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

352 Physics—XI  = 100 or = 100 or v = 400 L  =  = 200 Hz. 8. The extension in a string (obeying Hooke’s law) is x. The speed of the sound in the stretched string is v. What will be the speed of the sound in the string, if the extension is increased to 1.5x? Ans. v =  As extension in the string obeys Hooke’s law,  Tx  v ...(1) Now, let v be the speed of sound in the string, when the extension is increased to 1.5x. Then, v  ...(2) From (1) and (2)  = = = 1.225 or v = 1.225v. 9. Water is being continuously poured into a vessel. Can you estimate the height of water level reached in the vessel simply by listening to the sound produced? Ans. The frequency of the sound produced by an air column is inversely proportional to the length of the air column. As the level of water in the vessel rises, the length of the air column in the vessel decreases. Frequency of sound increases. The shrillness of sound increases. From the shrillness of sound, we can have a rough estimate of the level of water in the vessel. 10. A body vibrating with a constant frequency sends waves 100 mm long through a medium A and 0.25 m long through a medium B. If the velocity of the waves in medium A is 80 cm s–1, calculate the velocity of waves in medium B. Ans. In medium A Velocity of waves,v = 80 cm s–1, wavelength,  = 100 mm = 10 cm frequency,  =  Hz = 8 Hz  In medium B Let v be the velocity of waves in medium B.  = 0.25 m,  = 8 Hz v =  = 8 × 0.25 m s–1 = 2 m s–1. 11. How far does the sound travel in air when a tuning fork of frequency 400 Hz makes 30 vibrations? Given : velocity of sound in air = 360 m s–1. Ans. Frequency of tuning fork,  = 400 Hz, Velocity of sound in air, v = 360 m s–1

Waves 353 Wavelength, =  m or  = 0.9 m  When tuning fork completes one vibration, the distance travelled by sound is 0.9 m. When the tuning fork makes 30 vibrations, then the distance covered is given by S = 0.9 × 30 m = 27 m. 12. Figure 7.7 shows different standing wave patterns on a string of linear mass density 4.0 × 10–2 kg m–1 under a tension of 100 N. The amplitude of antinodes is indicated in each figure. The length of the string is 2.0 m. 0.1 m 0.08 m 0.04 m 0.01 m Fig. 7.7 (i) Obtain the frequencies of the fundamental mode and its different harmonics shown in the figure. (ii) Write down the transverse displacement y as a function of x and t for each mode. (Take the initial configuration of the wire in each mode to be as shown by the dark lines in the figure). (iii) Plot on a graph paper the wave patterns obtained by superposing the four stationary patterns at different times. Does the resulting pattern have nodes and antinodes at fixed locations on the string apart from the two end points? Can the pattern be characterised by a definite wavelength? Ans. (i) fundamental : 12.5 Hz; second harmonic : 25 Hz ;third harmonic : 37.5 Hz ; fourth harmonic : 50 Hz. (ii) (a)   (b) y = 0.08 sin x sin 50 t (c)    (d) y = 0.01 sin 2x sin 100 t (iii) No, No.

354 Physics—XI 13. The figure shows a snapshot of three waves travelling along a string. The phases for the waves are given by (a) 2x – 4t (b) 4x – 8t and (c) 8x – 16t. Which phase corresponds to which wave in Fig. 7.8? Ans.   [Note that  = kx – t]     or 1 23 Fig. 7.8 (a) Maximum wavelength. Clearly, it is wave 2. (b) Intermediate wavelength. Clearly, it is wave 3. (c) Minimum wavelength. Clearly, it is wave 1. 14. Here are the equations of three waves : 1. y(x, t) = 2 sin (4x – 2t), 2. y(x, t) = sin (3x – 4t), 3. y(x, t) = 2 sin (3x – 3t). Rank the waves according to their (a) wave speed and (b) maximum transverse speed, greatest first. Ans. (a) 2, 3, 1 2.    3.    (b) 3, then 1 and 2 tie. 3. Transverse speed, = – 6 cos (3x – 3t); maximum transverse speed = 6 unit 1. maximum transverse speed = 4. 2. Maximum transverse speed = 4. 15. What is the percentage increase in the intensity of sound if the intensity level in- creases by one decibel? Given : antilog10 0.1 = 1.2589. Ans. Intensity level in decibel is given by L= L+1 = Subtracting, 1= 

