98 + Interview Questions in SOUND-WAVES-AND-DOPPLER-EFFECT IN GENERAL AWARENESS Page 2 InterviewSolution

51.	Taking the composition of air to be `75%` of nitrogen and `25%` of oxygen by weight, calculate the velocity of sound through air.
Answer» The molecular weight of a mixture is given by `(m_1+m_2+m_3)/(M)=(m_1)/(M_1)+(m_2)/(M_2)+(m_3)/(M_3)+`…. `(75+25)/(M)=(75)/(28)+(25)/(32)` or `M=28.9g` `c=sqrt(gamma(RT)/(M))=sqrt((1.4xx8.3xx1000xx273)/(28.9))=331.3(m)/(s)` Here `gamma=1.4rarr` as both the gases are diatomic

Discussion

52.	A source `S` of sound wave of fixed frequency `N` and an observer `O` are located in air initially at the space points `A` and `B`, a fixed distance apart. State in which of the following cases, the observer will NOT see any Doppler effect and will receive the same frequency `N` as produced by the source.A. Both the source `S` and observer `O` remain stationary but a wind blows with a constant speed in an arbitrary direction.B. The observer remains stationary but the source `S` moves parallel to and in the same direction and with the same speed as the wind.C. The source remains stationary but the observer and the wind have the same speed away from the source.D. The source and the observer move directly against the wind but both with the same speed.
Answer» Correct Answer - A::D In both case (a) and (d) the source and observer are relatively at rest, thus neither of htem is approaching or separating from each other. Effectively, it is the medium that moves in each of these cases. The received (apparent) frequency differs from the emitted frequency if and only if the time required for the wave to travel from the source to observer is different fror different wavefronts. With a uniform steady motion of the medium, past the observer and source, the transit time from source to observer is the same for all wavefronts. Hence it follows that apparent freqeuncy is equal to the true emitted frequency. Thus there is no Doppler effect. In cases (b) and (c ), Doppler effect will be observed as the source and observer have a relative speed and so they will approach or recede from each other.

Discussion

53.	Which of the following statements are incorrect?A. Wave pulses in strings are transverse waves.B. Sound waves in air are transverse waves of compression and rarefaction.C. The speed of sound in air at `20^@C` is twice that at `5^@C`.D. A 60dB sound has twice the intensity of a 30 dB sound.
Answer» Correct Answer - B::C::D `vpropsqrtT` `60dB=10log((I_1)/(I_0))` `30dB=10log((I_2)/(I_0))` `30=10log((I_1)/(I_2))implies(I_1)/(I_2)ne2`

Discussion

54.	A source of sound of frequency `f_1` is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency `f(Hz)` is plotted against time `t(sec)` . The speed of sound in air is `300(m)/(s)`. Find `f_1` (`g=10(m)/(s)`).A. `0.5xx10^3Hz`B. `2xx10^3Hz`C. `0.25xx10^3Hz`D. `0.2xx10^3Hz`
Answer» `f=((V+V_0)/(V))f_1=f_1+f_1(V_0)/(V)` `V_0="gt"` So, `f+f_1+((f_1g)/(V))t` Slope of graph`=(f_1g)/(V)` `((2)(10^-3)-f_1)/(30)=((f_1)(10))/(300)` or `f_1=10^3Hz`

Discussion

55.	A source of sound of frequency `f_1` is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency `f(Hz)` is plotted against time `t(sec)` . The speed of sound in air is `300(m)/(s)`. Find `f_1` (`g=10(m)/(s)`).A. `0.5xx10^3Hz`B. `2xx10^3Hz`C. `0.25xx10^3Hz`D. `0.2xx10^3Hz`
Answer» `f=((V+V_0)/(V))f_1=f_1+f_1(V_0)/(V)` `V_0="gt"` So, `f+f_1+((f_1g)/(V))t` Slope of graph`=(f_1g)/(V)` `((2)(10^-3)-f_1)/(30)=((f_1)(10))/(300)` or `f_1=10^3Hz`

Discussion

56.	A source of sound of frequency `f_1` is placed on the ground. A detector placed at a height is released from rest on this source. The observed frequency `f(Hz)` is plotted against time `t(sec)` . The speed of sound in air is `300(m)/(s)`. Find `f_1` (`g=10(m)/(s)`).A. `0.5xx10^3Hz`B. `2xx10^3Hz`C. `0.25xx10^3Hz`D. `0.2xx10^3Hz`
Answer» `f=((V+V_0)/(V))f_1=f_1+f_1(V_0)/(V)` `V_0="gt"` So, `f+f_1+((f_1g)/(V))t` Slope of graph`=(f_1g)/(V)` `((2)(10^-3)-f_1)/(30)=((f_1)(10))/(300)` or `f_1=10^3Hz`

Discussion

57.	In sport meet the timing of a 200 m straight dash is recorded at the finish point by starting an accurate stop watch on hearing the sound of starting gun firen at the starting poing. The time recorded will be more accurateA. In winterB. in summerC. in all seasonsD. none of these
Answer» Correct Answer - B Time recorded in summer is more accurate. The velocity of sound is directly proportional to the squre root of absolute temperature. Hence, the sound of the gun fired at the starting point will reach the finishing point quicker in summer than in winter. The lapse of time due to the time taken by the sound in reaching the finish point will be less in summer and hence the time recorded will be more accurate in summer than in winter.

