320 + Interview Questions in SOUND-WAVES IN GENERAL AWARENESS Page 2 InterviewSolution

51.	In a stationary wave system, all the particles of the mediumA. have zero displacement simultaneously at some instantB. have maximum displacement simultaneously at some instantateC. are at rest simultaneously at some instantD. sreach maximum velocity simultaneously at some instant
Answer» Correct Answer - A::B::C::D

Discussion

52.	Why is a stationary wave so named?
Answer» Correct Answer - A sationary wave is so named because there is no propagation of energy.

Discussion

53.	A metal string is fixed between rigid supports. It is initially at negligible tensin. Its Young modulus is `Y`, density `rho` and coefficient of thermal expansion is `alpha`. If it is now cooled through a temperature `=t`, transverse waves will move along it with speedA. `Ysqrt(alphat//rho)`B. `alphatsqrt(Y//rho)`C. `sqrt(Y alpha t//rho)`D. `t sqrt(Yalpha//rho)`
Answer» Correct Answer - C

Discussion

54.	A man of mass `50 kg` is runing on a plank of mass `150 kg` with speed of `8 m//s` relative to plank as shows in the figure (both were initially at rest and velocity of man with respect to ground any how remains constant). Plank is placed on smooth horizontal surface. The man, while runing, whistle with frequency `f_(o)`. `A` detected `(D)` placed on plank detects frequency. The man jumps off with same velocity (w.r.t. to groung) from point `D` and slides on the smooth horizontal surface [Assume coefficient of friction between man and horizontal is zero]. The speed of sound in still medium is `330 m//s`. Answer the following questions on the basis of above situations. The frequency of sound detected by `D`, after man jumps off the plank isA. `(332)/(324) f_(o)`B. `(330)/(322) f_(o)`C. `(328)/(336) f_(o)`D. `(330)/(338) f_(o)`
Answer» Correct Answer - C `underset(6 m//s)(larr_S)` `underset(2 m//s)(O rarr)` `f_(2) = f_(o) ((nu - nu_(o))/(nu + nu_(s)))` `= f_(o) ((330 - 2)/(330 + 6)) = ((328)/(336)) f_(o)`

Discussion

55.	A man of mass `50 kg` is runing on a plank of mass `150 kg` with speed of `8 m//s` relative to plank as shows in the figure (both were initially at rest and velocity of man with respect to ground any how remains constant). Plank is placed on smooth horizontal surface. The man, while runing, whistle with frequency `f_(o)`. `A` detected `(D)` placed on plank detects frequency. The man jumps off with same velocity (w.r.t. to groung) from point `D` and slides on the smooth horizontal surface [Assume coefficient of friction between man and horizontal is zero]. The speed of sound in still medium is `330 m//s`. Answer the following questions on the basis of above situations. The frequency of sound detected by detector `D`, before man jumps of the plank isA. `(332)/(324) f_(o)`B. `(330)/(322) f_(o)`C. `(328)/(336) f_(o)`D. `(330)/(338) f_(o)`
Answer» Correct Answer - A `O_underset(Plank)(rarr v)` `8-v_underset(Man)(larr)S` From conservation of linear momentum, `50 (8 - nu) = 150nu` :. `nu = 2 m//s` `8 - nu = 6 m//s` `f_(1) = f_(o) ((nu + nu_(o))/(nu - nu_(s)))` `= f_(o) ((330 + 2)/(330 - 6)) = (332)/(324) f_(o)`

Discussion

56.	A man of mass `50 kg` is runing on a plank of mass `150 kg` with speed of `8 m//s` relative to plank as shows in the figure (both were initially at rest and velocity of man with respect to ground any how remains constant). Plank is placed on smooth horizontal surface. The man, while runing, whistle with frequency `f_(o)`. `A` detected `(D)` placed on plank detects frequency. The man jumps off with same velocity (w.r.t. to groung) from point `D` and slides on the smooth horizontal surface [Assume coefficient of friction between man and horizontal is zero]. The speed of sound in still medium is `330 m//s`. Answer the following questions on the basis of above situations. Choose the correct plot between the frequency the frequency detected by detector usrsus position of the man relative to detector.A. B. C. D.
Answer» Correct Answer - A `f_(1) gt f_(o)` but `f_(1) =` constant. Similarly, `f_(2) lt f_(o)` but `f_(2) =` constant