Waves 355 or =  or = 1.26 or     = (1.26 – 1) × 100 = 0.26 × 100 = 26.   16. If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased? Ans. 50 = 10 log10 60 = 10 log10  60 – 50 = 10     or log10 = 1 = log10 10  or = 10 or   = 10 or = . 17. Densities of oxygen and nitrogen are in the ratio 16 : 14. At what temperature the speed of sound in oxygen will be the same as at 15°C in nitrogen? Ans. = , = Equating,  = or = or = =       or 14 × 273 + 14t = 16 × 288 = 4608 or t = 56.1°C 18. At normal temperature and pressure, the speed of sound in air is 332 m s–1. What will be the speed of sound in hydrogen (i) at normal temperature and pressure, (ii) at 819°C temperature and 4 atmospheric pressure? Given : air is 16 times heavier than hydrogen. Ans. (i) Let va and vh represent the speeds of sound in air and hydrogen respectively. va =  and 

356 Physics—XI Now, = But   = or    = 1328 m s–1 (ii) Pressure has no effect on the velocity of sound. =    or v819 = 2 × v0 = 2 × 1328 m s–1 = 2656 m s–1 19. A progressive wave of frequency 550 Hz is travelling with a velocity of 360 m s–1. How far apart are the two points 60° out of phase? Ans.  = 550 Hz, v = 360 m s–1,     If phase difference is 360°, then the path difference is . But the phase difference is  given to be 60°. The corresponding path difference is . So, the required distance is  i.e.,  Required distance = = 10.9 cm  20. A sinusoidal wave travelling along a string is described by : y (x, t) = 0.00327 sin (72.1x – 2.72t) All the numerical constants in this equation are in SI units. Calculate the amplitude, wavelength, period, frequency and speed of the wave. Ans. Comparing the given equation with   A = 0.00327 m = 3.27 mm Now, k = 72.1 rad m–1 or    or      8.72 cm  Again,  = 2.72 or   or  = 2.31 s Now,   = 0.433 Hz Again,    = 3.77 cm s–1

Waves 357 21. Two harmonic waves have the same displacement amplitude of 4 × 10 –5 cm and their angular frequencies are 500 rad s–1 and 5000 rad s–1 . Calculate (i) particle velocity amplitude, and (ii) particle acceleration amplitude. Ans. (a) displacement amplitude, A = 4 × 10–5 cm angular frequency,  = 500 rad s–1 (i) particle velocity amplitude = A = 4 × 10–5 × 500 cm s–1 = 2 × 10–2 cm s–1 (ii) particle acceleration amplitude = A2 = 4 × 10–5 (500)2 cm s–2 = 10 cm s–2 (b) displacement amplitude, A = 4 × 10–5 cm angular frequency,  = 5000 rad s–1 (i) particle velocity amplitude = A = 4 × 10–5 × 5000 cm s–1 = 0.2 cm s–1 (ii) particle acceleration amplitude = A2 = 4 × 10–5 × (5000)2 cm s–2 = 103 cm s–2 22. A steel wire 0.72 m long has a mass of 5.0 × 10–3 kg. If the wire is under a tension of 60 N, what is the speed of transverse waves in the wire? [NCERT Solved Example] Ans. Mass per unit length of wire, =   = 6.9 × 10–3 kg m–1 Tension, T = 60 N Speed of wave on the wire, v = =   m s–1 = 93.25 m s–1  23. A 100 cm long wire of mass 40 g supports a mass of 1.6 kg as shown in Fig. 7.9. Find the fundamental frequency of the portion of the string between the wall and the pulley. Take g = 10 m s–2. 80 cm 20 cm 1.6 kg Fig. 7.9 Ans. T = 1.6 kg wt = 1.6 × 10 = 16 N  =   = 0.04 kg m–1 L = (100 – 20) cm = 0.8 m = = Hz = 12.5 Hz  24. A wire of length 1.5 m under tension emits a fundamental note of frequency 120 Hz. (a) What would be its fundamental frequency if the length is increased by half under the same tension? (b) By how much should the length be shortened so that the frequency is increased three-fold?

