DC Pandey Waves And Thermodynamics

340 — Waves and Thermodynamics Solution Thermal resistance, R = l (as lx = ly and Ax = Ay) ∴ KA Rx = K y Ry K x = 0.46 0.92 =1 2 So, if Rx = R then Ry = 2R CEDB forms a balanced Wheatstone bridge, i.e. TC = TD and no heat flows through CD. ∴ 1= 1 + 1 RBE R + R 2R + 2R or RBE = 4 R 3 The total resistance between A and E will be RAE = RAB + RBE = 2R + 4 R = 10 R 3 3 ∴ Heat current between A and E is H = (∆T )AE = (60 – 10) = 15 RAE (10/3) R R Now, if TB is the temperature at B, H AB = (∆T )AB or RAB 15 = 60 – TB R 2R or TB = 30° C Ans. Further, H AB = HBC + HBD or 15 = 30 – TC + 30 – TD [TC = TD = T (say)] RR 2R or 15 = (30 – T ) + (30 – T ) 2 Solving this, we get T = 20° C Ans. or TC = TD = 20° C V Example 22 A hollow sphere of glass whose external and internal radii are 11 cm and 9 cm respectively is completely filled with ice at 0°C and placed in a bath of boiling water. How long will it take for the ice to melt completely? Given that density of ice = 0.9 g / cm3 , latent heat of fusion of ice = 80 cal / g and thermal conductivity of glass = 0.002 cal / cm- s°C. Solution In steady state, rate of heat flow (result is taken from Example 4) H = 4πKr1r2∆T r2 – r1

Chapter 22 Calorimetry and Heat Transfer — 341 Substituting the values, H = (4) (π ) (0.002) (11) (9) (100 – 0) (11 – 9) or dQ = 124.4 cal/s dt This rate should be equal to, L dm dt ∴  ddmt  = dQ /dt = 124.4 = 1.555 g /s L 80 Total mass of ice, m = ρice (4πr12) = (0.9) (4) (π ) (9)2 = 916 g ∴ Time taken for the ice to melt completely Ans. t = m = 916 = 589 s (dm/dt) 1.555 V Example 23 A point source of heat of power P is placed at the centre of a spherical shell of mean radius R. The material of the shell has thermal conductivity K. Calculate the thickness of the shell if temperature difference between the outer and inner surfaces of the shell in steady state is T. Solution Consider a concentric spherical shell of radius r and thickness dr as shown in figure. In steady state, the rate of heat flow (heat current) through this shell will be H = ∆T = (– dθ)  R = KlA R dr (k) (4πr2) or H = – (4πKr2) dθ dr Here, negative sign is used because with increase in r, θ decreases dθ r1 P dr r r2 . ∴ ∫ ∫r2 dr= – 4πK θ2 dθ This equation gives, H In steady state, r1 r2 θ1 ∴ Thickness of shell, H = 4πKr1r2 (θ1 – θ2) (r2 – r1 ) H = P, r1r2 ≈ R2 and θ1 – θ2 = T r2 – r1 = 4πKR2T Ans. P

342 — Waves and Thermodynamics V Example 24 A steam cylindrical pipe of radius 5 cm carries steam at 100°C. The pipe is covered by a jacket of insulating material 2 cm thick having a thermal conductivity 0.07 W/m-K. If the temperature at the outer wall of the pipe jacket is 20°C, how much heat is lost through the jacket per metre length in an hour? K = 0.07 W/m-K 7 cm dr r 5 cm Solution Thermal resistance per metre length of an element at distance r of thickness dr is dR = dr  R = KlA K (2πr) ∴ Total resistance ∫R = r2 = 7 cm dR r1 = 5 cm ∫= 1 7.0 × 10–2 m dr 2πK 5.0 × 10–2 m r =1 ln  57 2πK = 1 ln (1.4) (2π ) (0.07) = 0.765 K/W Heat current, H = Temperature difference Thermal resistance = (100 – 20) = 104.6 W 0.765 ∴ Heat lost in one hour = Heat current × time Ans. = (104.6) (3600) J = 3.76 × 105 J

Exercises LEVEL 1 Take cice = 0.53 cal/g-°C, cwater = 1.0 cal/g-°C, (Lf )water = 80 cal/g and (Lv )water = 529 cal/g unless given in the question. Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : Specific heat of any substance remains constant at all temperatures. Reason : It is given by s = 1 ⋅ dQ m dT 2. Assertion : When temperature of a body is increased, in radiant energy, number of low wavelength photons get increased. Reason : According to Wien’s displacement law λm ∝ 1. T 3. Assertion : Warming a room by a heat blower is an example of forced convection. Reason : Natural convection takes place due to gravity. 4. Assertion : A conducting rod is placed between boiling water and ice. If rod is broken into two equal parts and two parts are connected side by side, then rate of melting of ice will increase to four times. Reason : Thermal resistance will become four times. 5. Assertion : A normal body can radiate energy more than a perfectly black body. Reason : A perfectly black body is always black in colour. 6. Assertion : According to Newton’s law, good conductors of electricity are also good conductors of heat. Reason : At a given temperature, eλ ∝ aλ for any body. 7. Assertion : Good conductors of electricity are also good conductors of heat due to large number of free electrons. Reason : It is easy to conduct heat from free electrons. 8. Assertion : Emissivity of any body (e) is always less than its absorptive power (a). Reason : Both the quantities are dimensionless. 9. Assertion : Heat is supplied at constant rate from one end of a conducting rod. In steady state, temperature of all points of the rod becomes uniform. Reason : In steady state, temperature of rod does not increase.

344 — Waves and Thermodynamics 10. Assertion : A solid sphere and a hollow sphere of same material and same radius are kept at same temperatures in atmosphere. Rate of cooling of hollow sphere will be more. Reason : If all other conditions are same, then rate of cooling is inversely proportional to the mass of body. Objective Questions 1. For an enclosure maintained at 2000 K, the maximum radiation occurs at wavelength λ m. If the temperature is raised to 3000 K, the peak will shift to (a) 0.5 λm (b) λm (c) 2 λm (d) 3 λm 3 2 2. A substance cools from 75°C to 70°C in T1 minute, from 70°C to 65°C in T2 minute and from 65°C to 60°C in T3 minute, then (a) T1 = T2 = T3 (b) T1 < T2 < T3 (c) T1 > T2 > T3 (d) T1 < T2 > T3 3. Two liquids are at temperatures 20°C and 40°C. When same mass of both of them is mixed, the temperature of the mixture is 32°C. What is the ratio of their specific heats? (a) 1/3 (b) 2/5 (c) 3/2 (d) 2/3 4. The specific heat of a metal at low temperatures varies according to S = aT 3 , where a is a constant and T is absolute temperature. The heat energy needed to raise unit mass of the metal from temperature T = 1 K to T = 2 K is (a) 3a (b) 15a (c) 2a (d) 13a 4 3 4 5. The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the North star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the sun and the north star is (a) 1.46 (b) 0.69 (c) 1.21 (d) 0.83 6. A solid material is supplied heat at a constant rate. The temperature of material is changing with heat input as shown in the figure. What does the slope of DE represent? Temperature E CD AB O Heat input (a) Latent heat of liquid (b) Latent heat of vaporization (c) Heat capacity of vapour (d) Inverse of heat capacity of vapour 7. Two ends of rods of length L and radius R of the same material are kept at the same temperature. Which of the following rods conducts the maximum heat? (a) L = 50 cm, R = 1 cm (b) L = 100 cm, R = 2 cm (c) L = 25 cm, R = 0.5 cm (d) L = 75 cm, R = 1.5 cm

Chapter 22 Calorimetry and Heat Transfer — 345 8. 1 g of ice at 0°C is mixed with 1 g of steam at 100°C. After thermal equilibrium is achieved, the temperature of the mixture is (a) 100°C (b) 55°C (c) 75°C (d) 0°C 9. A wall has two layers A and B each made of different materials. The layer A is 10 cm thick and B is 20 cm thick. The thermal conductivity of A is thrice that of B. Under thermal equilibrium temperature difference across the wall is 35°C. The difference of temperature across the layer A is (a) 30°C (b) 14°C (c) 8.75°C (d) 5°C 10. A wall has two layers A and B each made of different materials. Both the layers have the same thickness. The thermal conductivity of material A is twice of B. Under thermal equilibrium the temperature difference across the layer B is 36°C. The temperature difference across layer A is (a) 6°C (b) 12°C (c) 18°C (d) 24°C 11. The end of two rods of different materials with their thermal conductivities, area of cross-section and lengths all in the ratio 1 : 2 are maintained at the same temperature difference. If the rate of flow of heat in the first rod is 4 cal/s. Then, in the second rod rate of heat flow in cal/s will be (a) 1 (b) 2 (c) 8 (d) 16 Subjective Questions 1. A thin square steel plate 10 cm on a side is heated in a black smith’s forge to temperature of 800° C. If the emissivity is 0.60, what is the total rate of radiation of energy? 2. A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is 27° C and its melting point is 327° C. Latent heat of fusion of lead = 2.5 × 104 J/ kg and specific heat capacity of lead = 125 J/ kg-K. 3. A ball is dropped on a floor from a height of 2.0 m . After the collision it rises upto a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Specific heat of the ball is 800 J/K. 4. A nuclear power plant generates 500 MW of waste heat that must be carried away by water pumped from a lake. If the water temperature is to rise by 10°C, what is the required flow rate in kg/s? 5. The emissivity of tungsten is 0.4. A tungsten sphere with a radius of 4.0 cm is suspended within a large evacuated enclosure whose walls are at 300 K. What power input is required to maintain the sphere at a temperature of 3000 K if heat conduction along supports is neglected? Take, σ = 5.67 × 10–8 Wm2-K4. 6. A pot with a steel bottom 1.2 cm thick rests on a hot stove. The area of the bottom of the pot is 0.150 m2. The water inside the pot is at 100°C and 0.440 kg are evaporated every 5.0 minute. Find the temperature of the lower surface of the pot, which is in contact with the stove. Take Lv = 2.256 × 106 J/ kg and ksteel = 50.2 W/ m-K 7. A carpenter builds an outer house wall with a layer of wood 2.0 cm thick on the outside and a layer of an insulation 3.5 cm thick as the inside wall surface. The wood has K = 0.08 W/ m-K and the insulation has K = 0.01 W/ m-K. The interior surface temperature is 19° C and the exterior surface temperature is – 10° C. (a) What is the temperature at the plane where the wood meets the insulation? (b) What is the rate of heat flow per square metre through this wall?

346 — Waves and Thermodynamics 8. A closely thermally insulated vessel contains 100 g of water at 0°C. If the air from this vessel is rapidly pumped out, intensive evaporation will produce cooling and as a result of this, water freeze. How much ice will be formed by this method? If latent heat of fusion is 80 cal/g and of evaporation 560 cal/g. [ Hint If m gram ice is formed, mLf = (100 – m) Lv ] 9. In a container of negligible mass 140 g of ice initially at –15° C is added to 200 g of water that has a temperature of 40°C. If no heat is lost to the surroundings, what is the final temperature of the system and masses of water and ice in mixture? 10. A certain amount of ice is supplied heat at a constant rate for 7 minutes. For the first one minute the temperature rises uniformly with time. Then, it remains constant for the next 4 minutes and again the temperature rises at uniform rate for the last two minutes. Calculate the final temperature at the end of seven minutes. (Given, L of ice = 336 × 103 J/ kg and specific heat of water = 4200 J/ kg-K) 11. Four identical rods AB, CD, CF and DE are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures T1,T2 and T3 respectively. Assuming no loss of heat to the atmosphere. Find the temperature at B, the mid-point of CD. T3 T2 FE CB D T1 A 12. The ends of a copper rod of length 1 m and area of cross-section 1 cm2 are maintained at 0° C and 100° C. At the centre of the rod there is a source of heat of power 25 W. Calculate the temperature gradient in the two halves of the rod in steady state. Thermal conductivity of copper is 400 Wm−1K–1. 13. A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically . A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper. LEVEL 2 Single Correct Options 1. A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is 1/4 that of first, the rate at which ice melts in g/s will be (a) 0.4 (b) 0.05 (c) 0.2 (d) 0.1

Chapter 22 Calorimetry and Heat Transfer — 347 2. Two sheets of thickness d and 3d, are touching each other. The temperature just outside the thinner sheet is T1 and on the side of the thicker sheet is T3. The interface temperature is T2. T1, T2 and T3 are in arithmetic progression. The ratio of thermal conductivity of thinner sheet to thicker sheet is (a) 1 : 3 (b) 3 : 1 (c) 2 : 3 (d) 3 : 9 3. A long rod has one end at 0°C and other end at a high temperature. The coefficient of thermal conductivity varies with distance from the low temperature end as K = K 0(1 + ax), where K 0 = 102 SI unit and a = 1 m−1. At what distance from the first end the temperature will be 100°C ? The area of cross-section is 1 cm2 and rate of heat conduction is 1 W. (a) 2.7 m (b) 1.7 m (c) 3 m (d) 1.5 m 4. Two rods are of same material and having same length and area. If heat ∆Q flows through them for 12 min when they are joined side by side. If now both the rods are joined in parallel, then the same amount of heat ∆Q will flow in (a) 24 min (b) 3 min (c) 12 min (d) 6 min 5. Three rods of identical cross-sectional area and made from the same metal A (T ) form the sides of an isosceles triangle ABC right angled at B as shown in the figure. The points A and B are maintained at temperatures T and 2 T respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be (b) T 90° C (a) T 2 +1 B (√2T ) 2 −1 (c) 3T (d) T 2 +1 3( 2 − 1) 6. A kettle with 2 litre water at 27°C is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J/s, when its lid is open. In how much time will water heated to 77°C with the lid open? (specific heat of water = 4.2 kJ/ ° C-kg) (a) 8 min 20 s (b) 6 min 2 s (c) 14 min (d) 7 min 7. The temperature of the two outer surfaces of a composite slab consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x respectively are T2 and T1(T2 > T1). The rate of heat transfer through the slab in steady state is AK (T2 − T1) f. x where, f is equal to x 4x T2 K 2K T1 (a) 1 (b) 1/2 (c) 2/3 (d) 1/3