Discussion

57.	When two waves with same frequency and constant phase differenc interfere,A. there is a gain of energyB. there is a loss of energyC. the energy is redistributed and the distribution changes with timeD. the energy is redistributed and the distribution remains constant in time.
Answer» Correct Answer - D

Discussion

58.	An organ pipe, open at both ends, containsA. longitudinal stationary wavesB. longitudinal travelling wavesC. transverse stationary wavesD. transverse travelling waves
Answer» Correct Answer - A

Discussion

59.	An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximumA. at the two endsB. at the middle of the pipeC. at distances L/4 inside the endsD. at distance L/8 inside the ends
Answer» Correct Answer - B

Discussion

60.	An open organ pipe of length L vibrates in its fundamental mode. The pressure variation is maximumA. endsB. middle of pipeC. `(L)/(4)` from centreD. `(3L)/(8)` from centre
Answer» Correct Answer - B

Discussion

61.	A closed organ pipe and an open organ pipe are tuned to the same fundamental frequency. The ratio of their lengths isA. `1 : 2`B. `2 : 1`C. `1 : 4`D. `4 : 1`
Answer» Correct Answer - A `(nu) / (2l_(o)) = (nu) / (4l_(c)) rArr l_(c) : l_(o) = 1:2`

Discussion

62.	Two organ pipes are emitting their fundamental notes, when each closed at end, give 5 beat/s if their foundamental frequencies are 250 Hz and 255 Hz. Then find the ratio of their lenghts.A. `49/50`B. `49/51`C. `50/51`D. `51/50`
Answer» Correct Answer - C

Discussion

63.	What is the amplitude of motion for the air in the path of a `60 dB` , `800 H_(Z)` sound wave? Assume that `rho_(air)= 1.29 kg//m^(3)` and `upsilon = 330 m//s`.
Answer» Correct Answer - A::C `L = 10log_(10)((I)/(I_(o)))` `60 = 10 log_(10) (I//10^(-12))` Solving, we get `I = 10^(-6) W//m^(2)` Now, using the result derived in above problem. `A = sqrt((2l)/(rho (2 pi f )^(2)nu))` `= sqrt((2 xx10^(-6))/(1.29 (2pi xx800)^(2)(330)))` `= 1.36 xx10^(-8) m` `= 13.6 nm`

Discussion

64.	Two narrow organ pipes, one open (length `l_(1)`) and the other cloed (length `l_(2)`) are sounded in their respective fundamental modes. The beat frequency heard is `5 Hz`. If now the pipes are sounded in their first overtones, then also the beat frequency heard is `5 Hz`. Then:A. `(l_(1))/(l_(2)) = (1)/(2)`B. `(l_(1))/(l_(2)) = (1)/(1)`C. `(l_(1))/(l_(2)) = (3)/(2)`D. `(l_(1))/(l_(2)) = (2)/(3)`
Answer» Correct Answer - B::C fundamental frequency of open pipe, `n_(0) = (V)/(2l_(1))` , closed pipe, `n_(C) = (V)/(4l_(2))` `(V)/(2l_(1)) - (V)/(4l_(2)) = 5` For first overtone , `n_(0) = (V)/(l_(1)) , n_(0) = (3V)/(4l_(2))` `(V)/(l_(1)) - (3V)/(4l_(2)) = 5` .......(2) on solving `(1)` and `(2) l_(1) = l_(2) rArr (l_(1))/(l_(2)) = (1)/(1)`