358 Physics—XI Ans. (a)    Both T and  are constants.  L = constant   or         or   or    = 80 Hz (b) Now,    or   or   The wire should be shortened by (1.5 – 0.5) m, i.e., 1.0 m. 25. A sitar wire is under a tension of 40 N and the length between the bridges is 70 cm. A 5 m sample of the wire has a mass of 1.0 g. Deduce the speed of transverse waves on the wire, frequency of the fundamental and the frequencies of the first two harmonics. Ans. T = 40 N,     , L = 70 cm = 0.7 m   = 447.2 m s–1    = 319.4    = 638.8 Hz ;   = 958.2 Hz  26. In a sonometer cord of length 24.7 cm, resonance occurs with a tuning fork of frequency 256 Hz. Deduce the velocity of the waves over the cord. At what length will resonance be observed with tuning fork of frequency 453 Hz? Ans.          = 126.46 m s–1    = 13.95 cm         Aliter.      = 13.95 cm ;    27. The length of the sonometer wire between two fixed ends is 100 cm. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio of 1 : 2 : 3?

Waves 359 Ans. Let L1, L2 and L3 be the lengths of the three segments. Then,  Let    be the fundamental frequencies of the three segments. Now,       Applying the law of lengths,             L2 = 27.27 cm  L1 = 54.54 cm and L3 = 18.18 cm 28. If the velocity of sound in air at 0°C be 332 m s–1, find the shortest length in an open pipe that will be thrown into resonant vibrations by a tuning fork of frequency 256 Hz when the temperature of air is 50°C. Ans. We know that     or   or     Now,   or   = 0.71 m  29. A pipe 30.0 cm long, is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as 330 m s–1. [NCERT Solved Example] Ans. Frequency of the nth harmonic for an open pipe,   where n = 1, 2, 3, ......        Now, 550 n = 1100 or n=2 Clearly, the given source shall excite second harmonic of the open pipe 30. An air column with a tuning fork of frequency 256 Hz, gives resonance at column lengths 33.4 cm and 101.8 cm. Deduce (i) the end correction, and (ii) the speed of sound in air. Ans. (i)   =  cm  = cm = cm = 0.8 cm

360 Physics—XI (ii)        v = ×  = 256 × 1.368 m s–1 = 350.2 m s–1 31. A vessel in the form of a long circular cylinder was being filled with water from a tap at a uniform rate. A man observed that there were resonances at intervals of 100 s with a tuning fork of frequency 300 Hz. Calculate the rate of supply of water in cm3 per second. Given : radius of cylinder = 0.10 m and velocity of sound = 330 m s –1. Ans. Between two consecutive resonances, difference in lengths of air column =  Now, v =  or      Rate of supply of water = =    = 1.73 × 10–4 m3 s–1 32. Two sonometer wires of the same material and cross section have lengths 50 cm and 60 cm and are stretched by tensions of 2.25 kg wt and 2.56 kg wt respectively. If the number of beats heard (when the two wires are both vibrating) be 12 per second, find the mass per unit length of the wire. Take g = 10 m s–2. Ans. Using    twice, we get             Now,       or    = 12     or   or     or  = 0.0193 g cm–1 33. Two perfectly identical wires are in unison. When the tension in one wire is increased by 1%, then on sounding together, 3 beats are heard in 2 second. What is the initial frequency of each wire? Ans.    where the letters have usual meanings        Dividing,     or  or   = 300 Hz 

Waves 361 34. Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz? [NCERT Solved Example] Ans. Increase in the tension of a string increases its frequency. Assume that the original frequency B of B is greater than the frequency A of A. Further increase in B would increase the difference B – A. This should increase the beat frequency. But the beat frequency is observed to decrease. This shows that B < A. But A – B = 5 Hz and A = 427 Hz  B = 422 Hz Note. In this example, if by increasing tension in B, the beat frequency were to increase to, say, 8 Hz, no conclusion would be possible as to whether B > A or B < A. 35. The frequency of a string is 375 Hz. Its mass per unit length is 4 × 10–3 kg m–1 and tension is 360 N. If its next frequency is 450 Hz, then find out the mass of the string. Ans. v =  =   m s–1 = 300 m s–1  = , 375 = , 450 = 450 – 375 = or l =  = m=2m  = or m = l = 4 × 10–3 × 2 kg = 8 × 10–3 kg. 36. The frequencies of two sonometer wires are each equal to 500 Hz. What percentage change in tension should be done in one wire so that 5 beats/s are produced? Ans.  =  or  = k log  = log k + log T Differentiating,  =  or   = = or  × 100 = × 100 = 2. 37. A tuning fork of frequency 480 Hz resonates with a tube, closed at one end, of length 16 cm and diameter 5 cm in fundamental mode. Calculate velocity of sound in air. Ans.  =  v or v =  For fundamental mode, n = 1, leff. = l + 0.3d     m s–1 = 336 m s–1.  v=