348 — Waves and Thermodynamics More than One Correct Options 1. A solid sphere and a hollow sphere of the same material and of equal radii are heated to the same temperature (a) both will emit equal amount of radiation per unit time in the beginning (b) both will absorb equal amount of radiation per second from the surroundings in the beginning (c) the initial rate of cooling will be the same for both the spheres (d) the two spheres will have equal temperatures at any instant 2. Three identical conducting rods are connected as shown in figure. Given that θa = 40° C, θc = 30° C and θd = 20° C . Choose the correct options. a b dc (a) temperature of junction b is 15°C (b) temperature of junction b is 30°C (c) heat will flow from c to b (d) heat will flow from b to d 3. Two liquids of specific heat ratio 1 : 2 are at temperatures 2 θ and θ (a) if equal amounts of them are mixed, then temperature of mixture is 1.5 θ (b) if equal amounts of them are mixed, then temperature of mixture is 4 θ 3 (c) for their equal amounts, the ratio of heat capacities is 1 : 1 (d) for their equal amounts, the ratio of their heat capacities is 1 : 2 4. Two conducting rods when connected between two points at constant but different temperatures separately, the rate of heat flow through them is q1 and q2 (a) When they are connected in series, the net rate of heat flow will be q1 + q2 (b) When they are connected in series, the net rate of heat flow is q1q2 q1 + q2 (c) When they are connected in parallel, the net rate of heat flow is q1 + q2 (d) When they are connected in parallel, the net rate of heat flow is q1q2 q1 + q2 5. Choose the correct options. (a) Good absorbers of a particular wavelength are good emitters of same wavelength. This statement was given by Kirchhoff (b) At low temperature of a body the rate of cooling is directly proportional to temperature of the body. This statement was given by the Newton (c) Emissive power of a perfectly black body is 1 (d) Absorptive power of a perfectly black body is 1

Chapter 22 Calorimetry and Heat Transfer — 349 Match the Columns 1. Match the following two columns. Column I Column II (a) Stefan’s constant (p) [ Lθ ] (b) Wien’s constant (q) [ ML2T–3 θ–2] (c) Emissive power (r) [ MT–3 ] (d) Thermal resistance (s) None of these 2. Heat is supplied to a substance in solid state at a constant rate. Its temperature varies with time as shown in figure. Match the following two columns. θ (°C) f de bc a t (s) Column I Column II (a) Slope of line ab (p) de (b) Length of line bc (q) cd (c) Solid + liquid state (r) directly proportional to mass (d) Only liquid state (s) None of these 3. Six identical conducting rods are connected as shown in figure. In steady state temperature of point a is fixed at 100°C and temperature of e at − 80° C. Match the following two columns. c Column I Column II (a) Temperature of b (p) 10°C ab de (b) Temperature of c (q) 40°C (c) Temperature of f (r) – 20°C f (d) Temperature of d (s) None of these 4. Three liquids A, B and C having same specific heats have masses m, 2m and 3m. Their temperatures are, θ, 2θ and 3θ respectively. For temperature of mixture, match the following two columns. Column I Column II (a) When A and B are mixed (b) When A and C are mixed (p) 5 θ (c) When B and C are mixed 2 (d) When A, B and C all are mixed (q) 5 θ 3 (r) 7 θ 3 (s) 13 θ 5

350 — Waves and Thermodynamics 5. Match the following two columns. Column II Column I (p) watt (q) J/kg-°C (a) Specific heat (r) J/s (b) Heat capacity (s) None of these (c) Heat current (d) Latent heat Subjective Questions 1. As a physicist, you put heat into a 500 g solid sample at the rate of 10.0 kJ/ min, while recording its temperature as a function of time. You plot your data and obtain the graph shown in figure. θ (°C) 50 40 30 20 10 – 5 1 2 3 4 t (min) (a) What is the latent heat of fusion for this solid? (b) What is the specific heat of solid state of the material? 2. A hot body placed in air is cooled according to Newton’s law of cooling, the rate of decrease of temperature being k times the temperature difference from the surroundings. Starting from t = 0, find the time in which the body will lose half the maximum temperature it can lose. 3. Three rods of copper, brass and steel are welded together to form a Y -shaped structure. The cross-sectional area of each rod is 4 cm2. The end of copper rod is maintained at 100 °C and the ends of the brass and steel rods at 80 °C and 60 °C respectively. Assume that there is no loss of heat from the surfaces of the rods. The lengths of rods are : copper 46 cm, brass 13 cm and steel 12 cm. (a) What is the temperature of the junction point? (b) What is the heat current in the copper rod? K (copper) = 0.92, K (steel) = 0.12 and K (brass) = 0.26 cal/ cm-s °C 4. Ice at 0°C is added to 200 g of water initially at 70°C in a vacuum flask. When 50 g of ice has been added and has all melted the temperature of the flask and contents is 40°C. When a further 80 g of ice has been added and has all melted the temperature of the whole becomes 10°C. Find the latent heat of fusion of ice. 5. A copper cube of mass 200 g slides down a rough inclined plane of inclination 37° at a constant speed. Assuming that the loss in mechanical energy goes into the copper block as thermal energy. Find the increase in temperature of the block as it slides down through 60 cm. Specific heat capacity of copper is equal to 420 J/ kg-K. (Take, g = 10 m/ s2) 6. A cylindrical block of length 0.4 m and area of cross-section 0.04 m2 is placed coaxially on a thin metal disc of mass 0.4 kg and of the same cross-section. The upper face of the cylinder is

Chapter 22 Calorimetry and Heat Transfer — 351 maintained at a constant temperature of 400 K and the initial temperature of the disc is 300 K. If the thermal conductivity of the material of the cylinder is 10 watt/m-K and the specific heat of the material of the disc is 600 J/ kg-K, how long will it take for the temperature of the disc to increase to 350 K? Assume for purpose of calculation the thermal conductivity of the disc to be very high and the system to be thermally insulated except for the upper face of the cylinder. 7. A metallic cylindrical vessel whose inner and outer radii are r1 and r2 is filled with ice at 0° C. The mass of the ice in the cylinder is m. Circular portions of the cylinder is sealed with completely adiabatic walls. The vessel is kept in air. Temperature of the air is 50° C. How long will it take for the ice to melt completely. Thermal conductivity of the cylinder is K and its length is l. Latent heat of fusion is L. 8. An electric heater is placed inside a room of total wall area 137 m2 to maintain the temperature inside at 20° C . The outside temperature is −10° C. The walls are made of three composite materials. The inner most layer is made of wood of thickness 2.5 cm the middle layer is of cement of thickness 1 cm and the exterior layer is of brick of thickness 2.5 cm. Find the power of electric heater assuming that there is no heat losses through the floor and ceiling . The thermal conductivities of wood, cement and brick are 0.125 W/ m°-C, 1.5 W/ m°-C and 1.0 W/ m°-C respectively. 9. A 2 m long wire of resistance 4 Ω and diameter 0.64 mm is coated with plastic insulation of thickness 0.66 mm. A current of 5 A flows through the wire. Find the temperature difference across the insulation in the steady state. Thermal conductivity of plastic is 0.16 × 10−2 cal/ s cm°-C. 10. Two chunks of metal with heat capacities C1 and C2 are interconnected by a rod of length l and cross-sectional area A and fairly low conductivity K . The whole system is thermally insulated from the environment. At a moment t = 0, the temperature difference between two chunks of metal equals (∆T )0. Assuming the heat capacity of the rod to be negligible, find the temperature difference between the chunks as a function of time . 11. A rod of length l with thermally insulated lateral surface consists of material whose heat conductivity coefficient varies with temperature as k = a / T, where a is a constant. The ends of the rod are kept at temperatures T1 and T2. Find the function T (x), where x is the distance from the end whose temperature is T1. 12. One end of a uniform brass rod 20 cm long and 10 cm2 cross-sectional area is kept at 100° C. The other end is in perfect thermal contact with another rod of identical cross-section and length 10 cm. The free end of this rod is kept in melting ice and when the steady state has been reached, it is found that 360 g of ice melts per hour. Calculate the thermal conductivity of the rod, given that the thermal conductivity of brass is 0.25 cal/ s cm° C and L = 80 cal/ g. 13. Heat flows radially outward through a spherical shell of outside radius R2 and inner radius R1. The temperature of inner surface of shell is θ1 and that of outer is θ2. At what radial distance from centre of shell the temperature is just half way between θ1 and θ2? 14. A layer of ice of thickness y is on the surface of a lake. The air is at a constant temperature – θ° C and the ice water interface is at 0°C. Show that the rate at which the thickness increases is given by dy = Kθ dt Lρy where, K is the thermal conductivity of the ice, L the latent heat of fusion and ρ is the density of the ice.



Answers Introductory Exercise 22.1 1. 7200 cal 2. 31.43 °C 3. 80°C 4. 12.96 m/s 5. 12 g 7. 52°C 6. 0°C, mw = 15 g, mi = 10 g Introductory Exercise 22.2 1. For heat flow to take place, there must be a temperature difference along the rod. 3. K W–1 4. Conduction 5. 145.5°C, 118.2°C 6. 4 W 7. 9 min Exercises LEVEL 1 Assertion and Reason 4. (c) 5. (c) 6. (d) 7. (a) 8. (d) 9. (d) 10. (a) 1. (d) 2. (a) 3. (b) Objective Questions 1. (c) 2. (b) 3. (d) 4. (b) 5. (b) 6. (d) 7. (b) 8. (a) 9.(d) 10.(c) 11. (c) Subjective Questions 1. 900 W 2. 500 m/s 3. 2.5 × 10−3°C 4. 1.2 × 104 kg/s 5. 3.7 × 104 W 6. 105°C 9. 0°C, mass of ice is 54 g and that of water is 286 g 7. (a) −8.1° C (b) 7.7 W/ m2 8. 87.5 g 10. 40°C 11. 3T1 + 2(T2 + T3 ) 12. 424°C/m, 212°C/m 13. 0.3 7 LEVEL 2 Single Correct Option 1. (c) 2. (a) 3. (b) 4. (b) 5. (c) 6. (a) 7. (d) More than One Correct Options 1. (a,b) 2.(b,d) 3.(b,d) 4. (b,c) 5. (a,d) Match the Columns (d) → s 2. (a) → s (b) → r (c) → s (d) → q (d) → r 4. (a) → q (b) → p (c) → s (d) → r 1. (a) → s (b) → p (c) → r (d) → s 3. (a) → q (b) → p (c) → p 5. (a) → q (b) → s (c) → p,r Subjective Questions 1. (a) 40 kJ/kg (b) 1.33 kJ/kg-° C 2. ln(2) / k 3. (a) 84°C (b) 1.28 cal/s 4. 90 cal/g 5. 8.6 × 10−3°C 6. 166 s 7. t = mL ln(r2 / r1) /100 πKl 8. 17647 W 9. 2.23°C = KA(C1 + C2 )  T2  x/l 2R1R 2 lC1C2  T1  R1 + R 2 10. ∆T = (∆T )0e −αt , where α 11. T = T1   12. 0.222 cal/cm-s-°C 13.