Discussion

65.	Two narrow cylindrical pipes `A and B` have the same length. Pipe `A` is open at both ends and is filled with a monoatomic gas of molar mass `M_(A)`. Pipe `B` is open at one end and closed at the other end, and is filled with a diatomic gas of molar mass `M_(B)`. Both gases are at the same temperature. (a) If the frequency of the second harmonic of the fundamental mode in pipe `A` is equal to the frequency of the third harmonic of the fundamental mode in pipe `B`, determine the value of `M_(B)//M_(B)`. (b) Now the open end of pipe `B` is also closed (so that the pipe is closed at both ends). Find the ratio of the fundamental frequency in pipe `A` to that in pipe `B`.
Answer» Correct Answer - A::B::C::D Frequency of second harmonic in pipe `A=` frequency of third harmonic in pipe `B` `:. 2((v_(A))/(2l_(AP))=3((v_(B))/(4l_(B))` or `(v_(A))/(v_(B))=(3)/(4)implies(sqrt(gamma_(A)RT_(A))/(M_(A)))/(sqrt((gamma_(B)RT_(B))/(M_(B))))=(3)/(4)` or `sqrt((gamma_(A))/(gamma_(B)))sqrt((M_(B))/M_(A))=(3)/(4)` (as `T_(A)=T_(B)`) `:. (M_(A))/(M_(B))=((25)/(21))((16)/(9))=(400)/(189)` (b) Ratio of fundamental frequency in pipe `A` and in pipe `B` is `(f_(A))/(f_(B))=(v_(A)//2l_(A))/(v_(B)//2l_(B))=(v_(A))/(v_(B))` (as `T_(A)=T_(B)`) `=sqrt((((gamma_(A)RT_(A))/(M_(A)))/(gamma_(B)RT_(B)))/(M_(B)))=sqrt((gamma_(A))/(gamma_(B)).(M_(B))/(M_(A)))` (as `T_(A)=T_(B)`) Substituting `(M_(B))/(M_(A))=(189)/(400)` from paert (a). we get `(f_(A))/(f_(B))=sqrt((25)/(21)xx(189)/(400))=(3)/(4)`

Discussion

66.	Assume end correction approximately equals to `(0.3) xx` (diameter of tube), estimate the approximate number of moles of ir present inside the tube (Assume tube is at `NTP`, and at `NTP, 22.4` litre contains `1` mole)A. `(100pi)/(36 xx 22.4)`B. `(10pi)/(18 xx 22.4)`C. `(10pi)/(72 xx 22.4)`D. `(10pi)/(60 xx 22.4)`
Answer» Correct Answer - A End correction `= (0.3) d = 1 cm` `d = (10)/(3) cm` vol. of tube `= (pi(d^(2))/(4))l = (pi)/(4)((10)/(3))^(2) xx 100 cm^(2)` (take `I = 0.99 m =n 1 m`) `= (10pi)/(36) lit` moles `= (10 pi)/(36 xx 22.4)` moles (`22.4 "lt"`. contains `1` mole `(10pi)/(36)"lt"` contains `(10pi)/(36 xx 22.4)` mole)

Discussion

67.	If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased ?
Answer» Correct Answer - A `beta_1=50dB, beta_2=60dB,` I_1=10^-7W/m^2, I_2=10^-6W/m^2` `[because beta=10log_10(I/I_0)` where `I_0=10^-12W/m^2]` Again, `I_2/I_1=(p_2/p_1)^2=(10^-6/10^-7)=10` `(where p=pressure amplitude) `:.(p_2/p_1)^2=10` `rarr P_2/P_1=sqrt10`

Discussion

68.	Assertion : Fundamental frequency of a narrow pipe is more. Reason : According to laplace end correction if radius of pipe is lass, frequency should be more.A. If both Asseration and Reason are true an dthe Reason is correct explanation of the Asseration.B. If both Asseration and Reason are true but Reason is not the correct explanation of Asseration.C. If Asseration is true , but the Reason is false.D. If Asseration is false but the Reason is true.
Answer» Correct Answer - A `f_(o) = (nu) /(2 (l + 1.2 r)) ` of an open pipe and `f_(o) = (nu) / (4 (l + 0.6 r))` for a closed pipe.

Discussion

69.	A small source of sound oscillates in simple harmonic motion with an amplitude of 17 cm. A detector is placed along the line of motion of the source. The source emits a sound of frequency 800 Hz which travels at a speed of `340 ms^-1` If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.
Answer» Correct Answer - C Given that `r=0.17m, f=800Hz, u=340m/s` Frequency band `=f_1+f_2=8 Hz` Where `f_1 and f_2` correspond to the maximum and minimum apprent frequencies.(Both will be the mean positioin, because the velocity is maximum). `Now, f_1=(34/(340+v_s))f` `f_2=(340/((340-v_s))f` `:.f_1-f_2=8` `=340[1/(340-v_s)-1/(340+v_s)]` =8 `rarr (2vs)/((340s^2-v_s^2)=8/(340xx800)` `rarr 340^2-vs^2=68000vs` solving for `v_s` we get `vs=1.695m/s` for SHM `v_s=rw` `rarr omega=(1.695/0.17)=10` So, T=(2pi)/w=pi/5=0.63 sec

Discussion

70.	The second overtone of an open pipe A and closed pipe B have the same frequencies at a given temperature. Both pipes contain air. The ratio of fundamental frequency of a to the fundamental frequency of B isA. `3 : 5`B. `5 : 3`C. `5 : 6`D. `6 : 5`
Answer» Correct Answer - B