362 Physics—XI 38. Two loudspeakers, as shown in Fig. 7.10, separated by a distance 1.5 m, are in phase. Assume that the amplitudes of the sound from the speakers is approximately same at the position of a listener, who is at a distance 4.0 m in front of one of the speakers. Speakers Listener Fig. 7.10 For what frequencies in the audio range (20 Hz to 20 kHz) does the listener hear minimum signal? Given that the speed of sound in air is 330 m s–1. Ans. The distance of the listener from the second speaker = (1.5 m × 1.5 m + 4 m × 4 m)1/2 = 4.27 m The path length difference = (4.27 – 4.0) m = 0.27 m For fully destructive interference, 0.27 m = (2n + l)/2 Hence   = 0.54 / (2n + 1) m The corresponding frequencies are given by  =   /0.54 s–1, = 661.1 (2n + 1) s–1, Therefore the frequencies in the audio range for which the listener would hear a minimum signal are 661 Hz, 1.983 kHz, 3.305 kHz, 4.627 kHz, 5.949 kHz,... 19.169 kHz. 39. A pilot of an aeroplane travelling horizontally at 198 km h–1 fires a gun and hears the echo from the ground after an interval of 3 second. If the speed of sound in air is 330 m s–1, find the height of the aeroplane from the ground. Ans. 488 m Hint. = 3 or x = 495 m t A MB h xx Fig. 7.11 AM =

Waves 363 40. A and B are two sources producing 8 beats/s when sounded together. A is in resonance with a water column of 37.5 cm, B is in resonance with another water column of 38.5 cm. Calculate the natural frequencies of A and B. Ans. A – B = ± 8 Hz  Water column is analogus to closed pipe.  l= or  = 4l = 4(38.5) cm = m  Hz, B =  Hz A =   Now, B – A = 8;  –  =8  or     =8 or 25v    =8        or v =   m s–1 = 462 m s–1  A =  Hz = 300 Hz, B =  Hz = 308 Hz.   41. A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length  is given. Find the total energy of the string. x=0 x=l Fig. 7.12 Ans. The equation of standing wave in a string is y = A sin kx cos t (y) = – A sin kx sin t or vmax. = A sin kx Energy =    =   =           42. A string 25 cm long and having a mass of 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m s–1, find the tension in the string. Ans.  =   kg m–1 = 0.01 kg m–1, l = 0.25 m   The frequency of first overtone of string, 2 = = ...(1) 

364 Physics—XI The fundamental frequency of a closed pipe, 1 = = Hz = 200 Hz  Number of beats heard per second = 8 | 2 – 1 | = ± 8 or 2 = 1 ± 8 or 2 = 200 ± 8 = 208 Hz or 192 Hz Here it is given that decreasing the tension, beat frequency is decreased, therefore 2 = 208 Hz From (1), 208 = or T = 27.04 N. 43. A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area 10 –6 m2 is rigidly fixed at both ends. The temperature of wire is lowered by 20°C. If transverse waves are set up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration. Young’s modulus of steel = 2 × 1011 N m–2, coefficient of linear expansion of steel = 1.21 × 10–5 °C–1. Ans. L = L  T = 2.42 × 10–4 m Y = ×  or  T = = 48.4 N Now, =  = 11 Hz 44. A copper wire is held at the two ends by rigid supports. At 30°C, the wire is just taut with negligible tension. Find the speed of transverse waves in this wire at 10°C. Given that  = 1.7 × 10–5 °C–1, Y = 1.3 × 1011 N m–2 and  = 9 × 103 kg m–3. Ans. L = L  T = 3.4 × 10–4 L T =  = 4.42 × 107 a Also,  = 1 × a ×  = 9 × 103 a Now, v =  = 70.08 m s–1 45. A pipe of length 1.5 m closed at one end is filled with a gas and it resonates in its fundamental mode with a tuning fork. Another pipe of same length but open at both ends is filled with air and it resonates in its fundamental mode with the same tuning fork. Calculate the velocity of sound at 0°C in the gas. Given : velocity of sound in air is 360 m s–1 at 30°C. Ans. Frequency of the fundamental note emitted by an open pipe, 1 = = 120 Hz