Discussion

58.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. find the intensity at both the surfaces.A. `I_1=12xx10^-6(W)/(m^2)`,`I_2=12xx10^-6(W)/(m^2)`B. `I_1=6xx10^-9(W)/(m^2)`,`I_2=12xx10^-9(W)/(m^2)`C. `I_1=6xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`D. `I_1=12xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`
Answer» Correct Answer - C `I=(P)/(A)` so, `I_1=(P)/(A_1)` and `I_1=I_2=(P)/(A_1)`

Discussion

59.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. If two persons (having almost same physique) `A` and `B` are standing at the location of surfaces 1 and 2, respectively, then who will hear a quiter sound?A. Both will hear same sound.B. A will bear a quiter soundC. B will hear a quiter soundD. information is not sufficient
Answer» Correct Answer - C The intensity of sound received by listener B is less so he hears quieter sound

Discussion

60.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. find the intensity at both the surfaces.A. `I_1=12xx10^-6(W)/(m^2)`,`I_2=12xx10^-6(W)/(m^2)`B. `I_1=6xx10^-9(W)/(m^2)`,`I_2=12xx10^-9(W)/(m^2)`C. `I_1=6xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`D. `I_1=12xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`
Answer» Correct Answer - C `I=(P)/(A)` so, `I_1=(P)/(A_1)` and `I_1=I_2=(P)/(A_1)`

Discussion

61.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. If two persons (having almost same physique) `A` and `B` are standing at the location of surfaces 1 and 2, respectively, then who will hear a quiter sound?A. Both will hear same sound.B. A will bear a quiter soundC. B will hear a quiter soundD. information is not sufficient
Answer» Correct Answer - C The intensity of sound received by listener B is less so he hears quieter sound

Discussion

62.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. If two persons (having almost same physique) `A` and `B` are standing at the location of surfaces 1 and 2, respectively, then who will hear a quiter sound?A. Both will hear same sound.B. A will bear a quiter soundC. B will hear a quiter soundD. information is not sufficient
Answer» Correct Answer - C The intensity of sound received by listener B is less so he hears quieter sound

Discussion

63.	In the figure shown below, a source of sound having power `12xx10^-6W` is kept at `O`, which is emitting sound waves in the directions as shown. Two surfaces are labelled as 1 and 2 having areas `A_1=2xx10^3m^2` and `A_2=4xx10^3m^2`, respectively Q. find the intensity at both the surfaces.A. `I_1=12xx10^-6(W)/(m^2)`,`I_2=12xx10^-6(W)/(m^2)`B. `I_1=6xx10^-9(W)/(m^2)`,`I_2=12xx10^-9(W)/(m^2)`C. `I_1=6xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`D. `I_1=12xx10^-9(W)/(m^2)`,`I_2=3xx10^-9(W)/(m^2)`
Answer» Correct Answer - C `I=(P)/(A)` so, `I_1=(P)/(A_1)` and `I_1=I_2=(P)/(A_1)`

Discussion

64.	The speed of sound in hydrogen gas at certain temperature is v `(m)/(s)` Find the speed of sound in a gaseous mixture containing 2 moles of oxygen and 1 mole of hydrogen gas, at the same temperature. Assume the gases do no react at the ordinary temperature.
Answer» For a mixture of non reacting gases, `(n_1)/(gamma_1-1)+(n_2)/(gamma_2-1)=(n_1+n_2)/(gamma_(mix)-1)` For given problem `n_1=2`,`n_2=2`,`gamma_1=(7)/(5)`,`gamma_2=(7)/(5)`,`gamma_(mix)=1.4`. `M_(mix)=(n_1m_1+n_2m_2)/(n_1+n_2)` Here, `m_1=32`,`m_2=2`,`M_(mix)=22`. `V_(mix)=sqrt((gamma_(mix)RT)/(M_(mix)))` and `V_H=sqrt((gamma_(H2)RT)/(M_(H2)))` gamma`V_(mix)=sqrt((gamma_(mix))/(M_(mix))xx(M_H)/(gammaH))xxV_H=(V)/(sqrt(11))(m)/(s)`