Discussion

71.	The two pipes are submerged in sea water, arranged as shown in figure. Pipe `A` with length `L_(A) = 1.5 m` and one open end, contains a small sound source that sets up the standing wave with the second towards resonant freiquency of that pipe. Sound from pipe `A` set up resonance in pipe `B`, which has both ends open. The resonance is at the second lowest resonant frequency of pipe `B`. The length of the pipe `B` in meter is :
Answer» Correct Answer - 2 For pipe `A`, second resonant frequency is third harmonic thus `f = (3V)/(4L_(A))` For pipe `B`, second resonant frequency is third harmonic thus `f = (2V)/(2L_(B))` Equating `(3V)/(4L_(A)) = (2V)/(2L_(B)) rArr L_(B) = (4)/(3)L_(A) , = (4)/(3).(1.5)`

Discussion

72.	Two tuning forks `A` and `B` sounded together give `8` beats per second. With an air resonance tube closed at one end, the two forks give resonances when the two air columns are `32 cm` and `33 cm` respectively. Calculate the frequenciec of forks.
Answer» Let the frequency of the first fork be `f_(1)` and that of second be `f_(2)`. We rgen have, `f_(1) = (nu)/(4 xx32)` and `f_(2) = (nu)/(4 xx33)` We also see that `f_(1) gt f_(2)` `:. f_(1) - f_(2) = 8` …(i) and `(f_(1))/(f_(2)) = (33)/(32)` ..(ii) ltbgt Solving Eqs. (i) and (ii), we get `f_(1) = 264 H_(Z)` and `f_(2) = 256 H_(Z)`

Discussion

73.	A siren creates a sound level of `60 H_(Z)` at a location `500 m` from the speaker. The siren is powered by a batter that delivers a total energy of `1.0 kJ`. Assuming that the efficiency of siren is `30%` , determine the total time the siren can sound.
Answer» Sound level (in `dB`) `L = 10 log_(10) ((I)/(I_(0)))` where, `I_(0) = 10^(-12) W//m^(2)` `L = 60 dB` Hence, `I = (10^(6)) I_(0) = 10^(-6) W//m^(2)` Intensity, `I = (Power)/(Area) = (P)/(4 pi r^(2))` rArr `P = I (4 pi r^(2))` `P = (10^(-6))(4 pi)(500)^(2) = 3.14 W` :. Time, `t = ((10 xx 10^(3))((30)/(100)))/(3.14)` `t = 95.5s`

Discussion

74.	A small speaker has a capacity of power 3 W. A microphone is placed at distance 2 m from the speaker. Fine the displacement amplitude of particles of air near to microphone. The frequenct of sound emitted bt speaker is 1.0 KHz (Density of air = 1.2 `kg//m^(3)` and speed of sound in air = 330 m/s)A. `2.76xx10^(-4)cm`B. `4xx10^(-4)cm`C. `10xx10^(-4)cm`D. `3.8xx10^(-3)cm`
Answer» Correct Answer - A `becauselp/(4pir^(2))=3/(4pixx2^(2))=3/(16pi)` But `l=2pi^(2) A^(2)v^(2)p_(o)c` or `A^(2)=l/(2pi^(2)v^(2)p_(o)c)` `therefore" "A=sqrt(((l)/(2pi^(2)v^(2)p_(o)c)))=sqrt((3)/((16pixx2pi^(2)xx10^(6)xx12xx330)))` `0.002764xx10^(-3)m=2.76xx10^(-4)cm`

Discussion

75.	If the intensity is increased by a factor of `20`, by how many decibels is the intensity level increased.
Answer» Let the initial intensity be `I` and the density level be `beta_(1)` and when the intensity is increased by `= 20` times, the intensity level increases to `beta_(2)`. Then `beta_(1) = 10 log(I//I_(0))` and `beta_(2) = 10 log (20I//I_(0))` Thus, `beta_(2) - beta_(1) = 10 log (20I//I)` `= 10 log 20` `= 13 dB`.