17. Wave Motion INTRODUCTORY EXERCISE 17.1 2. (a) T = 2π , λ = 2π 1. ∂2y = − ω 2a sin ωt ωk ∂t2 (b) Find ∂y and then substitute the given values of and ∂2y = 0 ∂t ∂x2 x and t. (c) and (d) Procedure is same. Since, ∂2y ≠ ∂2y (constant) INTRODUCTORY EXERCISE 17.3 ∂t2 ∂x2 Hence, the given equation does not represent a 1. (a) coefficient of x and t are of same sign. wave equation. (b) v = 314 = 10 m/s ⇒ λ = 2π , f = ω 31.4 k 2π 2. Speed of wave = Coefficient of t = c (c) vmax = ωA Coefficient of x b 2. (a) The velocity of the particle at x at time t is 3. The converse is not true means if the function can v = ∂y = (3.0 cm) (−314 s−1) cos [(3.14 cm−1) be represented in the form y = f (x ± vt), it does ∂t not necessarily express a travelling wave. As the x − (314 s−1)t ] essential condition for a travelling wave is that the vibrating particle must have finite displacement = (−9.4 ms−1)cos[(3.14cm−1)x − (314 s−1)t ] value for all x and t. 4. (a) A has the dimensions of y. The maximum velocity of a particle will be v = 9.4 ms−1 x and t are dimensionless. aT (b) The acceleration of the particle at x at time t is (b) v = coefficient of t a = ∂v = −(9.4 ms−1)(314 s−1) sin[(3.14 cm−1) coefficient of x ∂t (c) x and t are of same sign x − (314 s−1)t ] aT (d) ymax = A = − (2952 ms−1)sin[(3.14 cm−1)x − (314 s−1)t ]. INTRODUCTORY EXERCISE 17.2 The acceleration of the particle at x = 6.0 cm at time t = 0.11 s is 1. Comparing the given equation with a = −(2952 ms−2)sin[6π − 11π ] = 0 y = sin (kx − ωt) we find, 3. (a) Wave velocity = Coefficient of t (a) Amplitude A = 5 mm (b) Angular wave number k = 1 cm−1 Coefficient of x = 1/0.01 = 5 m/s (c) Wavelength, λ = 2π = 2π cm k 1/ 0.05 (d) Frequency, ν = ω = 60 Hz Since, coefficient of t and coefficient of x are 2π 2π of same sign. Hence, wave is travelling in = 30 Hz negative x-direction. π or v = − 5 m/s (e) Time period, T = 1 = π s ν 30 (b) Velocity of particle, (f) Wave velocity, v = νλ = 60 cm s−1 vP = ∂y = 2 cos  x + 0.t01 ∂t 0.05 Substituting x = 0.2 and t = 0.3, we have vP = 2cos(34) = 2(−0.85) = − 1.7 m/s

Chapter 17 Wave Motion — 355 4. (a) ∆φ = 2π ∆x 6. (a) v = T = mg …(i) λ µµ ∴ ∆x = λ∆φ = v (∆φ) = 1.5 × 9.8 = 16.3 m/ s (b) 2π 2πf 0.055 = (350) (π /3) (b) λ = v = 16.3 = 0.136 m (2π ) (500) f 120 = 0.116 m (c) From Eq. (i), we can see that v ∝ m. If m is ∆φ = (ω)∆t doubled, then v will become 2 times. Hence, = (2πf ) (∆t) from the relation λ = v ⋅ λ will also become = (2π ) (500)(10− 3) f = π or 180° 2 times. As f remains unchanged. INTRODUCTORY EXERCISE 17.4 INTRODUCTORY EXERCISE 17.6 1. fmax = c 1. I = P = 1.0 = 1 W/m2 λ min 4 πr2 4π (1)2 4π fmin = c 2. Suppose power of line source is P. Then, λ max at distance r, surface area is 2πrl. 2. λ = v for both ∴ I=P= P f S 2πrl INTRODUCTORY EXERCISE 17.5 or I ∝ 1 r 1. The tension in the string is F = mg = 10 N. The Further, I ∝ A2 mass per unit length is ∴ A∝ 1 µ = 1.0 g cm−1 = 0.1 kg m−1. The wave velocity is, r therefore, v = Fµ = 10 N = 10 ms−1. The 3. Energy density, u = 1ρω2A2 0.1 kg m−1 2 time taken by the pulse in travelling through 50 cm Total energy = (u) (V ) is, therefore 0.05 s. = 1 (ρV ) (2πf )2 A2 2 2. v = T = 2π 2 mf 2A2 µ = (2)(π )2(0.08)(120)2(0.16 × 10−3)2 3. TCD = 3.2 g,TAB = 6.4 g = 581 × 10− 3J and v = T µ = 0.58 mJ 4. v = T , µ = m and t = d 4. Power = 1ρω2A2Sv µl v 2 5. v= T= T But, ρS = µ ρS ρ (πd2/ A) and v = T µ = 4T (T = mg) ∴ ρ Sv = µT ρπd 2 Hence, Power = 1 (2πf )2 A2 Tµ 4 × 20 2 (8920) (π ) (2.4 × 10− 3)2 = 2π 2 f 2 A2 Tµ = = (2) (π )2 (60)2 (0.06)2 80× 5 × 10− 2 = 22 m/s = 512W

356 — Waves and Thermodynamics 5. (a) As derived in above problem, power 6. P = 1 ρω2A2Sv …(i) = 2π 2 f 2A2 Tµ 2 But, ρS = µ = (2) (π )2 (200)2 (10− 3)2 60 × (0.006) and v = T = T = 0.47 W µ ρS (b) Energy density, u = 1ρω2A2 ∴ ρS = T or ρSv = T 2 v2 v Total energy = (u) (V ) =  1 ρω 2 A 2 (lS ) Substituting in Eq. (i), we have 2 P = 1 (2πf )2 A2 T 2v But, ρS = µ = 2π 2 f 2A2T ∴ Total energy = 1 µl (2πf )2 A2 v = (2) (π )2 (100) 2 (0.0005)2 (100) 2 100 = 2π 2 f 2A2µl = 0.049 J = 49 mJ = (2) (π )2 (200)2 (10− 3)2 (0.006) (2.0) = 9.4 × 10− 3J = 9.4 mJ Exercise LEVEL 1 Objective Questions 1. λ = 2π = 2π = 18 m Assertion and Reason k π/9 3. If both waves ωt and kx are of opposite signs, then 2. ∆φ = 2π ∆x = k∆x both are travelling in positive directions. λ 4. If f is doubled (which is source dependant), then = (10π × 0.01) (10) = π λ will automatically become half, so that speed remains same. Because speed is only medium 3. At t = 0, ymax = 1 at x =0 dependant. 2 5. Longitudinal or sound wave cannot travel in At t = 2 s, ymax = 1 at x = 2 m 2 vacuum. 7. Electromagnetic wave which can travel with or ∴ In 2 s, ymax has travelled 2 m in positive direction without medium. ∴ v = + 2 = + 1m/s 8. Just by observation we cannot say that µ2 > µ1 and 3 hence v2 > v1. 4. Wave velocity and particle velocity are two If their densities are different, then µ2 may be less different things. than µ1 also.  v 9. At mean position kinetic energy is maximum and 5. ∆φ = 2π (∆x) = 2π φ(∆x)  as λ = f  there is a maximum stretch in string. Because one λv side particles are moving up and the other side = 2π (25) (16 − 10) = π 300 particles are moving down. Hence, potential energy is also maximum. 6. v = ω = T 10. In vP = − v ∂y , kµ ∂x ∴ T = µ  ωk  2 = (1.3 × 10−4)  310 2 vP is negative and ∂y (the slope) is also negative. ∂x Hence, the velocity v should also be negative. = 0.12 N

Chapter 17 Wave Motion — 357 7. Q ω = 2π = 2π = 2πN (vP )max = ωA From these two expressions, we can see that T (t/N ) t = (2π ) (150) (vP )max = (kA) V 60 6. Mass per unit length, µ = m = (5π ) rad /s l Subjective Questions v = T = Tl = 500 × 2 1. (b) λ = 2π = 2π µm 0.06 k 2π /28 = 129.1 m/s = 28 cm 7. v = T as mass per unit length = ρS (c) f = ω = 2π /0.036 ρS 2π 2π = 27.8 Hz ∴ v= 0.98 (d) v = f λ = 778.4 cm/s 9.8 × 10 3 × 10− 6 = 7.8 m/s = 10 m/s Since, ωt and kx have opposite signs. Hence, wave is travelling in positive direction. 8. Since, wave is travelling along positive 2. (a) Put t = 0, x = 2cm x-direction. Hence, coefficient of t and coefficient of x should have opposite signs. Further, (b) λ = 2π = 2π = 16 cm k (30π /240) v = Coefficient of t Coefficient of x (c) Wave velocity = Coefficient of t Coefficient of x ∴ 2 = Coefficient of t Coefficient of x ∴ Coefficient of t = 2 (coefficient of x) =1 = 2 × 1= 2 SI units 1/ 240 ∴ y = (x − 10 2 = 240 cm/s 2t)2 + (d) f = ω = 30π = 15 Hz 9. (a) λ = 2 cm 2π 2π 2 3. At t = 0, y is maximum at x = 0. ⇒ λ = 4 cm At t = 2s, y is maximum at x = 1m. Hence, in 2 s f = v = 40 = 10 Hz wave has travelled 2 m in positive x-direction λ4 ∴ v = + 1 m/s (b) ∆φ = 2π ∆x =  2π  (2.5) = 5π 2 λ 4 4 = + 0.5 m/s (c) ωt = θ 4. Since, coefficient of t and x are opposite signs, or (2π /t) = θ then wave is travelling along negative x-direction. ∴ t = θ = π /3 Further, speed of wave = Coefficient of t 2πf (2π ) (10) Coefficient of x =1s = 2 = 2 m/s 60 1 (d) At P, particle is at mean position. So, Amplitude = maximum value of y v = maximum velocity = ωA 5. Wave speed, = 10 = 2 m = 2πfA 5 = (2π )(10)(2) = (40π ) cm/s v=ω = 125.7 cm/s = 1.26 m/s k and maximum particle speed,

358 — Waves and Thermodynamics Further, vP = − (v)  ∂∂xy …(i) = 2π (60π ) (8) Sign of v1, the wave velocity is given positive. Sign of ∂y, slope of y - x graph is also positive. = 4.2 × 10− 3 s ∂x = 4.2 ms Hence, from Eq. (i) particle velocity is negative. 12. (a) Q T = 2π = 2π = 0.02 s ω π /0.01 ∴ vP = − 1.26 m/s = 20 ms 10. (a) Wave velocity = Coefficient of t λ = 2π = 2π Coefficient of x k π/2 = (1/0.01) = 5 m/s = 4.0 cm (1/ 0.05) ∂y Since, coefficient of t and coefficient of x are (b) vP = ∂x of same sign. Hence, wave is travelling in  − π msm  x t s 0.01 2.0 cm 0.01 negative x-direction. = cos π − or v = − 5 m/s (b) Particle velocity, vP = ∂y =  − π ms  cos π  x − t ∂t 10  2.0 cm 0.01 s = 2 cos  x + 0.t01 Put x = 1 cm and t = 0.01 s in above equation 0.05 We get, Substituting x = 0.2 and vP = 0 (c) and (d) Putting the given values in the t = 0.3, we have above equation, we get the answers. vP = 2 cos 34 13. (a) k = 2π = 2π = 0.157 rad /cm = (2) (− 0.85) λ 40 = − 1.7 m/s T = 1 = 1 = 0.125 s f8 11. (a) ω = 2πf = 2π  v  = 2π  01.24 λ ω = 2π = 2π = 50.3 rad /s T 0.215 = (60 π ) rad /s v = ω = 320 cm/s k = 2π = 2π = (5π ) m−1 k λ 0.4 (b) Amplitude is 15 cm. Since, wave is travelling along + ve At t = 0, x = 0, y = + A. Hence, equation x-direction, ωt and kx should have opposite should be a cos equation. Further, wave is sign. Further at t = 0, x = 0 the string has zero travelling in positive x-direction. Hence, ω t displacement and moving upward (in positive and kx should have opposite signs. direction). Hence at x = 0, we should have A sin ωt not − A sin ωt. Therefore, the correct 14. (a) Tx = (mL − x ) g = µ (L − x) g expression is where, µ = mass per unit length y = A sin (ωt − kx) or y = 0.05 sin (60πt − 5πx) (b) Putting x = 0.2 m and ∴ vx = Tx = g (L − x) µ t = 0.15 s in above equation we have, y = − 0.035 4 m (b) vx = − dx = g (L − x) = − 3.54 cm ∴ dt g (L − x) dx (c) In part (b), y = A/ 2 t x=0 From y = A to y = 0, time taken is 2 ∫ ∫dt = − t = T = 2π 0 x=L 8 ω×8 Solving we get, t=2 L g

Chapter 17 Wave Motion — 359 15. (a) T cos dθ components are cancelled. T sin dθ 2. ω = 2πf = (2π ) (100) components provide the necessary centripetal Q force to Pθ. = (200 π ) rad /s dθ dθ v =ω T P dθ dθ Q k T ∴ k =ω =ω µ vT ∴ 2T sin dθ = mPQ Rω2 = (200π ) 3.5 × 10− 3 35 = 2π m− 1 At zero displacement, For small angles sin dθ ≈ dθ vP = ωA = − v ∂y ∴ 2T dθ = [µ (2R) dθ ] Rω2 ∂x Solving the equation, we get v  ∂∂xy  T  (slope)  µ  ∴ |A| = =   T = v = Rω µ ωω 16. ∫M = L dx = kL2  3.5 35 3  (π /20)  × 10−  ∴ kx =  02 200π 2M k = L2 = 0.025 m 3. ω = 2π = 2π Q vx = T= T= T T 0.25 µx kx (2M /L2) x = 8π rad /s  T  x− 1/ 2 = dx v=ω = L  k  2M  dt ω 8π  530 m−1 v 0.48 ∴ k = = = π t = 1 2M L 1/ 2 dt x ∫ ∫∴ dx y = A sin (ωt − kx) 0 L T0 or t =  2 2M  L3/ 2 = A sin  8πt − 50 πx   3  3L T  Put y = 3 cm, t = 1 s, x = 0.47 m = 2 2ML showing we get A = 6 cm 3T T= T 4. v = ρS ρ (πd2/4) LEVEL 2 Single Correct Option ∴ v∝ T d2 1. v=ω 5. E ∝ ω2A2 E ∝ f 2 A2 Q or k ∴ k = ω = 600π E is same v 300 ∴ = 2π m− 1 or fA = constant A∝1 ∴ y = 0.04 sin (600 πt − 2πx) f Now, put t = 0.01 s and x = 0.75 m