Discussion

65.	The velocity of sound in hydrogen at `0^@C` is `1200(m)/(s)`. When some amount of oxygen is mixed with hydrogen, the velocity decreases to `500(m)/(s)`. Determine the ratio of `H_2` to `O_2` by volume in this mixture, given that the density of oxygen in 16 times that of hydrogen.
Answer» Given that `1200=sqrt((gammaP)/(rho_H))` ..(i) ltbr. Let there be x volume of `H_2` and y volume of `O_2` then, `(x+y)rho_(mix)=xrho_h+yrho_0=xrho_H+16yrho_H` `impliesrho_(mix)=((x+16y))/(x+y)rho_H` `500=sqrt((gammaP(x+y))/((x+16y)rho_H))` ..(ii) Dividing Eq. (ii) by Eq. (i) `(12)/(5)=sqrt((x+16y)/(x+y))` or `(x)/(y)=(2.2)/(1)`

Discussion

66.	Find the molecular weight for a gas in which the velocity of sound is `1260(m)/(s)` at `0^@C` and whose `gamma` is 1.4.
Answer» `C=sqrt((gammaP)/(rho))` `1260=sqrt((1.4P)/(rho))` `(P)/(rho)` of the gas at `0^@C=11.34xx10^5` `P=(rho)/(M)RT` is the perfect gas equation in terms of density `(P)/(rho)=(RT)/(M)` `M=(8.3xx273)/(11.34xx10^5)=2xx10^-3kg=2g`

Discussion

67.	When a person wears a hearing aid, the sound intensity level increases by 30 dB. The sound intensity increases byA. `e^2`B. `10^3`C. `30`D. `10^2`
Answer» Correct Answer - B `L_2-L_1=30dB` `10dBlog((I_2)/(I_0))-10dBlog((I_1)/(I_0))=30dB` `log_(10)((I_2)/(I_1))=3implies(I_2)/(I_1)=10^3` Hence the sound intensity increases by `10^3`.

Discussion

68.	The difference in the speeds of sound in air at `-5^@C`, 60 cm pressure of mercury and `30^@C`, 75 cm pressure of mercury is (velocity of sound in air at `0^@C` is `332(m)/(s)`)A. `15.25(m)/(s)`B. `21.35(m)/(s)`C. `18.3(m)/(s)`D. `3.05(m)/(s)`
Answer» Correct Answer - B Velocity of sound is not affected by the change inpressure of air. Velocity of sound at `1^@C`, `v_1=(332+0.61t)(m)/(s)` At `-5^@C`,`v_(-5^@C)=(332-0.61xx5)(m)/(s)` At `30^@C`,`v_(30^@C)=(332+0.61xx30)(m)/(s)` `v_(30^@C)-v_(-5^@C)=(0.61xx35)(m)/(s)` `=21.35(m)/(s)`

Discussion

69.	A sonic source, located in a uniform medium, emits waves of frequency `n`. If intensity, energy density (energy per unit volume of the medium) and maximum speed of oscillations of medium particle are, respectively, `I`, `E` and `u_0` at a point, then which of the following graphs are correct?A. B. C. D.
Answer» Correct Answer - A::C::D If intensity at a point is `I`. Then energy density at that point is `E=(I)/(v)`, where `v` is wave propagation velocity. It means that `EpropI`. Hence, the graph between `E` and `I` will be a straight line passing through the origin. Therefore, (a) is correct and (b) is wrong. Intensity is given by `I=2pi^2n^2a^2rhov` Hence, `E=2pi^2n^2a^2rho` It means that `Epropn^2` Hence, the graph between `E` and `n` will be a parabola passing through origin, having increasing slope and symmetric about E-axis, Hence, option (d) is correct. Particle maximum velocity is `u_0=aomega=2pina` `impliespina=(u_0)/(2)` Hence, `E=(1)/(2)rhou_0^2` It means that graph between `E` and `u_0` will be a parabola, have increasing slope and will be symmetric about E-axis. Hence, option (c ) is also correct.

Discussion

70.	A sonic source, located in a uniform medium, emits waves of frequency `n`. If intensity, energy density (energy per unit volume of the medium) and maximum speed of oscillations of medium particle are, respectively, `I`, `E` and `u_0` at a point, then which of the following graphs are correct?A. B. C. D.
Answer» Correct Answer - A::C::D If intensity at a point is `I`. Then energy density at that point is `E=(I)/(v)`, where `v` is wave propagation velocity. It means that `EpropI`. Hence, the graph between `E` and `I` will be a straight line passing through the origin. Therefore, (a) is correct and (b) is wrong. Intensity is given by `I=2pi^2n^2a^2rhov` Hence, `E=2pi^2n^2a^2rho` It means that `Epropn^2` Hence, the graph between `E` and `n` will be a parabola passing through origin, having increasing slope and symmetric about E-axis, Hence, option (d) is correct. Particle maximum velocity is `u_0=aomega=2pina` `impliespina=(u_0)/(2)` Hence, `E=(1)/(2)rhou_0^2` It means that graph between `E` and `u_0` will be a parabola, have increasing slope and will be symmetric about E-axis. Hence, option (c ) is also correct.