Waves 365 Frequency of the fundamental note emitted by closed pipe, 2 =  =   As 1 = 2 ,  120 = or v = 720 m s–1 If the velocity of sound in the gas at 0°C = v0 , then  or  or v0 = 683.4 m s–1  46. A column of air at 51°C and a tuning fork produce 4 beats s–1, when sounded together. As the temperature of air column is decreased, the number of beats per second tends to decrease and when temperature is 16°C, the two produce 1 beat s–1. Find the frequency of tuning fork. Ans. The frequency of air column,  =  or  v Now,  or  = 1.06  51 > 16      Frequency of air column must be greater than the frequency of the tuning fork. Let f be the frequency of tuning fork,  51 = f + 4 and 16 = f + 1 or  = 1.06  On solving, f = 49 Hz 47. The speed of longitudinal waves in water is 1400 m s–1. Calculate (i) the bulk modulus of water (ii) the intensity of the wave if the amplitude of the wave is 0.01 mm and its frequency is 512 Hz (iii) the energy density. Ans. (i) v= B  or or B = v2 B = 1400 × 1400 × 1000 N m–2 = 1.96 × 109 N m–2 (ii) I = 222a2v =        × 1400 × 1000 W m–2   = 725 W m–2 (iii) Energy density =   = 0.52 J m–3 48. A thin steel wire has been stretched so that its length increases by 1%. Calculate the frequency of one metre length of the wire. Given : Young’s modulus for steel = 20 × 1010 N m–2 and density of steel = 7800 kg m–3. Ans. Y =   

366 Physics—XI = =    = = 253.18 Hz 49. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? Ans. Tension, T = 200 N; Length, l = 20.0 m ; Mass, M = 2.50 kg Mass per unit length,  =   Wave velocity, v=   or    v= Time, t=  = 0.5 s. 50. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top? Given that the speed of sound in air is 340 m s–1? (g = 9.8 m s–2) Ans. Time after which the splash is heard at the top is equal to the sum of the time t1 taken by the stone to fall down and the time t2 taken by the sound to travel from bottom to top. Using  we get (... u = 0 and a = g) or    Again,    Total time, t1 + t2 = (7.82 + 0.88) s = 8.7 s. 51. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20°C = 343 m s–1? Ans. v = 343 m s–1, l = 12.0 m; M = 2.10 kg mass/length  =   Now, v = or T = v2   = 343 × 343 × 0.175 N = 2.06 × 104 N.

Waves 367 52. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all : (a) y = 2 cos (3x) sin (10t) ; (b) y = 2 x – vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t Ans. (a) It represents a stationary wave. (b) It is unacceptable function for any wave. So, it does not represent either a travelling wave or a stationary wave. (c) It represents travelling harmonic wave. (d) It is superposition of two stationary waves. 53. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string? Ans. Mass of wire, M = 3.5 × 10–2 kg Linear density,  = mass/length = = 4.0 × 10–2 kg m–1  Length of wire, l =      m = 0.875 m In the fundamental mode,  = 2l = 2 × 0.875 m = 1.75 m (a) speed of transverse waves, v =  = 45 × 1.75 m s–1 = 78.75 m s–1 (b) v =  or T = v2 = 4 × 10–2 (78.75)2 N = 248.06 N. 54. A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel? Ans.      or           Fig. 7.13

368 Physics—XI 55. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B? Ans. Let A and B be the frequencies of strings A and B respectively. Number of beats/second = 6  B = A ± 6 = 324 ± 6 = 330 Hz or 318 Hz Since the frequency is directly proportional to square root of tension, therefore on decreasing the tension in the string A, its frequency A will be reduced i.e., number of beats will increase if B = 330 Hz. This is not so because number of beats becomes 3.  It is concluded that the frequency B = 318 Hz because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with B = 318 Hz. 56. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) wavelength, (ii) frequency, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to or 0.05 Hz? Ans. (a) The pulse does not have a definite wavelength or frequency but has a definite speed of propagation (in a non-dispersive medium). (b) No, the frequency of the note produced by whistle is not or 0.05 Hz. The frequency of the note will depend on the length of the whistle and other factors. 57. The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. 7.14. The velocity of two waves superimposing to form stationary waves is 360 m s–1 and their frequencies are 256 Hz. B B D D x in m Displacement A A C C E t=0 t=? x in m Fig. 7.14 (a) Calculate the time at which the second curve is plotted. (b) Mark nodes and antinodes on the curve. (c) Calculate the distance between A and C.