360 — Waves and Thermodynamics 6. The wave pulse is travelling along positive x-axis. (c) f = ω = b 2π 2π Hence, at and bx should have opposite signs. Further, wave speed (d) λ = 2π = 2π kc v = Coefficient of t Coefficient of x 2. vP = ∂y = (4π ) cm / s cos [πt + 2πx ] ∂t ∴ 4 = Coefficient of t 1 aP = ∂2y = (− 4π2) cm / s2 sin [πt + 2πx ] ∂t2 ∴ Coefficient of t = 4 s− 1 More than One Correct Options Now substitute the values of t and x. v = Coefficient of t 3. For velocity, vP = − v ∂y Coefficient of x ∂x 1. Q = 6.2 = 3.1 m/s ∂y = Slope of y- x graph. 2 ∂x A = 2 = 0.1 m Sign of v is not given in the question. 20 Hence, direction of vP cannot be determined. For particle acceleration, 3. y = A sin (πx + πt) aP ∝ − y vP = ∂y = πA cos (πx + πt) i.e. aP and y are away in opposite directions ∂t If y = 0, then aP = 0 aP = ∂2y = − π2A sin (πx + πt) 4. Energy density = 1ρω2A2 ∂t2 2 Now, substitute t = 0 and given value of x. = Energy per unit volume Since, ωt and kx are of same sign, hence the wave Power = energy transfer per unit is travelling in negative x-direction. Time = 1 ρω2A2Sv 4. Speed of wave 2 = Coefficient of t = b = b Intensity = energy transfer per unit per unit area Coefficient of x 1 = 1ρω2A2v = P 2S λ = 2π = 2π = a k (2π /a) Wave number = 1 λ 5. Speed of wave = Coefficient of t 5. vP = ∂y, aP = ∂2y Coefficient of x ∂t ∂t2 = 1/b = a If ωt and kx are of same sign, then wave travel in 1/a b negative x -direction. λ = 2π = 2π = a If they are of opposite signs, then wave travels in k 2π /a positive direction. T = 2π = 2π = b Subjective Questions ω 2π /b 1. (a) vP = − v  ddxy 6. y - t graph is sine graph. Therefore, v - t graph is cos graph and a - t is − sine graph as As vP and (slope)P are both positive, v must be negative. Hence, the wave is moving in vP = ∂y and aP = ∂2y ∂t ∂t2 negative x-axis. Match the Columns (b) y = A sin (ωt − kx + φ) …(i) 1. (a) Wave speed = Coefficient of t = b k = 2π = π cm−1 Coefficient of x c λ2 (b) Maximum particle speed = ωA = (b) (a) A = 4 × 10−3 m = 0.4 cm

Chapter 17 Wave Motion — 361 At t = 0, x = 0, slope dy = + ve λ= v = 0.1 = 2π m dx f 100  10  ∴ vP = − v(slope) = + ve 2π Further at t = 0, x = 0, y = + ve ∴ φ=π y= A cos  2π  (vt − x) λ 4 = 0.02 cos 100(0.1t − x) Ans. Further, 20 3 = − v tan 60° = 0.02 cos (10t − 100x) m ∴ v = − 20 cm/s The distance between two successive maxima f = v = 5 Hz = λ = 2π = 0.0628 m Ans. λ 100 ∴ ω = 2πf = 10π 4. (a) Dimensions of A and Y are same. ∴ y = (0.4 cm) sin 10πt + π x + π  Ans. Similarly, dimensions of a and x are same. 2 4 (b) As the wave is travelling towards positive (c) P = 2π 2A2 f 2µv x-axis, there should be negative sign between term of x and term of t. ∴ Energy carried per cycle E = PT = P = 2π 2A2 f µv Further, speed of wave f v = Coefficient of t Coefficient of x Substituting the values, we have Ans. E = 1.6 × 10−5 J ∴ Coefficient of t = (v) × coefficient of x 2. (a) v = T 5. From the given figure, we can see that ρS (a) Amplitude, A = 1.0 mm (b) Wavelength, λ = 4 cm = 64 (c) Wave number, k = 2π = 1.57 cm−1 ≈ 1.6 cm−1 12.5 × 103 × 0.8 × 10−6 λ = 80 m/s Ans. (d) Frequency, f = v = 20 = 5 Hz (b) ω = 2πf = 2π (20) = 40π rad/s λ4 k = ω = 40π = π m−1 6. v = T or v ∝ 1 v 80 2 ρS ρ ∴ y = (1.0 cm) cos (40πs−1)t −  π m−1 x ∴ v1 = ρ2 2 v2 ρ1 Ans. ρ1  vv21 2  21 2 1 ρ2  4 (c) Substituting x = 0.5 m and t = 0.05 s, we get ∴ = = = y = 1 cm Ans. T1 2 µ1 7. v1 = (d) Particle velocity at time t. vP = ∂y = 4.8 = 20 m/s ∂t 1.2 × 10−2 = − (40π cm/s) sin (40π s−1 )t −  π m−1 x  T2 = 7.5 2  µ2 1.2 × 10−2 v2 = Substituting x = 0.5 m and t = 0.05 s, we get vP = 89 cm/s Ans. = 25 m/s 3. keff = 2k = 1.0 N/m Pulses will meet when xA = xB or 20t = 25(t − 0.02) f = 1 keff = 10 s−1 2π m 2π ∴ t = 0.1 s v = 0.1 m/s, A = 0.02 m and xA or xB = 20 × 0.1 = 2 m

362 — Waves and Thermodynamics 8. (a) P = 1ρω2A2Sv T mω2 x 2 dT x dx ∫ ∫or− = 0 L x=L or A = 1 2P mω2  x 2 L2  ω ρSv   …(i) ∴ −T = − L  2 2 Here, ρS = µ = mass per unit length or T = mω2 (L2 − x2) = 6 × 10−3 kg/m 2L 8 v= T ω = 2πf = 2πv µ λ mω2 (L2 − x2) = 2π × 30 = 2L 0.2 m/L Substituting these values in Eq. (i), we have A = 0.2 2 × 50 × 8 = ω L2 − x2 2π × 30 6 × 10−3 × 30 2 = 0.0707 m or dx = ω L2 − x2 dt 2 = 7.07 cm (b) P ∝ vω2 t 2 L dx dt = or P ∝ v(v2) ∫ ∫∴ 0 ω 0 L2 − x2 or P ∝ v3 2 sin −1  Lx  L ω 0 When wave speed is doubled, then power will ∴ t= become eight times. 9. − dT = (dm)xω2 =  m dx xω 2 = 2⋅π = π Ans. L ω 2 2ω

18. Superposition of Waves INTRODUCTORY EXERCISE 18.1 INTRODUCTORY EXERCISE 18.3 1. I max  I1 /I2 + 1 2 1. Wall will be a node (displacement). Therefore, = Imin  I1 /I2 − 1 shortest distance from the wall at which air particles have maximum amplitude of vibration 2. (a) Amax = A1 + A2 (displacement antinode) should be λ /4. Amin A1 − A2 Here, λ = v = 330 = 0.5 m f 660 I max  Amax  2 I min  Amin  (b) = ∴ Desired distance is 0.5 = 0.125 m. 4 3. Path difference of λ is equivalent to a phase 2. 5 f − 2 f = 54 4 difference of π . ∴ 3 f = 54 f = 18 Hz 2 3. f∝ T √2 A0 3A0 ∴ f1 = T1 f2 T2 or 220 = 2.2g 260 (2.2 + M ) g 45° √2 A0 Solving we get, Anet = 5A0 M = 0.873 kg I ∝ A2 4. (a) f1 = 250 = 5 ⇒ Inet = 25 I0 f2 300 6 So, f1 is 5th harmonic and f2 is 6th harmonic. INTRODUCTORY EXERCISE 18.2 (b) f1 = (5) = v = 2.5 T /µ 2l l 1. It is either 0 or π. µ f12 l2 (0.036) (250)2 (1)2 v = ω = 40π ∴ T = (2.5)2 = (2.5)2 2. (a) k π/3 = 120 cm/s = 360 N (b) Distance between adjacent nodes 5. 6 T /µ  = 2 T /µ  =λ=π= π  2l1   2l2  2 k (π /3) ∴ l1 = 3 = 3 cm l2 (c) vP = ∂y = (− 200 π) sin πx sin 40πt INTRODUCTORY EXERCISE 18.4 ∂t 3 Now, substitute the given values of x and t. 1. At mean position, energy is in the form of kinetic 3. In this case, node points will also oscillate, but energy. standing waves are formed. 2. (a) Assuming tension to be same 4. λ = 2π = 2π = 15.7 m v∝ 1 µ k 0.4 f = ω = 200 = 31.8 Hz µ RHS = µ RHS 2π 2π 4 v = f λ = 500 m/s ∴ vRHS = 2vLHS = 20 cm/s

364 — Waves and Thermodynamics (b) Ar =  v2 − vv11 Ai Ar =  v2 − vv11 Ai = 2 × 10−3m  v2 +  v2 + 3 ∴ Ar = 20 − 10 = 1 At =  2v2  Ai = 8 × 10−3m Ai 20 + 10 3  v1 + v2  3 At =  2v2  = 2 × 20 = 4 In second medium, speed becomes two times. Ai  v1 + v2  20 + 10 5 Therefore, λ also becomes two times. So, k 3. Speed of wave in first medium is, remains one-half, value of ω will remain Coefficient of t unchanged. Further, second medium is rarer Coefficient of x v1 = medium (v2 > v1). Hence, there is no change in phase angle anywhere. = 50 = 25 m/s 2 ∴ yr = 2 × 10−3 cos π (2.0x + 50t) 3 v2 = 50 m/s Ai = 2 × 10−3m and yt = 8 × 10−3 cos π (x − 50t) 3 Exercises LEVEL 1 9. A A Assertion and Reason 1. At x = 0, y = y1 + y2 = A sin ωt + A cos ωt. So, it AA is neither a node nor an antinode. Resultant amplitude is A. So, intensity will remain same. 2. They are called stationary because net energy 10. Phase difference may be different but it should transfers from any section is zero, if amplitudes of remain constant with time. constituent waves are zero. 3. At =  2v2  Ai Objective Questions  v1 + v2  2. 6 mm 6 mm If v1 > v2 or 2 is rarer, then At > Ai A 4. ⇒ 4 mm 12 mm 8 mm NANANAN A = (8)2 + (6)2 = 10 mm 1122 33 4 3. (i) Two waves must travel in opposite directions. l = 3λ (ii) At x = 0, y = y1 + y2 should be zero at all 2 times. 5. f = η  2vl 6. f1: f2 : f3 = 1: 2 : 3 As η increases, frequency increases. Hence, λ λ = v or λ ∝ 1 decreases. ff 6. Amplitudes may be different also. 7. ∴ λ1 : λ2: λ3 = 1: 1 : 1 2 3 X=0 8. All frequencies are integral multiples of 35 Hz. 9. 3  2vl = 300 λ λ 8 4 Energy lying between λ and λ will be more. ∴ v = 200 l = (200) (1) 84 = 200 m/s

Chapter 18 Superposition of Waves — 365 10. These are multiples of 30 Hz. Hence, fundamental (b) Since, reflected waves come in the same frequency medium, we can say that P ∝ A2 f0 = 30 Hz Now, f0 = v Pr  Ar  2  − A/ 3 2 1 2l Pi  Ai  A 9 = = = ∴ v = 2 f0 l ∴ Fraction of power transmitted = 2 × 30× 0.8 =1− 1= 8 = 48 m/s 99 11. ∆φ =  2π  (∆x) 3. A = (10)2 + (20)2 + 2(10) (20) cos 60° λ Q  2π   2π  = 26.46 cm  v/ f  vT = (∆x ) = (∆x) 20 A =  300 2×π0.04  (16 − 10) 60°  φ 10 =π tan φ = 20 sin 60° 10 + 20 cos 60° 12. v = ω = T = 0.866 k ρS ∴ φ = 40.89°  ω  2 (8000 × 10−6)  310 2 = 0.714 rad k ∴ T = ρS = 4. Find y1 and y2 at given values of x and t and then = 7.2 N simply add them to get net value of y. Subjective Questions 5. v= T= 16 = 20 m/s µ 10−3 )/ (10−2 ) 1. AR = 4 2 cm (0.4 × = 5.66 cm AR (a) t = d = 2l = 0.4 m v v 20 4 = 0.02 s 4 (b) At half the time pulse is at other end and it gets 2. v1 a phase difference of π. Hence, the shape is as 2 shown below. v2 = (a) Ar =  v2 − vv12 A 6. (a) For fixed end.  v1 + 1 cm/s =  (v1 /2) − v1  A O  v1 + v1 /2  2 cm 2 cm =− A 3 1 cm/s At =  2v2  A By the superposition of these two pulses we  v1 + v2  will get the resultant. But only to the left of point O, where string is actually present. =  2 × v1/2  A  v1 + v1 /2 (b) For free end = 2A O 3 2 cm 2 cm s

366 — Waves and Thermodynamics 7. y = y1 + y2 = (6.0 cm) sin (πx) cos (0.6πt) 12. Fundamental frequency = (490− 420) Hz ∴ Ax = 6.0 cm sin (πx) …(i) f0 = 70 Hz In parts (a), (b) and (c) substitute the given But, f0 = v = T /µ 2l 2l values of x and find the displacement amplitude at these locations. ∴ l = T /µ = (450)/ 0.005 2 f0 2× 70 (d) At antinodes = 2.142 m Ax = maximum = ± 6.0 cm ∴ πx = π , 3π , 5π 13. λ = v = 0.5 m 22 2 f or x = 0.5 cm, 1.5 cm, 2.5 cm 1 234 8. (a) Distance between successive antinodes l = 4  λ2 = 1 m = λ = π = π = 2cm 2 k (π /2) (b) Amax = 2A = (2π ) cm 14. (a) f4 = 4 f0 If x = 0 is taken as node, we shall taken sin ∴ f0 = f4 = 400 = 100 Hz equation for Ax 4 4 ∴ Ax = Amax = sin kx (b) f7 = 7 f0 = 700 Hz = (2π cm) sin kx Put k = π and x = 0.5 15. Fundamental frequency ∝ 1 2 l 9. v= T= 20 = 47.14 m/s 1 1 1 µ 9 × 10− 3 l1 : l3 f1 : f2 : f3 = : l2 v 47.14 f1 = 2l = 2 × 30 = 1: 2: 3 = 0.786 Hz ∴ l1 : l2 : l3 = 1 : 1 : 1 = 6 : 3 :2 1 2 3 Next three frequencies are 2 f1, 3 f1 and 4 f1 l1 =  161 (1) m = 6 m 10. (a) f = v = T /µ 11 2l 2l l2 =  131 (1) m = 3m ∴ T = 4l2 f 2 µ 11 = (4 ) (0.7)2 (220)2  1.2 × 10− 3  121 2   11  0.7  l3 = (1) m = m = 163 N 16. f ∝ 1 (b) f3 = 3 f1 = 3 × 220 = 660 Hz l 11. Fundamental frequency, f1 = l2  f1  f = v = T /µ ∴ f2 l1 or l2 =  f2  l1 2l 2l =  112846 (90 cm) = 60 cm = (50)/(0.1 × 10− 3 /10− 2) 2 × 0.6 17. (a) λ = 15 cm = 58.93 Hz 2 Let, n th harmonic is the highest frequency, then ∴ λ = 30 cm (58.93) n = 20000 k = 2π =  1π5 cm− 1 ∴ n = 339.38 λ Hence, 339 is the highest frequency. ω = 2π = 2π s− 1 T 0.075 ∴ fmax = (339) (58.93) Hz = 19977 Hz