Discussion

71.	A point source emits sound waves with an average power output of 80.0 W (a) Find the intensity 3.00 m from the source. (b) find the distance at which the intensity of the sound is `1.00xx10^-8(W)/(m^2)`
Answer» a. Imagine a small loudspeaker sending sound out at an average rate of 80.0 W uniformly in all directions. You are standing 3.00 m away from the speakers. As the sound propagates, the energy of the sound waves is spread out over and ever expanding sphere. We evaluate the intensity from a given equation, so we categorize this example as substitution problem. Because a point emits energy in the form of spherical waves, use Eq. (ii) to find the intensity: `I=(p_(avg))/(4pir^2)=(80.0W)/(4pi(3.00m)^2)=0.707(W)/(m^2)` This intensity is close to the threshold of pain. b. solve for r from above equation and use the given value for I: `r=sqrt((p_(avg))/(4piI))=sqrt((80.0W)/(4pi(1.00xx10^-8(W)/(m^2)))=2.52xx10^4m`

Discussion

72.	An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?A. `5%`B. `20%`C. `0%`D. `0.5%`
Answer» Correct Answer - B `f=f_0((v_S+v_0)/(v_S))` `=f_0[(v+(v)/(5))/(v)]` `=(6)/(5)f_0` Hence, percentage increase is `[((6)/(5)f_0-f_0)/(f_0)]xx100=20%`

Discussion

73.	Explain why the speed of sound through a gas cannot be greater than the r.m.s. speed of the molecules of the gas.
Answer» In gases, the molecules move about at random. The oscillations produced by a sound wave are superimposed on this random thermal motion. An impulse given to one molecule is passed on to another molecule after the first one has moved through the empty space between them and collided with the second. The average of the speeds in the space between molecules is the average speed which is close to the rms speed hence the speed of a sound wave can never be greater than the rms speed of the molecules of a gas.

Discussion

74.	Two trains A and B simultaneously start moving along parallel tracks from a station along same direction. A starts with constant acceleration `2(m)/(s^2)` from rest, while B with the same acceleration but with initial velocity of 40 m/s. Twenty seconds after the start, passenger of A hears whistle of B. If frequency of whistle is 1194 Hz and velocity of sound in air is 322 m/s, calculate frequency observed by the passenger.
Answer» Since both the trains move in same direction with same acceleration, therefore, their relative acceleration is zero. Initial velocity of train A is zero while that of train B is 40 m/s. Therefore, train A of observer is behind the train B or source when whistle is heard by passenger of A. Let this sound be produced at time t. Considering motion of train B (sound source) up to this instant, `u=40(m)/(s)`,`a=2(m)/(s^2)`,`time=t` `v=v_b=?` `s=s_B=?` ..(i) We have `v=u+at`, `v_B=40+2t` `s=ut+(1)/(2)at^2` `s_B=40t+t^2` But this sound is heard by the observer at `t=20`s. Considering his motion (motion of train A) up to this instant, `u=0`,`a=2(m)/(s^2)`,`t=20s` `v=v_B=?`,`s=s_B=?` We have, `v=u+at` `v_A=40(m)/(s)` `s=ut+(1)/(2)at^2` `s_A=400m` When sound was produced, train B (source) was at a distance `s_B` from initial point while observer receives this sound at `t=20s` and at a distance `s_A` from initial point. It means that sound waves travel a distance `(s_B-s_A)` in air and take time `(20-t)` to travel it. `s_B-s_A=(20-t)v` ..(iii) Where `v=322(m)/(s)` (speed of sound). Substituting values of `s_A` and `s_B` in Eq.(iii), `t=18s` When sound was produced source was moving with velocity `v_B=(40+2t)=76(m)/(s)` (away from observer) while observer receive these sound waves at `t=20s`, when he was moving with velocity `v_A=40(m)/(s)` (towards the source). Hence, the observed frequency is `n=n_0(v+v_A)/(v+v_B)` Where `n_0=1194Hz` (natural frequency of source) and so `n=1086Hz`.

Discussion

75.	A family ice show is held at an enclosed arena. The skaters perform to music with level 80.0dB. This level is too loud for your baby, who yells at 75.0 dB. (a) What total sound intensity engulfs you? (b) what is the combined sound level ?
Answer» Resist the temptation to say 155 dB. The decibel scale is logarithmic, so addition does not work that way. We must figure out the intensity of each sound, add the intensities, and then translate back to a level. We have `beta=10log((I)/(10^(-12)(W)/(m^2)))` `I=[10^((beta)/(10))]10^(-12)(W)/(m^2)` a. for your baby, `I_b=(10^((75.0)/(10)))(10^(-12)(W)(m^2)=sqrt(10)xx10^_5(W)/(m^2)` For the music, `I_m=(10^((80.0)/(10)))(10^(-12)(W)/(m^2))=10.0xx10^-5(W)/(m^2)` The combined intesity is `I_(total)=I_m+I_b` `=10.0xx10^-5(W)/(m^2)+3.16xx10^-5(W)/(m^2)` `=13.2xx10^-5(W)/(m^2)` b. The combined sound level is then `beta_(total)=10log((I_(total))/(10^(-12)(W)/(m^2)))` `=10log((1.32xx10^(-4)(W)/(m^2))/(10^(-12)(W)/(m^2)))=81.2dB` The result is only a little louder than 80 dB.