Waves 369 Ans. (a) The second curve represents particles passing simultaneously through mean position. This happens earliest at t =   s = 9.8 × 10–4 s  (b) Nodes are at points A, B, C, D, E. Antinodes are at points A, B, C, D. (c) Distance between A and C=     58. If c is rms speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases. Ans. c = , v =  or =  =    Since  is constant and independent of temperature,  = constant and independent of temperature For diatomic gases,  = 1.4  = 1.46 = constant 59. Given below are some functions of x and t to represent the displacement of an elastic wave. (i) y = 5 cos (4x) sin (20t) (ii) y = 4 sin (5x – t/2) + 3 cos (5x – t/2) (iii) y = 10 cos [(252 – 250) t] cos [(252 + 250) t] (iv) y = 100 cos (100 t + 0.5x) State which of these represent (a) a travelling wave along –x direction (b) a stationary wave (c) beats (d) a travelling wave along +x direction. Given reasons for your answers. Ans. (a) Eqn. (iv) represents a wave travelling along –x direction. (b) Eqn. (i) represents a stationary wave because the terms containing x and t are independent. (c) Eqn. (iii) represents beats involving sum and difference of two frequencies 252 Hz and 250 Hz. (d) Eqn. (ii) represents a wave travelling along +x direction.

370 Physics—XI LONG ANSWER TYPE QUESTIONS (5 MARKS) 1. Write the difference between longitudinal and transverse waves. Ans. Distinction between longitudinal and transverse waves Longitudinal Waves Transverse Waves 1. The particles of the medium vibrate along 1. The particles of the medium vibrate at right the direction of propagation of the wave. angles to the direction of propagation of the wave. 2. The longitudinal waves travel in the form 2. The transverse waves travel in the form of of alternate compressions (condensations) alternate crests and troughs. One crest and and rarefactions. One compression and one one trough constitute one wave. rarefaction constitute one wave. 3. These waves can be formed in any medium 3. These waves can be formed in solids and (solid, liquid or gas). on the surfaces of liquids only. 4. When longitudinal waves propagate, there 4. When transverse waves propagate, there are are pressure changes in the medium. no pressure changes in the medium. 5. These waves can be only indirectly 5. These waves can be directly represented represented by a sine curve. by a sine curve. 6. These cannot be polarised. 6. These can be polarised. 2. Derive Newton’s formula for velocity of sound in air. What is Laplace’s correction? Discuss the changes brought about by him Ans. Newton’s formula for the velocity of sound waves in air Newton assumed that sound waves travel in air under isothermal conditions, i.e., temperature remains constant. So, the changes in pressure and volume obey Boyle’s law.  PV = constant Differentiating, PdV + VdP = 0 or PdV = –VdP or  = (isothermal) elasticity Bi Now, v= =  which is Newton’s formula for the velocity of sound waves in air or in a gas. Let us apply this formula to calculate the velocity of sound in air at NTP. At NTP, density  of air = 1.293 kg m–3 and pressure, P = 0.76 m of Hg column = 0.76 × 13600 × 9.8 Nm–2 (... P = hdg and dHg = 13600 kg m–3)       This value is nearly 16% less than the experimental value of 332 m s–1. This discrepancy could not be satisfactorily explained by Newton.

Waves 371 Laplace’s correction Laplace, a French mathematician, suggested that sound waves travel in air under adiabatic conditions and not under isothermal conditions as suggested by Newton. He gave the following two reasons for this. (i) When sound waves travel in air, the changes in volume and pressure take place rapidly. (ii) Air or gas is a bad conductor of heat. Due to both these factors, the compressed air becomes warm and stays warm whereas the rarefied air suddenly cools and stays cool. For adiabatic changes in pressure and volume, PV  = constant On differentiation,     or   =  ,  where Ba is adiabatic elasticity. Now, v= =   which is Laplace’s corrected formula for velocity of sound waves in air or gas. Again,         This result agrees very well with the experimental value of 332 m s–1. This establishes the correctness of Laplace’s formula. 3. What do you mean by progressive wave? Derive the equation of progressive wave. Ans. A progressive wave is one which travels in a given direction with constant amplitude, i.e., without attenuation. Let us consider transverse wave motion. However, the treatment is valid for longi- tudinal wave motion also. Let a plane wave originate at O as shown in Fig. 7.15 Let it proceed from left to right in an elastic medium. The particles of the medium shall execute SHM of the same amplitude and time period about its mean position. Let us count time from the instant the particle at O crosses its mean position in the positive direction of Y-axis. The displacement y of the particle at any time t is given by y(0, t) = A sin t where A and  represent the amplitude and angular frequency respectively of simple har- monic motion executed by the particle at O. Since the disturbance is handed over from one par- B ticle to the next therefore there is a gradual fall in O phase from left to right, i.e., in the direction of mo- tion. Let the phase of particle at P lag behind the P phase of particle at O by . Then, the displacement of particle at P at any time t is given by x y(x, t) = A sin (t – ) ...(1) Fig. 7.15. Plane progressive wave