Discussion

76.	If the intensity of sound is doubled, by how many decibels does the sound level increase ?A. `1 dB`B. `2 dB`C. `3 dB`D. `4 dB`
Answer» Correct Answer - C `L_(2)-L_(1)=10 log_(10) (I_(2)/I_(1))=10 log_(10) (2)` `=10xx0.3=3 dB`

Discussion

77.	A sound has an intensity of `2 xx 10^(-8) W//m^(2)`. Its intensity level in decibels is `(log_(10) 2 = 0.3)`A. `23`B. `3`C. `43`D. `4.3`
Answer» Correct Answer - C `L = 10 log_(10)((I)/(I_(0))) = 10 log_(10) ((2 xx 10^(-8))/(10^(-12)))` `= 10 log_(10) (2 xx 10^(4)) = 10[log_(10) 2 + log_(10) 10^(4)]` `= 10[0.3 + 4] = 43 dB`

Discussion

78.	Find the displacement ammplitude amplitude of amplitude of particles of air of density `1.2 kg//m^(3)`,if intensity, frequency and speed of sound are `8 xx 10^(-6) w//m^(2), 5000 Hz` and `330 m//s` respectively.
Answer» The relation between the intensity of sound and the displacement amplitude is `I = (P_(0^2))/(2rhov)`, `P_(0) = sqrt(2rhovI)` `s_(0) = (P_(0))/(Bk) = (sqrt(2rhogI))/(rhov^(2)pif)v` `= sqrt((I)/(rhov2)) (1)/(pif)` or, `s_(0) = 6.4 nm`.

Discussion

79.	If the intensity increased by a factor of 20, byhow may decibels is the sound level increased?
Answer» Let the initial intensity be I and te sound level be `beta_1`. When the intensity is increased to 20 I, the level increases to `beta_2`. `Then beta_1=10log(I/I_0)` and `beta_2=10log(20I/I)` `Thus, beta_2-beta_1=10log(20I/I)` `=10log 20` `=13dB`.

Discussion

80.	Two sound waves one in air and the other in fresh water are equal in intensity. (a) Find the ratio of pressure amplitude of the wave in water to that of the wave in air. (b) If the pressure amplitudes of the waves are equal then what will be the ratio of the intensities of the waves. [`V_(sound) = 340 m//s` in air & density of air `= 1.25 kg//m^(3), V_(sound) = 1530 m//s` in water, density of water `= 1000 kg//m^(3)`]
Answer» Correct Answer - (a) `(P_(0_(w)))/(P_(0_(a))) = 60` , (b) `(P_(w))/(P_(a)) = (1)/(3600)`

Discussion

81.	If the pressure amplitude in a sound wave is tripled, then by what factor the intensity of sound wave is increased?A. `3`B. `6`C. `9`D. `sqrt(3)`
Answer» Correct Answer - C `I=p_(0)^(2)/(2 rho v)` `I_(2)/I_(1)=[((p_(0))_(2))/((p_(0))_(1))]^(2)=(3)^(2)=9`

Discussion

82.	The speed of sound in hydrogen is `1270 ms^(-1)` at temperature T. the speed at the same T in a mixture of oxygen and hydrogen mixed in a volume ratio 1:4 will be ?A. 520 m/sB. 585 m/sC. 615 m/sD. 635 m/s
Answer» Correct Answer - D `1270=sqrt((gammaP)/(rho))` Now, density becomes `4 rho`. So, `sqrt((gammaP)/(4rho))=sqrt((rP)/(rho))xx1/2=1/2xx1270=635 m//s`

Discussion

83.	The ratio of velocity of sound in hydrogen and oxygen at STP isA. `16:1`B. `8:0`C. `4:1`D. `2:1`
Answer» Correct Answer - C Velocity of sound, `v=sqrt((lamdaRt)/(M))` `" "(v_(H))/(v_(o))=sqrt((M_(o))/(M_(1)))-sqrt((16)/1)=4:1`

Discussion

84.	The value of adiabatic constant `gamma` for oxygen and nitrogen is same. The speed of sound in oxygen is 470 m/s at STP. The speed of sound in nitrogen at STP isA. 340 m/sB. 580 m/sC. 502 m/sD. None of these
Answer» Correct Answer - C `c=sqrt(((gammaRT)/(M)))` `therefore (c_(2))/(c_(1))=sqrt(((M_(1))/(M_(2))))or (C_("nitrogen"))/(C_("oxygen"))=sqrt(((32)/(28)))=sqrt((8/7))` `because" "C_("nitrogen")=sqrt((8/7))C_("oxygen")=sqrt((8/7))xx470` `" "=502.45m//s ~~502m//s`