Discussion

76.	Calculate the sound level (in decibels) of a sound wave that has an intensity of `4.00mu(W)/(m^2)`.
Answer» We expect about 60 dB we use the definition of the decibel scale. We use the equation `beta=10dBlog((I)/(I_0))` Where `I_0=10^(-12)(W)/(m^2)` `beta=10log((4.00xx10^-6(W)/(m^2))/(10^(-12)(W)/(m^2)))=66.0dB` This could be the sound level of a chamber music concert, heard from a few rows back in the audience.

Discussion

77.	A source of sound is travelling with a velocity of `30(m)/(s)` towards a stationary observer. If actual frequency of source is 1000 Hz and the wind is blowing with velocity `20(m)/(s)` in a direction at `60^@C` with the direction of motion of source, then the apparent frequency heard by observer is (speed of sound is `340(m)/(s)`)A. 1011 HzB. 1000 HzC. 1094 HzD. 1086 Hz
Answer» Correct Answer - C `f=((v+v_m)/(v+v_m-v_(source)))1000` `=((340+20cos60^@)/(340+20cos60^@-30))1000` `=1094Hz`

Discussion

78.	An isotropic stationary source is emitting waves of frequency n and wind is blowing due north. An observer A is on north of the source while observer B is on south the source. IF both the observers are stationary, thenA. frequency received by A is greater than nB. frequency received by B is less than nC. freqeuency received by A equals to that received by BD. frequencies received by A and B cannot be calculated unless velocity of waves in still air and velcity of wind are known
Answer» Correct Answer - C Since source and both the observer are stationary, therefore no change will be observed by the two observer. It means both the observers will receive waves with natural frequency, which is equal to n.

Discussion

79.	A source S emitting sound of 300 Hz is fixed of block A which is attached to free end of a spring `S_A` as shown in the figure. The detector D fixed on block B attached to the free end of spring `S_B` detects this sound. The blocks A and B are simultaneously displaced towards each other through of 1.0 m and then left to vibrate. Find the maximum and minimum frequencies of sound detected by D if the vibrational frequency of each block is 2 Hz (velocity of sound is 340 m/s).
Answer» The motion of A and B is sychronized. Hence the maximum frequencies detected by D will be `f_(max)=((v+v_D)/(v-v_S))f` (when S and D are approaching each other while at their equilibrium position.) `=((340+Aomega)/(340-Aomega))f=((340+12.56)/(340-12.56))xx300=323Hz` Similarly, the apparent frequency will be minimum when the source and the detector recede from each other with maximum speed `(=Aomega)` `f_(min)=((v-Aomega)/(v+Aomega))f=((340-12.56)/(340+12.56))xx300=278.6Hz`

Discussion

80.	A car emitting sound of frequency 500 Hz speeds towards a fixed wall at `4(m)/(s)`. An observer in the car hears both the source frequency as well as the frequency of sound reflected from the wall. If he hears 10 beats per second between the two sounds, the velocity of sound in air will beA. `330(m)/(s)`B. `387(m)/(s)`C. `404(m)/(s)`D. `340(m)/(s)`
Answer» Correct Answer - D The frequency that the observer receives directly from the source has frequency `n_1=500Hz`. As the observer and source both move towards the fixed wall with velocity `u`, the apparent frequency of the reflected wave coming from the wall to the observer wii have frequency `n_2=((V)/(V-u))500Hz` Where `V` is the velocity of sound wave in air. The apparent frequency of this reflected wave as heard by the oberver will then be `n_3=((V+u)/(V))n_2=((V+u)/(V))((V)/(V-u))500=((V+u)/(V-u))500` It is given that the nubmer of beat per second is `n_3-n_1=10` `(n_3-n_1)=10=((V+u)/(V-u))500-500` `=500[(V+u)/(V-i)-1]` `10=(2xxuxx500)/(V-u)` Hence, `10V=1000u+10u=1010u` putting `u=4(m)/(s)` we have `V=(1)/(10)[4040]=404(m)/(s)`

