disturbed and again by moving a distance x, destructive interference occurs so that 2x = . Similarly if the distance moved between successive constructive and destructive interference is x then 2x Now by having value of x, speed of sound is given by v = n 2 2 . Kundt's tube : It is the used to determine speed of sound in different gases. It consists of a glass tube in which a small quantity of lycopodium powder is spread. The tube is rotated so that powder starts slipping. Now rod CD is rubbed at end D so that stationary waves form. The disc C vibrates so that air column also vibrates with the frequency of the rod. The piston P is adjusted so that frequency of air column become same as that of rod. 18 E
JEE-Physics So resonance occurs and column is thrown into stationary waves. The powder sets into oscillations at antinodes while heaps of powder are formed at nodes. clamped at P2 P1 the middle rod R lycopodium power Let n is the frequency of vibration of the rod then, this is also the frequency of sound wave in the air column in the tube. For rod : rod rod For air : air air 2 2 Where is the distance between two heaps of powder in the tube (i.e. distance between two nodes). If v and air air v air v rod vrod are velocity of sound waves in the air and rod respectively, then n = air rod Therefore, v air air air Thus knowledge of v determines v . v rod rod rod air rod (ii) RESONANCE TUBE Construction : The resonance tube is a tube T (figure) made of brass or glass, about 1 meter long and 5 cm in diameter and fixed on a vertical stand. Its lower end is connected to a water reservoir B by means of a flexible rubber tube. The rubber tube carries a pinch-cock P. The level of water in T can be raised or lowered by water adjusting the height of the reservoir B and controlling the flow of water from B to T or from T to B by means of the pinch-cock P. Thus the length of the air–column in T can be changed. The position of the water level in T can be read by means of a side tube C and a scale S. \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Determination of the speed of sound in air by resonance tube First of all the water reservoir B is raised until the water level in the tube T rises almost to the top of the tube. Then the pinch–cock P is tightened and the reservoir B is lowered. The water level in T stays at the top. Now a tuning fork is sounded and held over the mouth of tube .The pinch–cock P is opened slowly so that the water level in T fal l s a nd t he l engt h of t he ai r–c ol umn i ncreas e s . At a p ar t ic ular l engt h o f air–co lum n i n T, a lou d sou nd is hear d. This is the first state of resonance. In this position the following phenomenon takes place inside the tube. (i) For first resonance = /4 1 (ii) For second resonance = 3/4 – = /2 = 2( – ) 2 2 1 2 1 If the frequency of the fork be n and the temperature of the air-column be toC, then the speed of sound at toC is given by v = n = 2n ( – ) t 2 1 The speed of sound wave at 0oC v = (v – 0.61 t) m/s. 0t E 19
JEE-Physics End Correction : In the resonance tube, the antinode is not formed exactly at the open but slightly outside at a distance x. Hence the length of the air -column in the first and second states of resonance are (l + x) and 1 ( + x) then + x = /4 ..........(i) 2 1 ( i ) For first resonance (ii) For second resonance + x = 3/4 ...........(ii)) 2 Subtract Equation (ii) from Equation (i) – = /2 2 1 = 2 ( – ) 21 Put the value of in Equation (i) + x = 2(2 1 ) 1 4 + x = 2 1 x 2 31 1 2 2 B E AT S When two sound waves of same amplitude travelling in same direction with slightly different y1 frequency superimpose, then intensity varies periodically with time. This effect is called Beats. Suppose two waves of frequencies f and f (<f ) 1 21 are meeting at some point in space. The y2 corresponding periods are T and T (>T ). If y1 & y2 1 21 the two waves are in phase at t=0, they will again be in phase when the first wave has gone through exactly one more cycle than the second. This will happen at a time t=T, the period of the beat. Let n be the number of cycles of the first wave in time T, then the number of cycles y of the second wave in the same time is (n–1). Hence, T = nT = (n–1) T 12 Eliminating n we have T T1 T2 1 1 1 f T f1 f2 T2 T1 1 1 f1 f2 T1 T2 The reciprocal of the beat period is the beat frequency Waves Inter ference On The Bases Of Beats : \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Conditions %Two equal frequency waves travel in same direction. Mathematical analysis If displacement of first wave y1 = a sin 1t (N1 , a) I N2a2 Displacement of second wave y = a sin 2t (N , a) 2 2 By superposition y = y + y 1 2 Equation of resulting wave y = a {sin 2N1t + sin 2 N t} 2 20 E