372 Physics—XI At B, which is one wavelength  apart from O, the phase difference is 2. In other words, particles at O and B have the same phase of vibration. At a distance , the phase changes by 2. At a distance x, the phase changes by       where x is the distance of P from O. From equation (1),       ...(2) Now,           here v is called the wave velocity or phase velocity. From equation (2),         or   ...(3)  Also,          or       ...(4)   Again, from equation (2),   ...(5)      4. A wave travelling along a string is described by, y(x, t) = 0.005 sin (80.0 x – 3.0 t), in which the numerical constants are in SI units (0.005 m, 80.0 rad m–1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s? [NCERT Solved Example] Ans. On comparing the given displacement equation with   we find (a) the amplitude of the wave is 0.005 m = 5 mm (b) the angular wave number k and angular frequency  are k = 80.0 rad m–1 and  = 3.0 rad s–1 We then relate the wavelength  to k through      = 7.85 cm  (c) Now we relate T to  by the relation T = 2/    = 2.09 s and frequency,  = 1/ T = 0.48 Hz

Waves 373 The displacement y at x = 30.0 cm and time t = 20 s is given by     = 5 mm 5. A displacement wave is represented by y = 0.25 × 10–3 sin (500t – 0.025 x), where y, t and x are in cm, second and metre respectively. Deduce (i) the amplitude (ii) the period (iii) the angular frequency (iv) the wavelength. Deduce also the amplitude of particle velocity and particle acceleration. Ans. The general equation of a displacement wave is          Comparing this equation with the given equation, we get (i) amplitude, A = 0.25 × 10–3 cm  (ii) Period,  (iii) angular frequency,  = 500 rad s–1 (iv)     = 80 m  Amplitude of particle velocity, i.e., maximum particle velocity = A = 0.25 × 10–3 × 500 cm s–1 = 0.125 cm s–1 Amplitude of particle acceleration i.e., maximum particle acceleration = A2 = 0.25 × 10–3 × (500)2 cm s–2 = 62.5 cm s–2 6. What are stationary waves? Explain various types of stationary waves. Ans. Whenever two progressive waves of the same wavelength and amplitude travel with the same speed through a medium in opposite directions and superpose upon each other, they give rise to waves called stationary or standing waves. When two sets of progressive wavetrains having the same frequency and amplitude travel along the same path but with exactly equal and opposite velocities, the result of their superposition is a set of waves which only expand and shrink but do not proceed in any direction. These waves are called standing or stationary waves. In these waves, there is no progressive propagation of energy in any direction. There is no propagation of the resultant wave in any direction. Only the amplitude of vibrations or velocity changes periodically with time. Stationary waves are of two types : (i) Transverse stationary waves. These are formed due to the superposition of two progressive transverse waves. Example. The stationary waves produced in the vibrating string of a sonometer. (ii) Longitudinal stationary waves. These are formed due to the superposition of two progressive longitudinal waves. Example. The stationary waves formed in the vibrating air columns in the pipes.