Discussion

81.	Two tuning forks with natural frequencies of `340 Hz` each move relative to a stationary observer. One fork moves away form the observer, while the other moves towards him at the same speed. The observer hears beats of frequency `3 Hz`. Find the speed of the tuning fork.
Answer» Let v be the speed of sound and `v_S` the speed of forks. The apparent frequency of fork which moves towards the observer is `n_1=(v)/(v-v_S)n` The apparent frequency of fork which moves away from observer is `n_2=(v)/(v+v_S)n` If x is the number of beats heard per second the `x=n_1-n_2impliesx(v)/(v-v_S)n-(v)/(v+v_S)n` or `x=(v+v_S-(v-v_S))/(v^2-v_S^2)(vn)` or `(2vv_Sn)/(v^2-v_S^2)=x` `2((v_S)/(v))n=x` or `v_S=(vx)/(2n)` Given `v=340(m)/(s)`,`x=3`,`n=340Hz` `v_S=(340xx3)/(2xx340)=1.5(m)/(s)`

Discussion

82.	A source S emitting sound of 300 Hz is fixed of block A which is attached to free end of a spring `S_A` as shown in the figure. The detector D fixed on block B attached to the free end of spring `S_B` detects this sound. The blocks A and B are simultaneously displaced towards each other through of 1.0 m and then left to vibrate. Find the maximum and minimum frequencies of sound detected by D if the vibrational frequency of each block is 2 Hz (velocity of sound is 340 m/s).
Answer» The motion of A and B is sychronized. Hence the maximum frequencies detected by D will be `f_(max)=((v+v_D)/(v-v_S))f` (when S and D are approaching each other while at their equilibrium position.) `=((340+Aomega)/(340-Aomega))f=((340+12.56)/(340-12.56))xx300=323Hz` Similarly, the apparent frequency will be minimum when the source and the detector recede from each other with maximum speed `(=Aomega)` `f_(min)=((v-Aomega)/(v+Aomega))f=((340-12.56)/(340+12.56))xx300=278.6Hz`

Discussion

83.	One train is approaching an observer at rest and another train is receding from him with the same velocity `4 m//s` . Both trains blow whistles of same frequency of `243 H_(Z)` . The beat frequency in `H_(Z)` as heard by observer is (speed of sound in air = `320 m//s`)A. 10B. 6C. 4D. 1
Answer» Correct Answer - B When the train is approaching `n_1=(v)/(v-v_s)xxn=(320)/(320-4)xx243=(80)/(79)xx243` When the train is receding, `n_2=(v)/(v+v_s)xxn=(320)/(324)xx243=(80)/(81)xx243` Beat frequency is `n=n_1-n_2=80xx243((1)/(79)-(1)/(81))=6Hz`

Discussion

84.	Due to point isotropic sound source, theintensity at a point is observed as 40 dB. The density of air is `rho=((15)/(11))(kg)/(m^3)` and velocity of sound in air is `330(m)/(s)`. Based on this information answer the following questions. Q. The pressure amplitude at the observation point isA. `3(N)/(m^2)`B. `3xx10^3(N)/(m^2)`C. `3xx10^-3(N)/(m^2)`D. `6xx10^-2(N)/(m^2)`
Answer» Correct Answer - C In the propagation of sound waves, let pressure amplitude be `trianglep_0` and displacement amplitude be `A`, then, `trianglep_0=BAK` where symbols have their usual meanings. We have `FL=10log((I)/(I_0))` `implies40=10log((I)/(10^(-12)))` `impliesI=10^-8(W)/(m^2)` `I=(trianglep_0^2)/(2rhov)` `impliestrianglep_0=sqrt(Ixx2rhov)` `=sqrt(10^-8xx2xx(15)/(11)xx330)(N)/(m^2)` `=3xx10^-3(N)/(m^2)`

Discussion

85.	Due to point isotropic sound source, the intensity at a point is observed as 40 dB. The density of air is `rho=((15)/(11))(kg)/(m^3)` and velocity of sound in air is `330(m)/(s)`. Based on this information answer the following questions. Q. The ratio of displacement amplitude of wave at observation point to wavelength of sound waves isA. `3.22xx10^-6`B. `3.22xx10^-12`C. `3.22xx10^-9`D. `1.07xx10^-10`
Answer» Correct Answer - C We have `trianglep_0=BAK` `impliestrianglep_0=rhov^2Axxk` `impliestrianglep_0=(rhov^2Axx2pi)/(lamda)` `(A)/(lamda)=(trianglep_0)/(rhov^2xx2pi)=3.22xx10^-9`

Discussion

86.	Statement I: A 80 dB sound has twice the intensity of a 40 dB sound. Statement II: Loudness of a sound of a certain intensity I is defined as `L("in "dB)=10log_(10)((I)/(I_0))`A. Statement I is true, Statement II is true: Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is true, Statement II is NOT a correct explanation for Statement I.C. Statement I is true, Statement II is falseD. Statement I is falce: Statement II is true
Answer» Correct Answer - D `L("in "dB)=10log_(10)((I)/(I_0))` `impliesI=I_10^((L)/(10))` `(I_(80))/(I_(40))=(10^8)/(10^4)=10^4`