Discussion

85.	A vibrating string of certain length `l` under a tension `T` resonates with a mode corresponding to the first overtone (third harmonic ) of an air column of length `75 cm` inside a tube closed at one end. The string also generates `4 beats//s` with a tuning fork of frequency `n` . Now when the tension of the string is slightly increased the number of beats reduces to `2` per second. Assuming the velocity of sound in air to `340 m//s` , the frequency `n` the tuning fork in `H_(Z)` is (a) `344` (b) `336` (c ) `117.3` (d) `109.3`
Answer» With increase in tension , frequency of vibrating string will increase. Since, number of beats are decreasing. Therefore , frequency of vibrating string or third harmonic frequency or closed pipe should be less than the frequency of tuning fork by `4` . `:.` Frequency of tuning fork = third harmonic frequency of closed pipe `+ 4` = `3 ((upsilon)/(4l)) + 4 = 3 ((340)/(4xx0.75)) + 4` = `344 H_(Z)` `:.` Correct option is (a).

Discussion

86.	Two trains moves towards each other with the same speed. Speed of sound is `340 m//s^(-1)`. If the pitch of the tone of the whistle of one when heard on the end other changes to `9//8` times, then the speed of each train is :
Answer» Correct Answer - `20 m//s`

Discussion

87.	Calculate the speed of longitudinal waves in the following gases at `0^(@) C` and `1 atm( = 10^(5) pa):` (a) oxygen for which the bulk modulus is `1.41 xx 10^(5)` pa and density is `1.43 kg//m ^(3)` . (b) helium for which the bulk modulus is `1.7xx 10^(5)` pa and density is `0.18 kg//m^(3)` .
Answer» Correct Answer - A::B::C::D (a) `nu_(o2)= sqrt (B/(rho))` `= sqrt((1.41xx10^(50))/(1.43)` `314 m//s` (b) `nu_(He) =sqrt(B/(rho))` `= sqrt((1.7xx10^(5))/(0.18)` `= 972 m//s`

Discussion

88.	A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air is `320(m)/(s)`) will be equal toA. 6B. 3C. 4D. 8
Answer» Correct Answer - A `f_(b)=(2V_(s))/Vxxf=6Hz`

Discussion

89.	A transverse sinusoidal wave of amplitude `a`, wavelength `lambda` and frequency `f` is travelling on a stretched string. The maximum speed of any point in the string is `v//10`, where `v` is the speed of propagation of the wave. If `a = 10^(-3)m` and `v = 10ms^(-1)`, then `lambda` and `f` are given byA. `lamda=2pixx10^(-3)m`B. `lamda=10^(-3)m`C. `f=10^(3)//(2pi)Hz`D. `f=10^(3)Hz`
Answer» Correct Answer - A::C

Discussion

90.	For a travelling harmonic wave `y=2.0 cos(10t-0.0080x+0.35)`, where x and y are in centimetres and t in seconds. What is the phase difference between oscillatory motion of two points separated by a distance of (a) 4m (b) 0.5 m (c ) `lambda//2` (d) `3lambda//4`
Answer» Correct Answer - A::B::C::D

Discussion

91.	How many frequencies below `1 kH_(Z)` of natural oscillations of air column will be produced if a pipe of length `1 m` is closed at one end? [ velocity of sound in air is `340 m//s`]A. `3`B. `6`C. `4`D. `8`
Answer» Correct Answer - B Fundamental frequency, `f_(o) = (nu) / (4l) = (340) / (4xx1) = 85 H_(Z)` Six frequencies can be produced below `1 kH_(Z)` . Those six frequency are , `f_(o)`, `3 f_(o)`, `5 f_(o)`, `7 f_(o)`, `9 f_(o)` and `11 f_(o)`. As `13 f_(o) = 1105 H_(Z) gt 1000 H_(Z)` or `1 kH_(Z)` .

Discussion

92.	A typical loud sound wave with a frequency of `1 Kh_(Z)` has a pressure amplitude of about `10` Pa (a) At `t = 0`, the pressure is a maximum at some point `X_(1)`. What is the displacement at that point at `t = 0`? (b) What is the maximum value of the displacement at any time and place/ Take the density of air to be `1.29 kg//m^(3)` and speed of sound in air is `340 m//s`.
Answer» Correct Answer - A::B::C (a) Displacement is zero when pressure is maximum. (b) `Delta p_(max) = BAK` `= (rho nu)^(2) A ((omega)/(nu)) = 2 pi f Arho nu` :. `A = (Deltap_(max))/(2 pi f rho nu)` `= (10)/((2 pi)(10^(3))(1.29)(340)` `= 3.63 xx 10^(-6) m`

