Title 16. Waves and Sound Med Ans.pdf ✅

1. (c) Ultrasonic waves are used in sonography. 2. (a) Light, X – rays and require medium for their waves. They do not require medium for their propagation. They can travel through vacuum, Seismic waves are mechanical waves. They require medium for their propagation. They do not travel in vacuum. 3. Ultrasonic waves produced by a vibrating quartz crystal are longitudinal. 4. (a) Sound waves travel fastest in solids. 5. (c) A transverse wave travels through a medium in the form of crests and troughs. 6. (d) In consecutive compressions and rarefactions there is no transfer of heat. 7. (a) In a transverse wave, the particles of medium vibrate in a directions perpendicular to the directions of the propagation. 8. (c) Resultant amplitude 3 4 9 16 5cm 2 2 = + = + = 9. (d) In the given figure, point C and E are in the same phase. 10. (a) During propagations of a plane progressive mechanical wave all the particles are vibrating with different phases. While all other statement are correct. 11. (d) The propagation constant of a wave is also called its angular wave number. 12. (d) The basic requirement for a wave functions to represent a travelling wave is that for all values of x and t wave functions must have a finite value. Out of the given functions for y no one satisfied this conditions. Therefore none can represent a travelling wave. 13. (a) The given equations of a stationary wave is x cos40 t 3 2 y 5sin   = ....(i) The standard equations of a stationary wave is y = 2a sinkx cost ....(ii) Comparing (i) and (ii), we get 3 2 k  = Or or 3cm 3 2 3 2  =  =  Separations between two adjacent nodes is 1.5cm 2 3 2 = =  14. (b) Since the two waves are identical, they have same amplitudes. The amplitude of the resultant wave is 2 A 2a cos  = Here, a = 10 mm,  =120 A = 2 (10 mm) 2 120 cos  mm 10mm 2 1 = 210 = 15. (d) The speed of sound in a gas does not depends upon pressure of the gas, till temperature remains constant i.e., speed remains the same whatever be the pressure. Therefore, graph (d) is correct. 16. (b) Here, A =100mm = 0.10m B = 0.25m 1 1 vA 80cms 0.80ms − − = = As the frequency of the wave remains same in the two media. B B B vA v  =   = 0.80 0.10 0.25 v vA A B B  =     = 1 VB 2ms−  = 17. (c) Velocity of the wave, v =  Or   = v 18. (a) Sound waves are mechanical waves. These waves require a medium for their propagations. They cannot propagate through vacuum. 19. (c) Speed of sound in air is M RT v  = Where T is the absolute temperature. Since  and M are constant.  v  T 273 273 t v v 0 t +  = 273 273 t v 3v 0 0 + = Squaring both sides, we get
or 2457 273 t or t 2184 C 273 273 t 9 = + =  + = 20. (a) Newton assumed that sound propagation in a gas taken under isothermal condition. 21. (b) Here, 6 6 1 3.2MHz 3.2 10 Hz 3.2 10 s −  = =  =  1 3 1 v 1.6kms 1.6 10 ms − − = =  Wavelength, 6 1 3 1 3.2 10 s v 1.6 10 ms − −   =   = 0.5 10 m 0.5mm 3  = − 22. (c) Here, Length, L=10m Mass, M 5g 5 10 kg −3 = =  Tension, T = 80N Mass per unit length of the wire is 4 1 3 5 10 kg m 10m 5 10 kg L M − − − =    = = Speed of the transverse wave on the wire is 4 1 5 10 kg m T 80N v − −  =  = 2 1 1 4 10 ms 400ms − − =  = 23. (d) 1 Mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is Volume of onemoleof air at STP Mass of onemoleof air  = 3 3 3 3 1.29kg m 22.4 10 m 29 10 kg − − − =   = At STP, P = 1 atm = 5 2 1.01 10 Nm−  According to Newton’s formula, the speed of sound in air at STP is 1 3 3 2 280ms 1.29 kgm P 1.01 10 Nm v − − − − =   =         = 24. (d) The give progressive wave is y = 5sin(100t − 2x) Particle velocity, 500 sin(100 t 2 x) dt dy vp = =   −  1 vPmax 500 ms− =  25. (a) rms speed of gas molecules is  3P vrms ..... (i) Speed of sound in the gas is   = P v ..... (ii) Divide (i) by (ii),we get  = 3 v vrms 26. (b) Speed of transverse wave on a stretched string is  = T v Where T is tension of the string and  is the mass per unit length of the string. 27. (b) The frequency of the wave in the string is equal to the frequency of the sound produced. 28. (a) Speed of sound wave in a fluid is  = B v Where B is the bulk modulus and  is the density of the medium. 29. (d) Given : Frequency of the wave,  =150Hz From the figure 20cm 2 5  =  Wavelength of the wave, cm 8cm 5 40  = = 8 10 cm 0.08m 2 =  = − Velocity of the wave, v =  1 2 1 (150s )(8 10 m) 12ms − − − =  = 30. (a) Speed of sound in water is 1 3 6 1450ms 10 B 2100 10 v − =  =  = Where B is the bulk modulus and  is the density of water. Here, B 2100MPa 2100 10 Pa 6 = =  3 3 10 kgm −  = 1 3 6 1450ms 10 2100 10 v − =   = 31. (d) The given transverse harmonic wave equations is        = + + 4 y 3sin 36t 0.018x ..... (i) As there is positive sign between t and x terms therefore the given wave is travelling in the negative x directions. The standard transverse harmonic wave equations is y = a sin(t + kx + ) ..... (ii)