Discussion

87.	A sound differs by 6 dB from a sound of intensity equal to `10(nW)/(cm^2)`. Find the absolute value of intensity of the sound.
Answer» Intensity level `=log_(10)((I_2)/(I_1))bels=10log((I_2)/(I_1))dB` here, `I_1=10xx10^-9xx10^4=10^-4(W)/(m^2)` `6=10log((I_2)/(10^-4))` or `I_2=(10^(0.6))/(10^4)=(3.98)/(10^4)=4.0xx10^-4(W)/(m^2)`

Discussion

88.	A person speaking normally produces a sound intensity of `40 dB` at a distance of `1 m`. If the threshold intensity for reasonable audibility is `20 dB`, the maximum distance at which he can be heard cleary is.A. 4 mB. 5 mC. 10 mD. 20 m
Answer» Correct Answer - B `40=10log(10)((I_1)/(I_0))` `(I_1)/(I_0)=10^4` Also, `20=10log_(10)((I_2)/(I_0))` `implies(I_2)/(I_0)=10^2` .(ii) `(I_2)/(I_1)=10^-2=(r_1^2)/(r_2^2)` `r_2=100r_1^2impliesr_2=10m`

Discussion

89.	Loudness of sound from an isotropic point source at a distace of `70cm` is `20dB`. What is the distance (in m) at which it is not heard.
Answer» Intensity from a point source varies with distance as `Iprop(1)/(r^2)` Let at distace `r_1=10m` Intensity is `I_1` Then given `20=10log((I_1)/(I_0))` .(i) let for `r=r_2`, sound level be zero. Then intensity at that point should be `I_2=I_0` And `(I_1)/(I_2)=((r_2)/(r_1))^2implies(I_1)/(I_0)=((r_2)/(r_1))^2` .(ii) From Eqs. (i) and (ii) we get `20=10log((r_2)/(r_1))^2implies20=20log((r_2)/(r_1))` `implies(r_2)/(r_1)=10impliesr_2=10r_1=7m`

Discussion

90.	Due to a point isotropic sonic source, loudness at a point is `L=60dB` If density of air is `rho=((15)/(11))(kg)/(m^3)` and velocity of sound in air is `v=33(m)/(s)`, the pressure oscillation amplitude at the point of observation is `[I_0=10^-12(W)/(m^2)`]A. `0.3(N)/(m^2)`B. `0.03(N)/(m^2)`C. `3xx10^-3(N)/(m^2)`D. `3xx10^-4(N)/(m^2)`
Answer» Correct Answer - B `60dB=10dBlog((I)/(I_0))` `impliesI=(10^6xx10^-12)(W)/(m^2)=10^-6(W)/(m^2)` `[I_0=10^-12(W)/(m^2)]` `I=((triangleP_m)^2)/(2rhov)` where `rho=(15)/((11kg)/(m^3))`,`v=330(m)/(s)` `(triangleP_m)^2=2rhovI=2xx(15)/(11)xx330xx10^-6` `impliestriangleP_m=0.03(N)/(m^2)`

Discussion

91.	As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listerner along with the source is moving towards the wall with the same speed i.e., `2(m)/(s)`, such that he (listener) remains between the source and the wall, number of beats heard by him will beA. 2B. 6C. 8D. 4
Answer» Correct Answer - D Frequency received from source directly by observer will remain same. Hence frequency received by observer is `512Hz` let `f_1` be the frequency reflected by wall Then, `f_1=(v)/(v-v_S)xxf=(340)/(338)xx512=515Hz` The frequency received by observer (reflected from wall) is `f_2=((v_v_0)/(v))f=(342)/(340)xx515=518Hz` Hence beats heard is `f_2-f_1=518-512=6`.

Discussion

92.	As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listener, along with the source, is moving towards the wall with the same speed i.e., `2(m)/(s)`, such that the source remains between the listerner and the wall, number of beats heard by the listerner per second will beA. 4B. 8C. 0D. 6
Answer» Correct Answer - B As no relative motion is there between observer and listener, hence frequency heard by observer is 512 Hz. He will observe frequency reflected from wall, `f_1=515Hz`. Hence, the wave reflected from wall will act as another source of frequency 515 Hz. Therefore, the frequency received by the observer from wall is `f_2=(v+v_0)/(v)f_1` `=(342)/(340)xx515=518Hz` Hence, beats observed is `f_2f=518-512=6`.

Discussion

93.	As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. If the listener is at rest and located such that the tuning fork is moving between the listener and the wall, number of beats heard by the listerner per second will be nearly
Answer» Correct Answer - D As the source is moving away from the listetner hence frequency observed by listerner is `f_1=(v)/(v+v_S)f=(340)/(340+2)xx512` `=(340)/(342)xx512=509Hz` The frequency reflected from wall (we can assume an observer at rest) is `f_2=(v)/(v-v_S)xxf` `=(340)/(338)xx512=515Hz` Therefore beats heard by observer `(L)` is `515-509=6`.