Discussion

93.	The pressure at a point varies from `99980 Pa` to `100020 Pa` due to simple harmonic sound wave. The amplitude and wavelength of the wave are `5 xx 10^(-6) m` and `40 cm` respectively. Find the bulk modulus of air
Answer» Correct Answer - `(P_(0)lambda)/(2piS_(0)) = (8 xx 10^(5))/(pi) N//m^(2)`

Discussion

94.	If the amplitude of a wave at a distance `r` from a point source is `A`, the amplitude at a distance `2 r` will beA. `2 A`B. `A`C. `A//2`D. `A//4`
Answer» Correct Answer - C `I prop (1)/(r^(2)) prop A^(2) rArr (1)/(r ) prop A` `A_(1)r_(1) = A_(2)r_(2) rArr Ar = A_(2). 2r rArr A_(2) = A//2`

Discussion

95.	Which of the following does not affect the apparent frequency in doppter effect?A. Speed of sourceB. Speed of observerC. Frequency of sourceD. Distance between source and observer
Answer» Correct Answer - D Doppler effect in Frequency depends upon relative velocity between source and observer

Discussion

96.	For a certain organ pipe, three successive resonance frequencies are observer at `425`, `595` and `765 H_(Z)` respectively. Taking the speed of sound in air to be `340 m//s`, (a) explain whether the pipe is closed at one or open at boyh ends. (b) determine the fundamental frequency and length of the pipe.
Answer» (a) The given frequencies are the ratio `5: 7: 9`. As the frequencies are odd multiple of `85 H_(Z)`, the pipe must be closed at one end.(b) Now, the fundamebtal frequency is the lowest, i.e.`85 H_(Z)`. ltbr. :. `85 = (nu)/(4l)` rArr `l = (340)/(4 xx85) = 1m`

Discussion

97.	Two sound waves emerging from a source reach a point simultaneously along two paths. When the path difference is `12 cm ` or `36 cm`, then there is a silence at that point. If the speed of sound in air be `330 m//s`, then calculate maximum possible frequency of the source.
Answer» Correct Answer - A::C `Deltax = lambda//2 = (nu)/(2 f)` `:. f = (nu)/(2 Deltax) = (330)/(2(0.12)) = 1375 H_(Z)`

Discussion

98.	A source of sound is in the shape of a long narrow cylinder radiating sound waves normal to the axis of the cylinder. Two points P and Q are at perpendicular distances of 9 m and 25 m from the axis. The ratio of the amplitudes of the waves at P and Q is :-A. `5:3`B. `sqrt(5):sqrt(3)`C. `3:5`D. `25:9`
Answer» Correct Answer - A

Discussion

99.	(a) A cylindrical metal tube has a length of `50 cm` and is open at both ends. Find the frequencies between `1000 Hz` and `2000 Hz` at which the air is `340 m//s`. (b) Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range `(20 - 20000 Hz)`. Speed of sound in air `= 340 m//s`. (c) Two successive resonance frequencies in an open organ pipe are `1944 Hz` and `2592 Hz`. Find the length of the tube. The speed of sound in air is `324 m//s`.
Answer» (a) For open pipe, fundamental frequency `f = (v)/(2l) = (340)/(2 xx 0.5) = 340 Hz` Possible frequencies of open pipe `= nf, n = 1,2,3,…` `340,680,1020,1360,1700,2040 Hz` Frequencies between `1000` and `2000 Hz` `= 1020 Hz,1360 Hz,1700 Hz` (b) `f = (v)/(2l)` `l = (v)/(f)` , l will be maximum if `f` is minimum `l_(max) = (340)/(2 xx 20) = 8.5 m` (c) The difference in successive frequencies of a pipe (open or closed) `= (v)/(2l)` `2592 - 1944 = (324)/(2l)` `l = (1)/(4) m = 0.25 m`

Discussion

100.	The effect of making a hole exactly at `(1//3^(rd))` of the length of the pipe from its closed end is such that :A. its fundamental frequency is an octave higher thanthe open pipe of same lengthB. its fundamental frequency is thrice of that before making a holeC. the fundamental frequency is `3//2` time of that before making a holeD. the fundamental alone is changed while the harmonics expressed as ratio of fundamentals remain the same
Answer» Correct Answer - B::D

Discussion

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