Chapter 18 Superposition of Waves — 367 Since, x = 0 is a node we will write sin T = 1 = 0.126 s equation. f ∴ Ax = Amax sin kx …(i) v = fλ = 1470 cm/s and y = Ax sin ωt (e) vmax = ω Amax (b) v = ω = (2π /0.075) = (50) (5.60) k (π /15) = 280 cm/s (f) Frequency and hence ω will become 8 times. = 400 cm/s = 4 m/ s (c) x = 0 3 ∴ ω′ = 50 × 8 = 133 rad /s x=0 3 P k = ω or k ∝ ω x v Hence, k will become 8 times. 15 cm 3 x = 7.5 − 3 = 4.5 cm k′ =  83 (0.034) = 0.0907 rad /cm From Eq. (i), 20. (a) λ = l or λ = 2l = 1.6 m Ax = (0.85 cm) sin  π × 4.5 2 15 v = f λ = 60 × 1.6 = 96 m/s = 0.688 cm (b) v = T µ 18. (a) f1 = v = 48 2l 3 = 16 Hz ∴ T = µv2 =  0.04  (96)2 0.8 λ1 = v = 3m f1 = 461 N (c) vmax = ω Amax (b) Second overtone means f3. = (2πf ) Amax f3 = 3 f1 and hence λ 3 = λ1 = (2π ) (60) (0.003) 3 = 1.13 m/s amax = ω2 Amax (c) Forth harmonic means f4. = (2πf )2 Amax = (2π × 60)2 (0.003) f4 =4 f1 and λ 4 = λ1 4 19. (a) = 426.4 m/s2 Third harmonic 21. (a) y = y1 + y2 (b) A = 5.6 = 2.8 cm = 0.2 sin (x – 3.0t) + 0.2 sin  x – 3.0t + π  2 2 (c) k = 0.034 cm−1 = A sin (x – 3.0t + θ) λ = 2π = 184.7 cm A2 k A 0.2 l = 3λ = 277 cm 2 θ 0.2 A1 (d) λ = 184.7 cm f = ω = 50 = 7.96 Hz 2π 2π

368 — Waves and Thermodynamics Here, A = (0.2)2 + (0.2)2 24. (a) nλ 1 = l and θ = π 2 4 ∴ n (0.54) = l  x π  2 4 ∴ y= 0.28 sin – 3.0t + Ans. or n= l …(i) Similarly, 0.27 …(ii) (b) Since, the amplitude of the resulting wave is 0.32 m and A = 0.2 m, we have (n+ 1)λ 2 = l 2 0.32 = (0.2)2 + (0.2)2 + (2) (0.2) (0.2) cos φ or (n + 1) (0.48) = l Solving this, we get 2 φ = ± 1.29 rad Ans. or (n + 1) = l 0.24 v T /µ 22. (i) fa = 2l = 2l From Eqs. (i) and (ii) we have, (100/4 × 10− 2) = n =8 2×2 n+1 9 (b) From Eq. (i), = 12.5 Hz l = (0.27) n fb = 3  2vl = 3 fa = 37.5 Hz = 0.27 × 8 (ii) ω = 2πf = 2.16 m (c) λ = l (Fundamental) ∴ ωa = 2πfa = 25π ωb = 2πfb = 75π 2 ∴ λ = 2l = 4.32 m v= T= 100 25. From fixed end there will be a phase change of π. µ 4 × 10− 2 Further, wave will start travelling in opposite = 50 m/s direction. Hence, ωt and kx both will now become position. From free end there is no change in k =ω v phase. 26. t1 = d1 = 4 + 4 = 8 s v 1 ∴ ka = ωa = 25π = π v 50 2 After one reflection from fixed boundary wave is inverted. kb = ωb = 75π = 3π v 50 2 d2 4 + 10 + 16 t2 = v = 1 = 20 s 23. (a) l = λ 1/4 After two times reflection from fixed boundaries ∴ λ 1 = 4l = 16 m , l = 3λ 2/4 wave pulse will again become upright. ∴ λ2 = 4l = 5.33 m 27. Let us plot at t = 3 s 3 l = 5λ 3/4 Y (cm) 4l 3 5 ∴ λ3 = = 3.2 m (b) v = T 1 X (cm) µ Net 456789 Y (cm) y = y1 + y2 4 = 400 = 100 m/s (0.16/ 4 ) Now, f1 = v = 100 = 6.25 Hz 1 X (cm) λ1 16 456789 Similarly, f2 and f3. Similarly, we can draw at other times.

Chapter 18 Superposition of Waves — 369 LEVEL 2 f2 = 2 f1 = 100 Hz Single Correct option f3 = 3 f1 = 150 Hz 1. f∝ T n1 = v = 25 Hz 4l Q ∴ f1 = T1 n2 = 3n1 = 75 Hz f2 T2 n3 = 5n1 = 125 Hz ∴ 3 = T + 2.5 7. f ∝1 2T Q l ∴ Solving, we get T = 2 N l1 : l2 : l3 = 1 1 1 ∴ : : f3 2. f = v = T /µ = T /ρS f1 f2 2l 2l l = 1 : 1 : 1 = 12 : 4 : 3 13 4 = T / ρ (πd2/4) l l1 =  12 +  (114 )  12 + 4 3 or f ∝ T = 72 cm dl 4  ∴ f1 =  T1   dd21  l2  l2 =  12 + 4 + 3 (114 )     l1  f2  T2  = 24 cm = 1  31  21 = 3 2 l3 = 3  (114 ) 2  12 + 4 3 + 3. f ∝ 1 ⇒ l = k = 18 cm lf 8. f ∝ T Now, l = l1 + l2+ l3 ∴ k =k +k +k f = T1 = W f0 f1 f2 f3 f /2 T2 W − V1 ∴ 1=1+ 1+ 1 2 = vρg = ρ f0 f1 f2 f3 Vρg − Vρwg ρ − ρw 4. y1 + y2 = 2A sin (ωt − kx)= y4 (say) Solving ρ = 4 ρw 3 Now, y4 and y3 produce standing waves where, Amax = 2 (Amplitude of constituent wave) f= W = Vρg = 2 (2A) = 4 A f /3 W − V2 Vρg − V ρl g 5. Tension in the string will be given by 9= ρ ρ − ρl T = YA∆l = YA  as ∆l = 1 lη  l η 8 32 Now, f ∝v ∴ ρl = 9 ρ = 27 ρω f1 = v1 = T /ρA ∴ Relative density of liquid = ρl f2 v2 Y /ρ ρw = T=1 = 32 = 1.18 YA η 27 6. f1 = v = T /µ 9. λ = 1.5 m 2l 2l 2 ∴ λ = 3.0 m = 100/0.01 = 50 Hz = k = 2π =  2π  m− 1 2 λ 3

370 — Waves and Thermodynamics Let us take antinode at x = 0, then 13. f5 = 5 f1 = 5 f2 2 f1 2 4mm 2mm ∴ f2 = 2 f5 5 –X X = 0 X = 2 × 480 5 Ax = Amax cos kx = 192 Hz  2π  14. I ∝ A2 3 ∴ (2 mm) = (4 mm) cos x Ir = 0.64 Ii ∴ Ar = 0.8 Ai ∴  2π  x = π 3 3 = 0.8A or x = 0.5 m Reflected from a denser medium. Hence, a phase The asked distance is 2x or 1.0 m. change of π will occur. Reflected wave will travel 10. f = 3Hz in opposite direction. Hence, ωt and kx will have positive signs. ∴ ω = 2πf = (6π ) rad /s 15. η  2vl B =  v  k = ω = 6π = (2π ) rad /m 2l v3 A y = A sin (kx − ωt)  T /ρS  =  v  = A sin (2πx − 6πt) η= 2l  2l B A  Putting y=± A or  4T /ρπ   4T /ρπ  …(i) η  =   2ld  B  2ld  A and x = 3 , we get (2π ) (3) − 6π t = π Given, TB = 2TA, lB = 2lA 2 dB = 2dA ∴ 6πt = 11 π and ρB = 2 ρA 2 Putting in Eq. (i), we get ∴ t = 11 s η=4 12 η = 4 means 4th harmonic or 3rd overtone. 11. ∆φ1 = π /2 ∆φ2 =  2π  (∆x) More than One Correct Options λ 1. f ∝ T =  2π  (1.5λ ) f2 = T2 λ f1 T1 ∴ f1 + 15 = 1.21T1 = 3π f1 T1 ∴ ∆φ net = ∆φ2 − ∆φ1 = 5π = 1.1 2 Solving we get f1 = 150 Hz 12. Fundamental frequency, v∝ T f0 = 450 − 400 ∴ v2 = T2 = 50 Hz v1 T1 f0 = v = T /µ 2l 2l ∴ l = T /µ = 490/0.1 =  1.21 T1  2 f0 2 × 50 or v2   v1  T1  = 0.7 m = 1.1 v1

Chapter 18 Superposition of Waves — 371 Hence, increase in v is 10% 3. Ax = − 1.6 sin ax λ =l 2 ∴ x = 0 is a node ∴ λ = 2l Second antinode is at a distance. ∴ Fundamental wavelength = 2λ is unchanged. x = λ + λ = 3λ = 3  2π  4 2 4 4 k   4. At =  2v2  Ai But, k = a v1 + v2 ∴ x = 3π If v2 > v1, At > Ai 2a 5. Match the Columns l = 4 λ = 2λ = 2  2π  1. v ∝ 1 2 k µ µ2 = 9µ1 v1 = 4π Hence, v2 = 3 k 6. Two identical waves should travel in opposite So let v1 = 3 units, then v2 = 1unit directions. (a) A1 =  v2 − v1  Ai  v1 + v2  7. y = y1 + y2 = (2A cos kx) sin ωt and A2 =  2v2  Ai = Ax sin ωt  v1 + v2  Here, Ax = 2A cos kx At x = 0, Ax is maximum or 2A. ∴ A1 = v2 − v1 = 3 − 1 = 1 So, it is an antinode. Next antinode will occur at A2 2v2 2 x = λ , λ … etc. (b) v1 = 3 2 v2 or x = π , 2π … etc. (c) I = 1 ρω2A2v kk 2 Comprehension Based Questions ρ is not given, so we cannot find I1/I2. (d) P = 1ρω2A2Sv 1. Reflected and incident rays are in the same 2 medium. Hence, I ∝ A2 But, Sρ = µ ∴ P = 1 ω2A2µv Ir has become 64% or 0.64 times of Ii 2 ∴ Ar = 0.8 Ai = 0.8 A or P ∝ A2µv (as ω → same) 2. y = yi + yr 2 = A sin (ax + bt + π /2) P1 =  A1   µ1   v1  P2  A2   µ2   v2  + 0.8 A sin (ax − bt + π /2) = 0.8 A sin  ax + bt + π  = (1)2  19 (3) = 1 2 3 + 0.8 A sin (ax − bt + π2   or P2 = 3  P1 2. + 0.2 A sin  ax + bt + π  2 (i) = − 1.6 sin ax sin bt + 0.2A cos (bt + ax) ∴ c = 0.2 (ii)

372 — Waves and Thermodynamics (a) Second overtone mode = f3 = 3 f1 So, percentage of energy transmitted will be Fifth harmonic mode = f5 = 5 f1 So, the ratio is 3/5, (100 − 51)% or 49%. Ans. (d) λ = v or λ ∝ 1 3. Amplitude at a distance x is A = a sin kx ff First node can be obtained at x = 0, ∴ λ 3 = f5 = 5 and the second at x = π/k λ 5 f3 3 At position x, mass of the element PQ is 3. (d) At =  2v2  Ai dm = (ρS )dx  v1 + v2  Its amplitude is A = a sin kx Hence mechanical energy stored in this element is If v2 > v1, i.e. 2 is rarer then At > Ai. NN 4. A = 2 A0 cos φ and I = 4 I0 cos2 φ x=0 p 2 2 k dx x = 5. Second overtone frequency means f3 = 3 f1 = 210 Hz (energy of particle in SHM) dE = 1 (dm)A2ω2 ∴ f1 = 70 Hz = fundamental frequency 2 Third overtone frequency = f4 = 4 f1 Second harmonic frequency = f2 = 2 f1 or dE = 1 (ρSA2ω2) dx 2 Subjective Questions = 1 (ρSa2ω2 sin2 kx) dx 2 1. (a) v1 = T /µ1 Therefore, total energy stored between two adjacent nodes will be v2 = F /4µ1 = 1 F /µ1 , 2 x = π/k ∫E = dE x=0 v3 = F / (µ1 /4) = 2 F /µ1 Solving this, we get ∴ t = t1 + t2 + t3 = L + L + L E = πSρω2a2 Ans. v1 v2 v3 4k = 7L µ1 Ans 4. l = λ 2F 2 2. Let ai and ar be the amplitudes of incident and or λ = 2l, k = 2π = π reflected waves. λl ai at a x=0 λ/2 x=l ar The amplitude at a distance x from x = 0is given by A = a sin kx Then, ai + ar = 6 (given) Hence, ai − ar Total mechanical energy at x of length dx is Now, dE = 1 (dm)A2ω2 ar = 5 2 ai 7 = 1 (µdx)(a sin kx)2(2πf )2 2 Er  ar  2 =  57 2 Ei  ai  or dE = 2π 2 µf 2 a2 sin2 kx dx …(i) = = 0.51 T  or percentage of energy reflected is v2  µ  π λ2 (4 l 2 ) l 100 × Er = 51%. Here, f = = and k = Ei