JEE-Physics 2 sin 2 t (N1 N2 ) cos 2t (N1 N 2 ) 2a cos 2t (N1 N2 ) (N1 N2 ) 2 2 2 2 y = a = sin 2t = A sin 2N't N1 N2 = 2a t (N – N) Frequency N' = N1 N 2 Amplitude A = 2a Cos 2t cos 1 2 2 2 • For max Intensity ( A = ± 2a ): – If cos (N – N ) t = ± 1 cos (N – N ) t = cos n , n = 0, 1, 2, ......... 1 2 1 2 n 123 (N1 – N2)t = n t = N1 N2 = 0, N , N , N ............. • For Minimum Intensity (A = 0) : cos (N – N ) t = 0 cos (N – N ) t = cos (2n + 1) n = 0, 1, 2 ............. 1 2 1 2 2 2n 1 135 (N – N ) t = (2n + 1) 2 t = 2(N1 N2) = 2 N , 2 N , 2 N ............. 1 2 GOLDEN KEY POINTS\\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 • When we added wax on tuning fork then the frequency of fork decreases. • When we file the tuning fork then the frequency of fork increases. Example A tuning fork having n = 300 Hz produces 5 beats/s with another tuning fork. If impurity (wax) is added on the arm of known tuning fork, the number of beats decreases then calculate the frequency of unknown tuning fork. Solution The frequency of unknown tuning fork should be 300 + 5 = 295 Hz or 305 Hz. When wax is added, if it would be 305 Hz, beats would have increases but with 295 Hz beats is decreases so frequency of unknown tuning fork is 295 Hz. Example A tuning fork having n = 158 Hz, produce 3 beats/s with another. As we file the arm of unknown, beats become 7 then calculate the frequency of unknown. Solution The frequency of unknown tuning fork should be 158 ± 3 = 155 Hz or 161 Hz. After filling the number of beats = 7 so frequency of unknown tuning fork should be 158 ± 7 = 165 Hz or 151 Hz. As both above frequency can be obtain by filing so frequency of unknown = 155/161 Hz. SPECIAL POINTS 1. Displacement and pressure waves A sound wave (i.e. longitudinal mechanical wave) can be described either in terms of the longitudinal displace- ment suffered by the particles of the medium (called displacement-wave) or in terms of the excess pressure generated due to compression and rarefaction (called pressure-wave). Consider a sound wave travelling in the x-direction in a medium. Suppose at time t, the particle at the undisturbed position x suffers a displacement y in the x-direction. The displacement wave then will be described by y = A sin (t – kx) ...(i) E 21
JEE-Physics Now consider the element of medium which is confined with in x and x+x in the undisturbed state. If S is the cross-section, the volume element in undisturbed state will be V = S x. As the wave passes the ends at x and x +x are displaced by amount y and y + y so that increase in volume of the element will be V = S y. This in turn implies that volume strain for the element under consideration V S y y ... (ii) V S x x So corresponding stress, i.e. excess pressure P B V a s B V P V P or P B y [From equation (ii)) ... (iii) V V V x Note: For a harmonic -progressive - wave from dy v pa , P B dy B vp a dx v dx v i.e. pressure in a sound wave is equal to the product of elasticity of gas with the ratio of particle speed to wave speed. y Ak cos(t kx) so T/2 T x x s But from equation (i) So P = AkB cos (t – kx) i.e. P= P cos (t – kx) ...(iv) po with 0 p From equation P0 = ABk ...(iv) x (i) and (iv) it is clear that • A sound wave may be considered as either a displacement wave y = A sin (t - kx) or a pressure wave P = P cos(t – kx) . 0 • The pressure wave is 900 out of phase with respect to displacement wave, i.e, displacement will be maximum when pressure is minimum and vice-versa. This is shown in figure • The amplitude of pressure wave:– P = ABk = Akv2 = vA [as v B / , v = /k] ...(v) 0 • As sound-sensors (e.g. ear or mic) detects pressure changes, description of sound as pressure-wave is pre- ferred over displacement wave. 2. Ultrasonic, Infrasonic and Audible (sonic) Sound : \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 Sound waves can be classified in three groups according to their range of frequencies. • Infrasonic Waves Longitudinal waves having frequencies below 20 Hz are called infrasonic waves. They cannot be heard by human beings. They are produced during earthquakes. Infrasonic waves can be heard by snakes. • Audible Waves Longitudinal waves having frequencies lying between 20-20,000 Hz are called audible waves. • Ultrasonic Waves Longitudinal waves having frequencies above 20,000 Hz are called ultrasonic waves. They are produced and heard by bats. They have a large energy content. • Applications of Ultrasonic Waves Ultrasonic waves have a large range of application. Some of them are as follows: (i) The fine internal cracks in metal can be detected by ultrasonic waves. (ii) Ultrasonic waves can be used for determining the depth of the sea, lakes etc. (iii) Ultrasonic waves can be used to detect submarines, icebergs etc. (iv) Ultrasonic waves can be used to clean clothes, fine machinery parts etc. (v) Ultrasonic waves can be used to kill smaller animals like rates, fish and frogs etc. 22 E