Comparing (i) and (ii) we get 1 1 a 3cm, 36rads ,k 0.018radcm − − = = =  Amplitude of wave, a = 3 cm Frequency of the wave, Hz 18 2 36 2  =  =    = Velocity of the wave 1 1 1 1 2000cms 20ms 0.018rad cm 36rads k v − − − − = = =  = 32. (c) Let 1 2 v , v be speed of S waves and P waves and 1 2 t ,t be the time by these waves to travel to reach the seismograph. Let the epicenter of the earthquake is located at a distance d from the seismograph. Then 1 1 2 2 d = v t = v t ...... (i) Here, 1 2 1 v1 4kms and v 8kms − − = = 1 2 1 2 4t = 8t  t = 2t ...... (ii) And t t 4min 240s 1 − 2 = = 2t 2 − t 2 = 240 [Using (ii)] Or t 240s 2 = Put in Eq. (ii), we get t 2 240s 480s 1 =  = From Eq. (i) We get D = (4 km 1 s − ) (480 s) = 1920km 33. (c) Velocity of sound in gas M RT v  = Where the symbols have their usual meanings, At the same temperature. M v   5 42 2 4 5 3 5 7 M M H H vH vH H2 He 2 2 2 2  =   =    = 34. (c) 517 Hz 35. (a) Velocity of a wave is independent of others. 36. (c) The given wave equations is y = Asin(t − kx) Wave velocity, k v  = .... (i) Particle velocity, A cos( t kx) dt dy vp = =   − Maximum particle velocity, (vP )max = A .... (ii) According to the given questions p max v = (v ) =   A k (Using (i) and (ii)) A 2 A or k 1 =   =         = 2 k  = 2A 37. (c) Here, 1 brass 1 vair 350ms , v 3500ms − − = = When a sound wave goes from one medium to another medium, its frequency (  ) remains the same. Frequency,   = = v Wavelength Velocity Since  remains the same in both the media, so brass brass air Vair V  =  Or air air air brass brass air 10 350 3500 v v  =   =   =  38. (c) Let the man M be at a distance x from hill H1 and y from hill H2 as shown in figure, Let y  x. The time interval between the original sound and echoes from H1 and H2 will be respectively v 2y and t v 2x t 1 = 2 = Where v is the velocity of sound. The distance between the hills is [1 2] 510m 3 340 [t t ] 2 v x y + = 1 + 2 = + = 39. (b) y a sin(kx t) 1 = − y a sin(kx t) 2 = + According to the principle of superposition, the resultant wave is y = y1 + y2 = a sin(ks −t) + a sin(kx +t) Using trigonometric identity sin(A+B) +sin(A−B) = 2sinAcosB We get y = 2a sinkx cost 40. (c) Here , L = 100cm 1 1m,v 5kms− = = 3 1 5 10 ms− =  As the rod is clamped at the middle, therefore the middle point is a node. IN the fundamental mode, the antinode is formed at each end as shown in figure.
Therefore, the distance two consecutive antinodes = L =1m But the distance between two consecutive antinodes is . 2  1m or 2m 2 =  =   The frequency of the fundamental mode is 2.5 10 Hz 2.5kHz 2m v 5 10 m s 3 3 1 =  =  =   = − 41. (d) When sound wave is refracted from air to water its frequency remains unchanged while all other given parameters changed. 42. (b) Here, Speed of sound, v = 1 330ms− Length of pipe, L 30cm 30 10 m −2 = =  In a open pipe (open at both ends), the frequency of its th n harmonic is 2L nv n = where n = 1,2,3,....... v 2L n n  = Let th n harmonic of open pipe resonate with 1.1kHz Source. 1.1kHz 1.1 10 Hz 3 n = =  2 330 2 30 10 1.1 10 n 2 3 =      = − 43. (c) Here, 100kHz 100 10 Hz 3  = =  5 5 1 10 Hz 10 s − = = 1 w 1 va 340ms , v 1500ms − − = = Frequency of both the reflected and transmitted sound remains unchanged. Wavelength of reflected sound. 34 10 m 10 s v 340ms 4 5 1 1 a a − − − = =    = 3.4 10 m 3.4mm 3 =  = − Wavelength of transmitted sound, 15 10 m 15mm 10 s v 1500ms 3 5 1 1 w w = =  =   = − − − 44. (a) The phenomenon of echo is an example of reflections by a rigid boundary. 45. (c) All the particles are oscillating with same amplitude. 46. (b) When the string vibrates in loops n, its frequency is 2L nv n = Where L is the length of the string and v is the velocity of the wave.  When the string fixed at its both ends vibrates in 1 loop, 2 loop, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4. 47. (b) Here, speed of sound, 1 v 340ms− = Length of pipe, L 17cm 17 10 m −2 = =  In a closed pipe (open at one end), the frequency of its th n harmonic is where n 1,3,5,7,....... 4L nv n = = Let th n harmonic of closed pipe resonates with 1.5kHz source. 1.5 10 Hz 3 n =  4L nv 1.5 103   = 340 1.5 10 4 17 10 n 3 −2     = N = 3 48. (d) End correction 2 L2 − 3L1 = 2 101.8 − 333.4 = 0.8cm 2 1.6 = = Speed of sound v 2 (L L ) =  2 − 1 = 2256(1.018−0.334) 1 2 256 0.684 350.2ms− =   = 49. (b) At the node the pressure change in maximum while the displacement is minimum. 50. (b) The distance between any two consecutive antinodes or nodes is . 2  Hence, options (a) is an incorrect statement. The distance between a node and adjoining antinode is . 4  Hence, options (b) is a correct statement. In the open end is an antinode. Hence, option (c) is an incorrect statement. In the closed end is a node. Hence, option (d) is an incorrect statement. 51. (c) An organ pipe of length L open at both ends vibrates with frequencies given by 2L nv n = where n = 1,2,3,4,......... The fundamental frequency corresponds to n = 1 and is given by