Chapter 18 Superposition of Waves — 373 Substituting these values in Eq. (i) and integrating i.e. the amplitude of transmitted wave will be 2.0 cm. it from x = 0 to x = l, we get total energy of string. The expressions of Ar and At are derived as E = π 2a2T Ans. below. 4l 5. Tension, T = 80 N Derivation Amplitude of incident wave, Ai = 3.5 cm Suppose the incident wave of amplitude Ai and Mass per unit length of wire PQ is angular frequency ω is travelling in positive 0.06 1 x-direction with velocity v1, then we can write 4.8 80 m1 = = kg/m yi = Ai sin ω[t − x/v1 ] …(i) and mass per unit length of wire QR is In reflected as well as transmitted wave, ω will not 0.2 1 change, therefore, we can write 2.56 12.8 m2 = = kg/m yr = Ar sin ω [t + x/v1 ] …(ii) and yt = At sin ω [t − x/v2 ] …(iii) P QR Now as wave is continuous, so at the boundary l1 = 4.8 m l2 = 2.56 m (x = 0). Mass = 0.06 kg Mass = 0.2 kg Continuity of displacement requires (a) Speed of wave in wire PQ is yi + yr = yi for x = 0 Substituting from Eqs. (i), (ii) and (iii) in the v1 = T /m1 = 80 = 80 m/s above, we get 1/80 Ai + Ar = At …(iv) and speed of wave in wire QR is Also at the boundary, slope of wave will be v2 = T /m2 = 80 = 32 m/s continuous, i.e. 1/12.8 ∂yi + ∂yr = ∂yt [for x = 0] δx δx ∂x ∴ Time taken by the wave pulse to reach from which gives Ai − Ar =  v1  At …(v)  v2  P to R is t = 4.8 + 2.56 =  4.8 + 23.526 s Solving Eqs. (iv) and (v) for Ar and At, we get the v1 v2 80 required equations, i.e. = 0.14 s Ans. Ar = v2 − v1 Ai and At = 2v2 Ai v2 + v1 v2 + v1 (b) The expressions for reflected and transmitted amplitudes (Ar and At ) in terms of v1, v2 and 6. When A is a node Suppose n1 and n2 are the Ai are as follows : complete loops formed on left and right side of point A. Then, Ar = v2 − v1 Ai v2 + v1 f1 = f2 and At = 2v2 Ai or n1  2vL1  = n2  2vL2  v1 + v2 Substituting the values, we get or n1 =  v2  = µ1 = 1 , 2 , 3 , K,etc n2  v1  µ2 3 6 9 Ar =  32 − 80 (3.5) = − 1.5 cm  32 + 80  as v ∝ 1    i.e. the amplitude of reflected wave will be  µ 1.5 cm. Negative sign of Ar indicates that there ∴ Possible frequencies are will be a phase change of π in reflected wave.    322 ×+ 3820 v1 , 2  2vL1  , 3v1 , K, etc.  v1 = T   2L 2L  Similarly, At = (3.5) = 2.0 cm µ

374 — Waves and Thermodynamics or 1 T,1 T, 3 T , K , etc. = 0.02 sin  πt − 8π  2L µL µ 2L 2µ 3 When A is an antinode Suppose n1 and n2 are y = y1 + y2 complete loops on left and right side of point A, = 0.06 sin πt cos 4π − 0.06 cos πt sin 4π n1 λ1 + λ1 = L + 0.02 sin πt cos 8π − 0.02 cos πt sin 8π 2 4 33 or = v1  n1 + 41 = 0.05 sin πt − 0.0173 cos πt Ans. L 2 f1 9. Resultant amplitude n2 λ2 + λ2 =L A/3 A 2 4 or f2 = v2  n2 + 41 A /2 L 2 Substituting f1 = f2, ⇒φ 2A /3 we get, 2n1 + 1 = 1 A/2 Ar 2n2 + 1 3 For n1 = 1, n2 = 4  2A  2  A2  2 n1 = 2, n2 = 7 3 n1 = 3, n2 = 10, etc. Ar = + ……… =5A Ans. Therefore, the possible frequencies are 6 v1  1 + 41 , v1  2 + 41 , v1  3 + 41 , K, etc. tan φ = − A/2 = − 3 L 2 L 2 L 2 2A/3 4 3 T, 5 T, 7 T , ..., etc. or φ = − tan−1 (3/4) Ans. 4L µ 4L µ 4L µ or L=λ 10. Speed of longitudinal waves in the rod, 44 7. v = Y = 1.6 × 1011 = 8000 m/s ρ 2500 ∴ L=λ At the clamped position nodes will be formed. Between the clamps integer number of loops will P be formed. Hence, n1 λ = 80 2 In the next higher mode, there will be total 6 loops or n1λ = 160 …(i) and the desired frequency is  62 (100) = 300 Hz RP Q S Ans. 8. k =ω =π v3 Q y1 = 0.06 (πt − kx) = 0.06 sin  πt − π × 12 Between P and R, P is a fixed end and R is the free 3 end. It means the number of loops between P and = 0.06 sin (πt − 4π ) R will be odd multiple of λ . Then, 4 Similarly, (2n2 − 1) λ = 5 22  πt π 8 y2 = 0.02 sin (πt − kx′ ) = 0.02 sin − 3 × or (2n2 − 1)λ = 20 …(ii)

Chapter 18 Superposition of Waves — 375 Also between Q and S, Energy density (energy per unit volume) of each wave will be (2n3 − 1)λ = 60 …(iii) From Eqs. (i) and (ii), we get u1 = 1 ρω 2 (8)2 = 32 ρω2 n1 = 160 = 8 2 2n2 − 1 20 …(iv) 1 ρω 2 (6)2 = 18 ρω2 and u2 = 2 and from Eqs. (i) and (iii). ∴ Total mechanical energy between two consecutive nodes will be n1 = 160 = 8 …(v) 2n3 − 1 60 3 E = (u1 + u2) V For minimum frequency n1, n2 and n3 should be = 50 π ρω2S least from Eqs. (iv) and (v). k We get, n1 = 8, n2 = 1, n3 = 2 (b) y = y1 + y2 λ = 20 = 20 cm [from Eq. (ii)] = 8 sin (ωt − kx) + 6 sin (ωt + kx) 2n2 − 1 = 2 sin (ωt − kx) + {6 sin (ωt + kx) = 0.2 m + 6 sin (ωt + kx)} v 8000 ∴ fmin = λ = 0.2 = 2 sin (ωt − kx) + 12 cos kx sin ωt = 40 kHz Ans. Thus, the resultant wave will be a sum of standing wave and a travelling wave. Next higher frequency corresponds to Energy crossing through a node per second n1 = 24, n2 = 2 = power or travelling wave and n3 = 5 Ans. ∴ P = 1ρω2(2)2Sv f = 120 kHz 2 11. (a) Distance between two nodes is λ/2 or π/k. The = 1ρω2(4)(S )  ωk  2 volume of string between two nodes is V =πs …(i) = 2ρω3S k k

19. Sound Waves INTRODUCTORY EXERCISE 19.1 2. v∝ T Q 1. ∆pmax= BAk  T2  ∴ v2 =   v1 B = ∆pmax = ∆pmax  T1  Ak A (2π /λ )  270  = (∆pmax ) λ v− =   (332) 2πA 3° C  273  = (14) (0.35) = 330.17 m/s (2π ) (5.5 × 10− 6)  273 + 30 (332) v30° C =  273  = 1.4 × 105 N/m2  2. λ min = v = 1450 = 349.77 m/s fmax 20000 The difference in these two speeds is = 0.0725 m = 7.25 cm approximately 19.6 m/s. λ max = v = 1450 = 72.5 m 3. v = fλ = B fmin 20 ρ Q 3. (a) Displacement is zero when pressure is ∴ B = ρ ( fλ )2 maximum. = (900) (250 × 8)2 (b) ∆pmax = BAk = 3.6 × 109 N/m2 = (ρv2) A  ω  = 2πfAρv 4. v = γRT v M ∴ A = ∆pmax 2πfρv = (7/5) (8.31) (273) (32 × 10− 3) = 10 (2π ) (103) (1.29) (340) = 315 m/s = 3.63 × 10− 6 m INTRODUCTORY EXERCISE 19.3 4. In the above problem, we have found that 1. (a) ∆pmax = BAk A = ∆pmax = (ρv2) (A)  ωv  2πfρv = (∆pmax )k  as v = ω and 2πf = ω ρω2 k = (2πfA ρv) Now substituting the value, we have = (2π ) (300) (6.0 × 10− 3)(1.2) (344) A = (12) (8.18) = 1.04 × 10− 5 m = 4.67 Pa (1.29) (2700)2 (b) I = v (∆p)m2 ax INTRODUCTORY EXERCISE 19.2 2B (as B = ρv2) 1. v ∝ T = (∆p)2max 2ρv ∴ v2 = T2 or T2 =  v2  2 = (2)2 (273) = (4.67)2 v1 T1  v1  (2) (1.2) (344) T1 = 1092 K = 0.0264 W/m2 = 819°C = 2.64 × 10− 2 W/m2

Chapter 19 Sound Waves — 377 (c) L = log10  I  For finest sound,   A = 2 × 10− 5  I0 (2π ) (500) (1.29) (345) = 10 log10  2.64 × 10− 2 = 1.43 × 10− 11 m  10− 12    = 104 dB INTRODUCTORY EXERCISE 19.4 2. L2 − L1 = 10 log10 I1 1. ∆x = λ /2 = v I2 2f Given, L2 − L1 = 9 dB Solving the equation, we get ∴ f = v = 330 = 1375 Hz 2∆x 2(0.12) I1 = 7.9 I2 2. (a) ∆φ = 2π (∆x) 3. I ∝ 1 λ r2 ∴ ∆x = λ∆φ = v (∆φ) I1  r2 2  33000 2 2π ( f ) (2π ) I2  r1 ∴ = = = 100 = (350) (π /3) (500) (2π ) I1 Now, L1 − L2 = 10 log10 I2 = 0.1166 m = 11.7 cm Substituting I1 = 100 (b) ∆φ =  2π  ∆t = (2πf ) ∆t I2 T We get, L1 − L2 = 20 dB = (2π ) (500) (10− 3) 4. (a) I = v (∆p)m2 ax = π or 180° 2B INTRODUCTORY EXERCISE 19.5 = (∆p)2max 1. Length of the organ pipe is same in both the cases. 2ρv v For finest sound, Fundamental frequency of open pipe is f1 = 2l I = (2 × 10− 5)2 2 × 1.29 × 345 and frequency of third harmonic of closed pipe will be = 4.49 × 10− 13 W/m2 f2 = 3  4vl L = 10 log10  I  Given that, f2 = f1 + 100  I0 or f2 − f1 = 100 4.49 × 10− = 10 log10  10−12 13  or 3  vl  −  21  vl  = 100   4   = − 3.48 dB ⇒ v = 100 Hz 4l Same formulae can be applied for loudest sound. ∴ v or f1 = 200Hz 2l (b) (∆p)max = BAk = (ρv2) (A)  ωv  Therefore, fundamental frequency of the open pipe is 200Hz. 2. First harmonic of closed pipe = Third harmonic of = 2πfρv open pipe ∴ A = (∆p)max ∴ v = 3  v  ⇒ ∴ l1 = 1 2πfρv 4l1  2l2 l2 6

378 — Waves and Thermodynamics 3. fc = v = 512 Hz INTRODUCTORY EXERCISE 19.6 4l 1. Frequency of first will decrease by loading wax fo = v = 2 fc = 1024 Hz 2l over it. Beat frequency is increasing. Hence, 4. (a) f1 = v f2 > f1 4l or f2 − f1 = 4 ∴ f1 = f2 − 4 = 256 − 4 ∴ l = v = 345 4 f1 4 × 220 = 252 Hz = 0.392 m 2. By putting wax on first tuning fork, its frequency  v  = 3  v  will decrease. (b) 5      4lc   2l0 Beat frequency is also decreasing. Hence, ∴ l0 = 6 lc =  65 (0.392) f1 > f2 5 ∴ f1 − f2 = 3 or f1 = 3 + f2 = 3 + 384 = 0.47 m = 387 Hz INTRODUCTORY EXERCISE 19.7 5. 1. Source is moving towards the observer x=0 f′ = f  v = 450  33033−033 (i) (ii) (iii)     v − vs   (a) In second figure, l = 3λ 4 f ′ = 500 Hz ∴ λ = l = 0.8 2. f1 =  v 43 3 f  = 0.267 m  v − vs  Displacement antinode is at x = λ = 0.267 m f1 = f  34034−034  = f  334006 4   and x = 3 λ = 0.8 m 4 and f2 = f  34034−017 = f  334203  (b) Pressure antinode is displacement node:  In third figure, ∴ f1 = 323 = 19 f2 306 18 Pressure antinode or displacement node is at x = 0, x = λ 3. Using the formula, f′ = f  v + v0 2 v and x = λ We get, 5.5 = 5  v + vA  …(i) v 6. (a) 400 = 5 and 6.0 = 5  v + vB  …(ii) 560 7 v Since, these are odd harmonics (5 and 7). Here, v = speed of sound Hence, pipe is closed. vA = speed of train A (c) Given frequencies are integer multiples of 80 Hz. Hence, fundamental frequency is 80 Hz. vB = speed of train B v = 80 Solving Eqs. (i) and (ii), we get vB = 2 4l ∴ l = v = 344 vA 320 320 = 1.075 m 4. Observer is stationary and source is moving. During approach, f1 = f  v   v − vs 