JEE-Physics • Shock Waves If the speed of the body in air is greater than the speed of the sound, then it is called supersonic speed. Such a body leaves behind a conical region of disturbance which spreads continuously. Such a disturbance is called a ' S h o c k Wave ' . T h i s wave c ar r i e s h u ge energ y. If i t str i ke s a b u i l d i ng , t hen t he bu i l d i ng m ay b e d am ag ed . 3. Sound intensity in decibels The physiological sensation of loudness is closely related to the intensity of wave producing the sound. At a frequency of 1 kHz people are able to detect sounds with intensities as low as 10–12W/m2. On the other hand an intensity of 1 W/m2 can cause pain, and prolonged exposure to sound at this level will damage a person's ears. Because the range in intensities over which people hear is so large, it is convenient to use a logarithmic scale to specify intensities. If the intensity of sound in watts per square meter is I, then the intensity level in decibels (dB) is given by I 10 log I0 where the base of the logarithm is 10, and I =10–12 W/m2 (roughly the minimum intensity that can be heard). 0 1 On the decibel scale, the pain threshold of 1 W/m2 is then 10 log 1012 120dB Example Calculate the change in intensity level when the intensity of sound increases by 106 times its original intensity. Solution Here I Final intensity 106 I0 Initial intensity I (106) Increase in intensity level, L = 10 log = 10 log = 10 × log 10 = 10 × 6 × 1 = 60 decibels. 10 I0 10 10 ECHO Multiple reflection of sound is called an echo. If the distance of reflector from the source is d then, 2d vt Hence, v = speed of sound and t, the time of echo. d vt 2 Since, the effect of ordinary sound remains on our ear for 1\\10 second, therefore, if the sound returns to the starting point before 1/10 second, then it will not be distinguished from the original sound and no echo will be heard. Therefore, the minimum distance of the reflector is, d min vt 330 1 16.5m 2 2 10 ACOUSTIC DOPPLER EFFECT (DOPPLER EFFECT FOR SOUND WAVES) \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 The apparent change in the frequency of sound when the source of sound, the observer and the medium are in relative motion is called Doppler effect. While deriving these expressions, we make the following assumptions (i) The velocity of the source, the observer and the medium are along the line joining the positions of the source and the observer. (ii) The velocity of the source and the observer is less than velocity of sound. Doppler effect takes place both in sound and light. In sound it depends on whether the source or observer or both are in motion while in light it depends on whether the distance between source and observer is increasing or decreasing. NOTATIONS n actual frequency n' observed frequency (apparent frequency) actual wave length ' observed (apparent) wave length v velocity of sound vs velocity of source v velocity of observer v wind velocity 0 w E 23
JEE-Physics Case I : Source in motion, observer at rest, medium at rest : n waves S OS n n n'=n (rest) (rest) v both source and observer at rest Suppose the source S and observer O are separated by distance v. Where v is the velocity of sound. Let n be the frequency of sound emitted by the source. Then n waves will be emitted by the source in one second. These n waves will be accommodated in distance v. So, wave length = total distance = v total number of waves n (i) Source moving towards stationary observer : Let the sources start moving towards the observer with velocity v . After one second, the n waves will be s crowded in distance (v – v ). Now the observer shall feel that he is listening to sound of wavelength ' and s frequency n' v S S' O S vS O vS n n' (moving) (rest) v–vS Now apparent wavelength ' = total distance = v vs total number of waves n v v v and changed frequency, n' = = = n v ' v v v s n s So, as the source of sound approaches the observer the apparent frequency n' becomes greater than the true frequency n. (ii) When source move away from stationary observer :- For this situation n waves will be crowded in distance v + v . s v+vS \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 S' S O vS S O vS n n' (moving) (rest) So, apparent wavelength ' = v v s and n Apparent frequency n' = v = v v = v vs ' n vs So n' < n vs n 24 E
JEE-Physics Case II : Observer in motion, source at rest, medium at rest Let the source (S) and observer (O) are in rest at their respective places. Then n waves given by source 'S' would be crossing observer 'O' in one second and fill the space OA (=v) v SO A S O n n'=n (rest) (rest) both source S and O at rest (i) Observer moves towards stationary source v + vo S A S vO O O' O n n' (rest) (moving) vo v When observer 'O' moves towards 'S' with velocity v , it will cover v distance in one second. So the oo observer has received not only the n waves occupying OA but also received additional number of n waves occupying the distance OO' (= v ). o So, total waves received by observer in one second i.e., apparent frequency (n') = Actual waves (n) + Additional waves (n) v vo v vo n v vo v v n n' = = v n = (so, n' > n) (ii) Observer moves away from stationary source :- For this situation n waves will be crowded in distance v – v . o vvo \\\\node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-6\\Wave Motion\\Eng\\Theory.p65 S O S O vo O O' n n' (rest) vo v When observer move away from source with v velocity then he will get n waves less than real number o of waves. So, total number of waves received by observer i.e. Apparent frequency (n') = Actual waves (n) – reduction in number of waves (n) n' = v vo = v vo = v vo = v vo n v (so n' < n) v n v n E 25
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