Discussion

94.	As shown if Fig. a vibrating tuning fork of frequency 512 Hz is moving towards the wall with a speed `2(m)/(s)`. Take speed of sound as `v=340(m)/(s)` and answer the following questions. Q. Suppose that a listener is located at rest between the tuning fork and the wall. Number of beats heard by the listener per second will beA. 4B. 3C. 0D. 1
Answer» Correct Answer - B The frequency heard directly from source is given by `f_1=((v)/(v-v_S))f` Here `v=340(m)/(s)`,`v_S=2(m)/(s)`,`f=512Hz` `f_1=(340)/(338)xx512=515Hz` the frequency of the wave reflected from wall will be same (no relative motion between wall and listener, so no change in frequency). Hence no beats are observed.

Discussion

95.	The sound level at a point is increased by 30 dB. What is factor is the pressure amplitude increased?
Answer» The sound level is dB is `L=10log_(10)((I)/(I_0))` If `L_1` and `L_2` are the sound levels and `I_1` and `I_2` are the coresponding intensities in the two cases, then `L_2-L_1=10[log_(10)((I_2)/(I_0))-log_(10)((I_1)/(I_0))]` `implies30=10log_(10)((I_2)/(I_1))` `implies(I_2)/(I_1)=10^3` As the intensity is proporional to the square of the pressure amplitude thus we have `(trianglep_2)/(trianglep_1)=sqrt((I_2)/(I_1))=sqrt(1000)=32`

Discussion

96.	The sound level at a distance of 3.00 m from a source is 120 dB. At what distance will the sound level be (a) 100 dB and (b) 10.0 dB?
Answer» A reduction of 20 dB means reducing the intensity by a factor of `10^2` so we expect the radial distance to be 10 times larger, namely 30 m. A further reduction of 90 dB may correspond to an extra factor of `10^(4.5)` in distance, to about `30xx30000m`, or about 1000 km. We use the difinition of the decibel scale and the inverse suqare law for sound intensity. From the difinition of sound level, `beta=10log((1)/(10^(-12)(W)/(m^2)))` We can compute the intensities corresponding to each of the levels mentioned as `I=[10^((beta)/(10))]10^(-12)(W)/(m^2)` They are `I_(120)=1(W)/(m^2)`,`I_(100)=10^(-2)(W)/(m^2)` and `I_(10)=10^(-11)(W)/(m^2)`. a. The power passing through any sphere around the source is `P=4pir^2l` If we ignore absoption of sound by the medium, conservation of energy requires that `r_(120)^(2)I_(120)=r_(100)^(2)I_(100)=r_(10)^(2)I_(10)` then, `r_(100)=r_(120)sqrt((I_(120))/(I_(100)))=(3.00m)sqrt((1(W)/(m^2))/(10^(-2)(W)/(m^2)))=30.0m` b. `r_(10)=r_(120)sqrt((I_(120))/(I_(10)))=(3.00m)sqrt((1(W)/(m^2))/(10^(-11)(W)/(m^2)))=9.49xx10^5m` At 949 km away the faint 10 dB sound would not be indetifiable among many other soft and loud sounds produced by other sources across the continent. And 949 km of air is such a thick screen that it may do a significant amound of sound absorption.

Discussion

97.	Seven grams of nitrogen is mixed with 12 g of oxygen in a tube and then sealed. Calculate the velocity of sound through the tube at `27^@C`.
Answer» `c=sqrt((gammaRT)/(M))` Since nitrogen and oxygen are both diatomic, `gamma=1.4` `(m_1+m_2)/(M)=(m_1)/(M_1)+(m_2)/(M_2)` `implies(7+12)/(M)=(7)/(28)+(12)/(32)` `impliesM=30.4` `c=sqrt((1.4xx8.3xx300)/(30.4xx10^-3))=338.6(m)/(s)`

Discussion

98.	A window whose area is `2m^2` opens on street where the street noise result in an intensity level at the window of 60 dB. How much acoustic power enters the window via sound waves. Now if an acoustic absorber is fitted at the window, how much energy from street will it collect in 5 h?
Answer» Loudness level is given as `L=10log((I)/(I_0))` Thus, `10log((I)/(I_0))=60` `implies(I)/(I_0)=10^6` `impliesI=(10^-12xx10^6)=10^-6(W)/(m^2)=1mu(W)/(m^2)` and the power of sound is given by `P=IS` `=1xx10^-6xx2=2muW` Thus energy is given by `E=Pxxt` `=2xx10^-6xx5xx60xx60=36xx10^-3J`

Discussion

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