Chapter 19 Sound Waves — 379 = 1000  32032−020 = 1000  320  = 941.18 Hz   320 +    20 = 1066.67 Hz |% change in frequency| =  f1 − f2 × 100   v  f1  During recede, f2 = f   v + vs  ≈ 12% Exercise LEVEL 1 10. Wavelength remains same. Assertion and Reason 1. Closed pipe Frequencies are, v , 3v , 5v 4l 4l 4l Open pipe Frequencies are v , 2  2vl , 3  2vl Objective Questions 2l They are never same. 1. Sound waves cannot travel in vacuum.  v  3. f0 = v and fc = v  −  2l 4l 2. f′ = f v  ⇒ f′ > f but f ′ = constant. vs ∴ f0 = 2 fc 3. Change in pressure is maximum at a point where 4. v = γRT ∝ T (γ = 1.4 for both) displacement is zero. MM 5. L = 10 log10 I ∴ TO2 = TN2 I0 MO2 MN2 Increase in the value of L is not linear with I.  MO2   MN2  ∆L = L2 − L1 = 10 log I2 ∴ TO 2 = TN 2 I1 ∆L = 3dB if I2 = 2I1 =  2328 (273 + 15) 6. It is independent of pressure as long as = 329 K = 56° C temperature remains constant. 5. 7. fA − fB = 4 Hz When A is loaded with wax fA will decrease. So, fA − fB will be less than 4 Hz till fA > fB. But fB − fA may be greater than 4 Hz when fB becomes greater than fA. 8. 450 = 3 750 5 Successive harmonics are odd 3 and 5. Hence, it is Third overtone a closed pipe. 6. f ∝ v and v ∝ T 9. f0 = v of an open pipe and 7. In air speed of sound is less, hence air is denser + 1.2 2 (l r) medium. f0 = 4 (l v r) for a closed pipe. 8. v = v ⇒ lc : l0 = 1: 2 + 0.6 2l0 4lc

380 — Waves and Thermodynamics 9. f ∝ v ∝ T (T → tension) 18. n = n1 − n2 = v − v λ1 λ2 f1 = T1  f2  2 ∴ f2 T2   ∴ ⇒ T2 = T1  f1 v = nλ 1λ 2 λ2 − λ1 = (10)  325260 2 = 6.4 kg 19. By putting wax on A. Its frequency will decrease. 10. O S' But beat frequency between A and B is also S decreasing. ∴ fA > fB Both S and S′ are moving toward observer. Hence, or fA − fB = 5 Hz …(i) fS = fS′ or fb = 0 ∴ fB = fA − 5 = 345 Hz = 10 = v v Now, fB ~ fC = 4 Hz 3 1 1.01 11. fb f1 − f2 = − fC is either 341 Hz or 349 Hz. Solving we get, v = 337 m/s If it is 341 Hz, then beat frequency with A will be  v   v  9 Hz.     12. f′= f = f = 0.5 f If it is 349 Hz, then beat frequency will be 1 Hz.  v + vs   v + v If wax is loaded on A, its frequency will decrease. 13. IR = Imax = ( I1 + I2 )2 To produce 6 beats/s with C it should become If I1 = I2 = I0, then either 347 Hz (if fC = 341 Hz) or it should become 343 Hz (if fC = 349 Hz). IR = 4I0 If it becomes 347 Hz, then only it produces 14. λ = 52 − 17 = 35 cm 2 beats/s with B, which is given in the question. 2 ∴ λ = 70 cm = 0.7 m ∴ fC = 341 Hz Ans. v = f λ = 500 × 0.7 20. λ = 122 − 40 = 82 cm = 350 m/s 2 15. f′ = f  v  ∴ Next resonance length = 122 cm + 82 cm   = 204 cm  v ± vs cos θ At θ = 90° ; f ′ = f 21. f ∝ T ∴ n1 = 0 f ′ = T ′ = 101 = 1.0049 f T 100 16. Fundamental frequency, f0 = v = 340 = 85 Hz f ′ = (1.0049) (200) 4l 4 ×1 ≈ 201 Hz Six frequencies can be produced below 1 kHz. ∴ fb = f ′ − f = 1 Hz Those six frequencies are, f0, 3 f0, 5 f0, 7 f0, 9 f0 and 11 f0. 22. λ = v = 340 = 1 m = 100 cm As 13 f0 = 1105 Hz > 1000 Hz or 1 kHz. f 340 λ = 25 cm 17. There is no relative motion between O and S ′. 4 OS 5 m/s S' Air column lengths required are, 5 m/s λ , 3λ , 5λ etc. 5 m/s 44 4 fb = fS′ − fS or 25 cm, 75 cm, 125 cm etc. Maximum we can take 75 cm. = f − fS = 180 − 180  355 − 5 ∴ Minimum water length  355 + 5  = 120 − 75 = 45 cm = 5 Hz

Chapter 19 Sound Waves — 381 23. Pressure node means displacement antinode or L2 − 10 = 10 log10 (16) or L2 = 22 dB 7λ = 105 cm 4 30. λ = 2π = 2π = 4 m λ = 15 cm 4 k π/2 Displacement antinodes are at a distance l = 5λ = 5 m 4 λ , 3λ , 5λ and 7λ from closed end or at a 44 4 4 31. λ = v = 330 = 0.55 m = 55 cm f 600 distance of 15 cm, 45 cm, 75 cm and 105 cm. 24. Number of moles ∝ Volume M = n1 MO2 + n2MH2 d n1 + n2 = (1) (32) + (1) (2) = 17 The desired distance, d = λ = 13.75 cm 2 4 Now, v = γRT ∝ 1 v v MM λ1 λ2 32. fb = f1 − f2 = − v1 = M2 = 2 = 332 − 332 ≈13 Hz v2 M1 17 0.49 0.5 25. f1 = f  v  = constant. But f1 > f 33. ( fb )A = fB − fA    v − vs   300 + 30 f2 = f  v = constant, but f2 < f = 300 300 − 300 = 30 Hz   v + vs  ( fb )B = fA − fB 26. f ∝ 1 = 300  30030−030 − 300 = 33.33 Hz  l  Both frequencies will becomes half. Hence, fb = f1 − f2 will also become half. 34. fa = f  v + v0 v 27. fb = 243  320  − 243  320  = 6 Hz  320 −   320 +   4   4  ∴ v0 = fa − 1 …(i) vf …(ii) 28. Closed pipe v − v0 Fundamental frequency is fr = f  v f1 = v = 320 = 80 Hz ∴ v0 = 1 − fr 4l 4 ×1 vf Other frequencies are 3 f1 : 5 f1 etc. or 240 Hz, 400 Hz etc. From Eqs. (i) and (ii), we get Open pipe Fundamental frequency is f = fa + fr 2 f1 = v = 320 = 100 Hz 2l 2 × 1.6 Subjective Questions Other frequencies are 2 f1 : 3 f1 : 4 f1 etc. or 200 Hz, 300 Hz and 400 Hz etc. So, then resonate at 400 Hz. 1. Speed of sound wave, 29. Resultant amplitude will become 4 time. v= B= (2 × 109 ) = 1414 m/s ρ 103 Therefore, resultant intensity is 16 times L2 − L1 = 10 log10 I1 Wavelength, λ = v = 5.84 m Ans. I2 f

382 — Waves and Thermodynamics 2. f′= f  v ± vm  = f 8. (a) v = fλ = B  v ±  ρ  vm  ∴ B = ρ ( fλ )2 3. v = γRT = (1300) (400 × 8)2 M = 1.40 × 8.31 × 300 = 1321 m/s = 1.33 × 1010 N/m2 2 × 10− 3 v=l= Y 4. L1 = 10 log10  I 1  (b) tρ  I0 ∴ 2   2  tl  1.5 = 10 log10  1100−−162 = 60 dB Y = ρ = (6400)  3.9 × 10− 4     I 2  = 9.47 × 1010 N/m2  I0 L2 = 10 log10  9. Y = 30 T = 30 F (as T = F and µ = ρA) = 10 log10  10− 9 = 30 dB ρ µ ρA  10− 12  ∴ F= Y  A 900 L1 = 2L2 10. Equal volume of different gases contains equal 5. f′ = f  v − vw − v0    number of moles at STP.  v − vw + vS  ∴ n∝V = 300  340 − 5 − 20  M = n1M1 + n2M2  340 − 5 + 10  n1 + n2 = (2) (2) + (1) (28) = 10.67 = 274 Hz 2+1 6. S v = γRT M ∴ v2 = T2 ⋅ M1 v1 T1 M2 dd  T2 M 1  or v2 =  T1 ⋅  v1 21  2d1 = v t1 M2 ∴ d1 = vt1 = 332 × 3/2  300 × 2  2 2 =  (1300)  273 10.67 = 249 m Similarly, d2 = vt2 = 332 × 5/2 ≈ 591 m/s 2 2 11. L1 = 10 log10  I    = 415 m  I0 ∴ Total distance = d1 + d2 = 664 m 100 = 10 log10  I 12 Next echo he will hear after time, 10− t = t2 + t1 = 5 + 3 = 4 s Solving we get, I = 10− 2 W/m2 2 2 P 7. v = γp = (5/3) × 0.76 × 13.6 × 103 × 9.8 Now, I = 4 πr2 ∴ ρ 0.179 P = I (4πr2) where, p = hρg = (10− 2)(4π )(40)2 ∴ v = 972 m/s = 201 W

Chapter 19 Sound Waves — 383 12. (a) L = 10 log10  I  16. L = 10 log10  I       I0  I0 60 = 10 log10  I 12 ∴ 102 = 10 log10  I 12  10− 10− Solving we get, Solving we get, I = 1.59 × 10− 2 W/m2 I = 10− 6 W/m2 (b) I = P I = P S 4 πr2 ∴ P = (I ) (S ) ∴ P = (I ) (4πr2) = (10− 6) (120 × 10− 4) = (1.59 × 10− 2) (4π ) (20)2 = 1.2 × 10− 8 W = 80 W 13. (a) L2 − L1 = 10 log10  I 2    17. I = 1 ρω2A2v  I1 2 ∴ 13 = 10 log10  I 2  (as ω = 2πf )   = 2ρπ 2 f 2A2v  I1 = (2) (1.29) (π )2(300)2 (0.2 × 10− 3)2 (330) Solving we get, I2 = 20 = 30.27 W/m2 I1 L = 10 log10 (I /I0) (b) From the above equation we can see that we do where, I0 = 10− 12 W/m2 not need L1 and L2 separately. Substituting the value we get, L =134.4 dB 14. (a) I = P = 5 πr2 (20)2 4 4π = 9.95× 10− 4 W/m2 B = 2.18 × 109 ρ 1000 (b) I = 1 ρω2A2v 18. (a) v = 2 = 1476 m/s ∴ λ = v = 1476 = 0.43 m ∴ A= 2I ρ (2πf )2 v f 3400 = 2 × 9.95 × 10− 4 I = 1ρω2A2v 1.29 (2π × 300)2 (330) 2 = 1.15 × 10− 6m A= 2I ρ (2πf )2v  I  15. L = 10 log10  I0 2 × 3 ×10− 6 (1000) (2π × 3400)2 (1476) 60 = 10 log10 (I /10− 12) = Solving, we get I = 10− 6 W/m2 = 9.44 × 10− 11m Now, using the result derived in above problem. (b) v = γp = (1.4) (105) ρ (1.2) A= 2I ρ (2πf )2 v = 341.56 m/s λ = v = 341.56 = 0.1 m = 2 × 10− 6 1.29 (2π × 800)2 (330) f 3400 = 1.36 × 10− 8 m A= 2I ρ (2πf )2 v = 13.6 nm