374 Physics—XI 7. Distinguish between stationary and progressive waves. Ans. Comparison between stationary and progressive waves Progressive Waves Stationary Waves 1. The disturbance progresses onwards ; 1. The disturbance is stationary, there being it being handed over from particle to no forward or backward movement of the particle. Each particle executes the wave. Each particle has its own vibration same type of vibration as the preceding characteristics. one, though at a different time. 2. The waves have the appearance of a sine/ 2. The waves are in the form of crests and cosine function, which shrink to a straight troughs, i.e., sine/cosine functions, line, twice in each vibration. It never which move onwards with a definite advances. velocity. 3. Every particle has a fixed allotted 3. Every particle has the same amplitude; amplitude. Some have zero amplitude which it attains in its own time (nodes) and some have maximum depending upon the progress of the amplitude (antinodes) always. Each wave. particle attains this at the same given moment. 4. The phase of every particle varies 4. All the particles in one-half of the waves continuously from 0 to 2. have a fixed phase and all the particles in the other half of the wave have the same 5. No particle remains permanently at phase in the opposite direction rest. Twice during each vibration, the simultaneously. particles are momentarily at rest. Different particles attain this position 5. There are particles which are at different times. permanently at rest (nodes) and all other particles have their own allotted maximum displacement, which they attain simultaneously. These particles are momentarily at rest twice in each vibration, all at the same time. 6. All the particles have the same 6. All the particles attain their individual maximum velocity which they attain allotted velocities depending upon their one after another, as the wave positions, simultaneously. Two particles advances. (nodes) in one waveform have zero velocities all the time. 7. There is a regular flow of energy across every plane along the direction of 7. There is no flow of energy at all, across propagation of the wave. The average any plane. Each particle has its own allotted energy in a wave is half potential and individual energy. They all attain their half kinetic. values of P.E. at one time and all energy becomes K.E. at another given time.

Waves 375 8. A sonometer wire made of silver alloy of density 10.8 g cm–3 and of diameter 0.25 mm is kept in a state of tension by hanging a load W at its end. When the distance between the bridges of the sonometer is 30 cm, it gives out a vibration whose fundamental frequency is 300 Hz. What is the changed value of the frequency in each of the following cases? (i) The distance between the bridges is increased to 90 cm. (ii) The load is increased to 9 W. (iii) The wire is replaced by a wire of steel of diameter 1.0 mm. (iv) The wire is replaced by a wire of the same diameter (i.e., 0.25 mm) but made of aluminium of density 2.7 g cm–3. Ans. (i)    , Here, T and  are constants.     or  = 100 Hz   (ii)    or   or 2 = 900 Hz     or  (iii)  =  or  or    or   or   = 75 Hz    (iv) or  or   = = 300 × 2 Hz = 600 Hz 9. What are beats? Explain the formation of beats analytically. Ans. When two sounding bodies of nearly the same frequency and same amplitude are sounded together, the resultant sound comprises of alternate maxima and minima. The phenomenon of alternate waxing and waning of sound at regular intervals is called beats. The number of beats heard per second is called beat frequency. It is equal to the difference in the frequencies of sounding bodies. Beats are heard only when the difference in frequencies of two sounding bodies is not more than ten. This is due to persistence of hearing. The time from each loud sound to the next loud sound is called one beat-period.

376 Physics—XI Analytical treatment of beats Consider two harmonic sound waves of nearly equal frequencies 1 and 2. The peri- odic dips in sound, called beats, will occur with a frequency equal to (1 – 2). Let a be the amplitude of each wave. Let us count time from the instant when the two sound waves are in the same phase. The displacements s1 and s2 at a point due to the two waves are given by   and   For the sake of simplicity, it is assumed here that there is no initial phase difference between the two wave trains. It is further assumed that the waves propagate over long distances so that the boundary effects can be neglected. Applying the principle of superposition of waves,  or     or s  a( cos 21t  cos 22 t) s  a 2 cos 21t  22 t cos 21t  22 t 2 2  or or s  2a cos 2 (1  2 ) t cos 2 (1  2 ) t 22 or s  2a cos 2  1  2  t cos 2  1  2  t   2    2  or s  A cos 2  1  2  t  2         is the amplitude of the resultant wave. It may be where       noted that the frequency of the resultant wave is the average of the frequencies 1 and 2 of the superposing wave trains. The amplitude A of the resultant wave is a function of time. A varies between + 2a and –2a. The amplitude A is maximum, i.e., + 2a or – 2a when       or      where n = 0, 1, 2,....... or (1  2 )t  n or (1  2 )t  n n 123 t   0, , , ,.... or 1  2 1  2 1  2 1  2 So, the time interval between two successive maxima is   Similarly, the amplitude A is minimum (zero) when      or     



ISBN: 978-93-93738-13-4 789393 738134 T11-8897-449-COMP.CBSE QB PHY T-II XI