384 — Waves and Thermodynamics = 2 × 3.0 × 10− 6 k = ω = 1297 = 3.93 rad /s 1.2 × (2π × 3400)2 (341.56) v 330 = 5.66 × 10− 9 m Now, x = 0 (the open end) is an antinode. Hence, we can write the equation, (c) Aair = 5.66 × 10− 9 = 60 Awater 9.44 × 10− 11 y = A cos kx sin ωt 23. λ = (84 − 50) cm = 34 cm 2 I = 1ρω2A2v Next length = 84 + λ = 118 cm 2 2 ρ and v are less in air. So, for same intensity A λ = 68 cm or 0.68 m should be large. v = fλ = (512) (0.68) 19. I = v (∆p)m2 ax ⇒ v = B = 348.16 m/s 2B ρ 24.  v  =  v2  2    ∴ B = ρv2  2l1  4l2 ∴ I = (∆p)m2 ax or T /µ =  v2  2ρv   l1  4l2 = (6 × 10−5)2 2 × 1.29 × 330  v2 l1  2   = 4.2 × 10− 12 W/m2 ∴ T =µ  4l2   I =  4 × 10− 3  340 × 40 2    L = 10 log10   0.4   4 × 100  I0 = 11.56 N = 10 log10  4.2 × 10− 12   10− 12  25. f = v   4 (l + 0.6 π ) = 6.23 dB ∴ v = 4 f (l + 0.6 π ) 20. (a) Fundamental, f1 = v = 344 = 382.2 Hz = 4 × 480 (0.16 + 0.6 × 0.025) 2l × 0.45 2 = 336 m/s First two overtones are 2 f1 and 3 f1. 26. (a) f= v 4l (b) Fundamental, f1 = v = 344 = 191.1 Hz 4l 4 × 0.45 ∴ l = v = 345 4 f 4 × 220 First two overtones are 3 f1 and 5 f1. = 0.392 m 21. λ = (45 −15) cm (b) 5  v  = 3  v  2  4lc   2l0 ∴ λ = 60 cm ∴ l0 = 6 lc = 6 (0.392) 5 5 v = fλ = (500) (0.6) = 0.470 m = 300 m/s Lowest frequency when open from both ends is v ⋅ 27. v1 = v2 ⇒ v1 = 0.4 2 (0.8l) 4l v2 2l = 300 = 250 Hz 28. (a) λ = v = 340 = 1.13 m 2 × 0.6 f 300 22. Fundamental frequency = v (b) Ahead 2l f ′ = f  v  = 300  340  f = 330 2 × 0.8 f = 206.25 Hz  v − vs   340 − 30 ω = 2πf = 1297 rad /s = 329 Hz

Chapter 19 Sound Waves — 385 λ′ = v = 340 = 1.03 m = 0.625  0.32  f ′ 329    0.32 + 0.245 Behind = 0.354 Hz λ = v = 0.32 = 0.904 m f′ = f  v  = 300  340      f ′ 0.354  v + vs   340 + 30 = 275.67 Hz 31. (a) f′= f  v + v0  λ′ = v = 340  v −   vs  f ′ 275.67  340 + 18 = 1.23 m = 262  340 − 30  29. (a) Possible fundamental frequencies of retuned = 302 Hz string is (440 ± 1.5) Hz. (b) f′ = f  v − v0  (b) f ∝ T  v + vs  ∴ f1 = T1 = 262  340 − 18  = 228 Hz f2 T2  340 + 30 ∴ T2 =  f2  2 32. 1 m/s  f1  T1 1 m/s O S S′ (i) T2 =  444410.5 2 fb = fS − fS′ T1 = 1.0068 T1 4= f  340  − f  340  % change = T2 − T1 × 100  340 −   340 +   1  1 T1 or 4 = f 1 − 3410 −1 − f 1 + 3410 − 1 = + 0.68% (ii) T2 =  443480.5 2 T1 = 0.9932 T1 Applying Binomial, we get 4 = f 1 + 3410 − f 1 − 3410 % change = T2 − T1 × 100 = f or f = 680 Hz T1 170 = − 0.68% 33. Beat frequency between P and Q is 7 or 3.5 Hz. 30. (a) λ = 0.12 2 Surface wave speed, v = 0.32 m/s On loading P with wax, its frequency will decrease and beat frequency is increasing. f′ = v = 0.32 = 2.66 Hz λ 0.12 f = 1 = 1 = 0.625Hz Q 5 m/s 5 m/s Q′ T 1.6 Now, using f′= f  v     v − vs  ∴ fQ > fP Here, vs = velocity of source or velocity of or fQ − fP = 3.5 …(i) duck Given, fQ′ − fQ = 5 ∴ 2.66 = 0.625  0.32  or fQ  332 + 5 − fQ =5  0.32 −   vs  332 − 5 Solving we get, vs = 0.245 m/s Solving this equation, we get (b) Behind the duck fQ = 163.5 Hz f′ = f  v v  Now, from Eq. (i) we have  +  fP = 160 Hz vs 

386 — Waves and Thermodynamics 34. fb = f1 − f2 = (135.4)2 (0.127) (8.31 × 400) 2 = 680  340  − 680  340   340 − vs   340 + vs  = 0.7 = 680 1 − 3v4s0 − 1 − 680 1 + 3v4s0 −1 Since, γ <| 1 So, we will have to substitute, Using Binomial expansion, we have M = 2 × 0.127 2 = 680  2 vs  ⇒ vs = 0.5 m/s The γ comes out to be 1.4 which is the γ of a  340  diatomic gas. 35. λ = 11.5 cm or λ = 23 cm or 0.23 m 39. λ = (100 − 60) × 10− 2 m 2 2 f = v = 331.2 = 1440 Hz ∴ λ = 0.8 m λ 0.23 Now, v = fλ = 440 × 0.8 36. λ = v = 330 = 1.5 m = 352 m/s f 220 40. When TA is increased, fA will increase. So, it will  2π   12.π5 become (600 + 6) = 606 Hz λ ∆φ1 = (∆x1) = (0.75) = π Now, f∝ T  2π   12.π5 TA  fA  2  660006 2 λ TB   ∆φ2 = (∆x2) = (3) = 4π ∴ =  fB  = Since, φ = ∆φ2 − ∆φ1 = 3π = 1.02 They will interfere destructively. 41. f ∝ 1 ∴ IR = ( I1 − I2 )2 …(i) l where, By decreasing the length, frequency of wire will I1 = P1 = 1.2 × 10− 3 increase. But beat frequency is decreasing. Hence, 4 πr12 (4π ) (0.75)2 its original frequency was (256 − 4) Hz or 252 Hz. Now, we have to make it 256 Hz for no beats. = 16.97 × 10− 5W/m2 f1 = l2 f2 l1 I2 = P2 = 1.8 × 10− 3 4 πr22 (4π ) (3)2  f1  ∴ l2 =   l1 = 1.59 × 10− 5 W/m2  f2 Substituting in Eq. (i), we get =  225526 (25) IR = 8.2 × 10− 5 W/m2 = 24.6 cm 37. SOD − SD = λ where, λ = v = 360 = 1 m ∴ ∆l = l1 − l2 = 0.4 cm f 360 42. If observer is moving, then ∴ 2 x2 + (2)2 − x = λ = 1 f′= f  v ± v0 = f 1 ± v0  4 v v Solving this equation, we get If source is moving, then x = 7.5 m f ′ = f  v   v m vs  38. v = fλ = (1000) (2 × 6.77 × 10− 2) Q = 135.4 m/s = γRT 1 m vs  − 1 1 vs  M v v = f = f ± ∴ γ = (135.4 )2  M  If vs < < v RT Both expressions come out to be of similar type.

Chapter 19 Sound Waves — 387 43. f′ = f  v  = 200  34034−080 LEVEL 2  −    v vs   Single Correct Option = 261.53 Hz ≈ 262 Hz 1. sin θC = 330 = 0.2357 λ′ = v = 340 = 1.3 m 1400 f ′ 261.53 ∴ θC = 13.6° 44. In front of locomotive, Since, i > θC, therefore only reflection will take place. f′ = f  v   v − vs  2. ∆pmax = BAk …(i) = 500  344  = 548 Hz v= B  344 −  ρ  30 ∴ λ′ = v = 344 = 0.628 m ∴ B = ρv2 f ′ 548 l = 3λ Behind the locomotive, 2  v  ∴ λ = 2l l  +  3 f′ = f  v vs  k = 2π = 2π = 3π = 500  34344+430 = 460 Hz λ 2l/3 l   = 3π = 1 m− 1 3.9 π 1.3 ∴ λ′ = v = 344 = 0.748 cm f ′ 460 Substituting in Eq. (i), we have 45. (a) The given frequencies are in the ratio 5 : 7 : 9. As A = ∆pmax = ∆pmax Bk ρv2k the frequencies are odd multiple of 85 Hz, the pipe must be closed at one end. = (0.01 × 105) × (200)2 × (1/1.3) (b) Now, the fundamental frequency is the lowest, 1.3 i.e. 85 Hz. = 0.025 m = 2.5 cm ∴ 85 = v 3. At =  2v2  Ai 4l   ⇒ l = 340 = 1 m Ans.  v1 + v2 4 × 85 ∴ At = 2 × 100 = 2 46. Let the frequency of the first fork be f1 and that of Ai 200 + 100 3 second be f2. 4. fb = f1 − f2 We then have, or 10 = 1700  340  − 1700  340   340 +   340 +  f1 = 4 v  v2  v1 × 32 = 1700 1 + 3v420 −1 − 1700 1 + 3v410 − 1 and f2 = 4 v × 33 Using Binomial therefore, we get We also see that f1 > f2 10 = 1700 1 − 3v420 − 1700 1 − 3v410 ∴ f1 − f2 = 8 …(i) …(ii) and f1 = 33 = (v1 − v2)  1374000 f2 32 Solving Eqs. (i) and (ii), we get ∴ v1 − v2 = 2 m/s f1 = 264 Hz Ans. 5. Velocity of source after 1 s, and f2 = 256 Hz vs = gt = 10 × 1 = 10 m/s

388 — Waves and Thermodynamics ∆f = f  v + v0  − f  v − v0  A1 = 2 and A2 = 1  v −   v + vs   vs    I max  A1 + A2  2 9   = =  300 −+120  330000+−120 Imin  A1 − A2  1 = 150  300 − 150    11. l = 3λ = 12 Hz 4 λ = 4l = 4 (34) = 45.33 cm ∴ 33 6. L = 7λ λ= v 4 f ∴ λ = 4L x=0 λ ∝v (as f = constant) 7 and v ∝ T (T = temperature) k = 2π ∴ λ∝ T λ ∴ λ 2 = T2 = 2π = 7π λ 1 T1 4L/ 7 2L x = 0 is a node ∴ Ax = a sin kx = a sin  7π   L7 = a  T2  2L T1  ∴ λ2 =  λ1  7. ∆xmax = 3m = λ =  16 + 273 (45.33) ∴ f = v = 330 = 110 Hz. For frequencies less  51 + 273 λ3 = 42.8 cm than 100 Hz, λ will be more than 3m and no maximum will be obtained. 12. f1 = f2 8. VS O W = 20 m/s N Q 176  333300−−2v2 = 165  330 + v 60° ∴  330 E f′= f  v v + w cos 60° vs  Solving this equation we get,  + w cos 60° −  v = 22 m/s = 500  30030+01+01−020  13. M = n1M1 + n2M2  n1 + n2 = 534 Hz = (2) (32) + (3) (48) = 41.6 5 9. 20 m/s 10 m/s S' S S f = v or f ∝ v 1 4l 2 1 10 m/s But, v = γRT M fb = fS′ − fS1 or v ∝ 1 = 500  340 + 20  − 500  340 + 20  M  −   +      ∴ f∝ 1 M 340 10 340 10 ∴ f2 = M1 f1 M2 ≈ 31 Hz 10. f1 = ω1 = (400π ) = 200 Hz  M 1  2π 2π or f2 =   f1 ∴  M2 f2 = ω2 = 404 π = 202 Hz 2π 2π  32 (200) = 175.4 Hz = fb = f2 − f1 = 2 Hz  48

Chapter 19 Sound Waves — 389 14. f = f0  v + v0 (b) First overtone frequency of closed pipe = 3 f1 v and first overtone frequency of open pipe = 103 1 + 1v0t (as v0 = gt) = 2 f2 = 2  5 f1 = 2.5 f1 4 Hence, f versus t graph is a straight line of slope (c) Fifteenth harmonic of closed pipe = 15 f1 104. Twelfth harmonic of open pipe v = 12 f2 = 12  5 f1 = 15 f1 4 ∴ 104 = slope = 100 v3 (d) Tenth harmonic of closed pipe = 10 f1 ∴ v = 300 m/s Eighth harmonic of open pipe 15. vp = − v  ∂∂xy , where v = + ve. = 8 f2 = 8  5 f1 = 10 f1 4 Q At E, ∂y or slope is positive. 5. f = v = γ RT /M ∂x 4 (l + 0.6r) 4 (l + 0.6 r) Hence, vp is negative. (d) v does not depend on pressure if temperature is At D, ∂y or slope is zero. Hence, vp is zero. kept constant. ∂x 6. During approach, More than One Correct Options f = f0  v   f = n  4vl  v − vs  1. (n = 1, 3, 5…) > f0 but f is constant. During receding. Q ∴ l = nv = n  330  f = f0  v 4f  4 × 264     v − vs  = (0.3125 n) m < f0 but f is again constant. = (31.25 n) cm Comprehension Based Questions Now, keep on substituting n = 1, 3, etc. 1. O v 8−v S 2. (a) Velocity of sound wave in air independent of Plank Man pressure if T = constant (b) v = γRT = γR (t + 273) From conservation of linear momentum, or MM v2 ∝ (t + 273) 50 (8 − v) = 150v ∴ v = 2 m/s (c) v = T /µ 8 − v = 6 m/s ∴ v∝ T f1 = f0  v + v0  v − vs  or v2 ∝ T  330 + 2 332 (d) f0 = v or v = f0  330 − 6 = 324 f0 2l 4l  ⇒ f0 ∝ 1 2. 6 m/s S O 2 m/s l  v − v0  4. Let f1 = fundamental frequency of closed pipe and f2 = f0  v + vs  f2 = fundamental frequency of open pipe.  330 − 2  332368  330 + 6 Given, 5 f1 = 4 f2 = f0  = f0 or f2 = 5 f1 3. f1 > f0 but f1 = constant. 4 Similarly, f2 < f0 but f2 = constant ∴ f2 > f1