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SOUND WAVES
EXERCISES- Q-81 to Q-89
81. A boy riding on a bicycle going at 12 km/h towards a vertical wall, whistles at his dog on the ground. If the frequency of the whistle is 1600 Hz and the speed of sound in air is 330 m/s, find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy.
ANSWER: (a) The speed of the boy u = 12 km/h
=12000/3600 m/s
=10/3 m/s
The frequency of the whistle ν = 1600 Hz.
The speed of sound in air V = 330 m/s
The apparent frequency received by the wall
ν' = Vν/(V-u)
= 330*1600/(330-10/3) Hz
= 330*1600*3/(990-10) Hz
= 330*1600*3/980 Hz
= 1616 Hz
(b) The apparent frequency is reflected by the wall which acts as a source now. The boy as an observer is moving. Hence the apparent frequency received by the boy = (V+u)𝜈'/V
= (330+10/3)1616/330 Hz
= 1000*1616/330*3 Hz
= 1632 Hz
82. A person standing on the road sends a sound signal to the driver of a car going away from him at a speed of 72 km/h. The signal traveling at 330 m/s in the air and having a frequency of 1600 Hz gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person.
ANSWER: The speed of the car u = 72 km/h
=72000/3600 m/s
=20 m/s
The frequency of the sound 𝜈 = 1600 Hz
The speed of the sound V = 330 m/s
Apparent frequency received by the car
𝜈' = (V-u)𝜈/V
= (330-20)1600/330 Hz
= 310*1600/330 Hz
= 1503 Hz
This is the frequency reflected by the car. The apparent frequency received by the person
= V𝜈/(V+u)
= 330*1503/(330+20) Hz
= 330*1503/350 Hz
= 1417 Hz
83. A car moves with a speed of 54 km/h towards a cliff. The horn of the car emits a sound of frequency 400 Hz at a speed of 335 m/s. (a) Find the wavelength of the sound emitted by the horn in front of the car. (b) Find the wavelength of the wave reflected from the cliff. (c) What frequency does a person sitting in the car hear for the reflected sound wave? (d) How many beats does he hear in 10 seconds between the sound coming directly from the horn and that coming after the reflection?
ANSWER: The frequency of the sound 𝜈 = 400 Hz
The speed of sound in air V = 335 m/s
The speed of car u = 54 km/h = 54000/3600 m/s
=15 m/s
(a) The apparent frequency in front of the car
𝜈' = V𝜈/(V-u) =335*400/(335-15) Hz
=418.75 Hz
Hence the wavelength = V/𝜈' = 335/418.75 m
= 0.80 m = 80 cm
(b) The same apparent frequency will be reflected by the cliff, hence the wavelength of the reflected frequency will be the same i.e. = 80 cm
(c) For the person sitting in the car, the apparent frequency reflected by the cliff is the frequency from the stationary source while the observer is moving at a speed of 15 m/s. Hence the frequency heard by him = (V+u)𝜈'/V
= (335+15)418.75/335 Hz
= 350*418.75/335 Hz
≈ 437 Hz
(d) The beat frequency reached the person in the car
= 437 -400 Hz
= 37 Hz
Since the beat frequency is more than 16 Hz which is beyond the normal person's perception, hence the person may not hear any beat.
84. An operator sitting in his base camp sends a sound signal of frequency 400 Hz. The signal is reflected back from a car moving towards him. The frequency of the reflected sound is found to be 410 Hz. Find the speed of the car. The speed of sound in air = 324 m/s.
ANSWER: The speed of sound in air V = 324 m/s.
Original frequency produced 𝜈 = 400 Hz. Let the speed of the car = u.
The frequency received by the car 𝜈' = (V+u)𝜈/V
The frequency 𝜈' is reflected by the car. Now for the reflected frequency, the source is approaching and the observer is stationary. The frequency of reflected sound received by the operator
𝜈" = V𝜈'/(V-u) = (V+u)𝜈/(V-u)
Putting the values,
410 = (324+u)*400/(324-u)
→410*324-410 u = 324*400 + 400 u
→810 u = 324(410-400) =324*10
→u = 324/81 m/s =4 m/s
85. Figure (16-E12) shows a source of sound moving along the X-axis at a speed of 22 m/s continuously emitting a sound of frequency 2.0 kHz which travels in air at a speed of 330 m/s. A listener Q stands on the Y-axis at a distance of 330 m from the origin. At t = 0, the source crosses the origin P. (a) When does the sound emitted from the source at P reach the listener at Q? (b) What will be the frequency heard by the listener at this instant? (c) Where will the source be at this instant?
The figure for Q-85
ANSWER: (a) The distance of the listener from the origin, d = 330 m. The speed of sound in air, V = 330 m/s. The time taken by the sound to reach the listener,
t = d/V =330/330 s = 1 second.
(b) Since the direction of the movement of the source is perpendicular to the line joining the source and the person, hence the speed of the source towards the listener will be zero. The same frequency of 2.0 kHz will be heard by the person.
(c) Since the time to reach the listener = 1 s and the speed of the source = 22 m/s. Hence at this instant, the source will move 22 m. Its position will be
at x = 22 m.
86. A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22m/s. A listener is situated on the ground at the position (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. The speed of sound in air = 330 m/s.
ANSWER: Assume that when the source crosses the origin, the sound received by the listener was produced at S on the Y-axis. Hence the time taken t by the source to reach the origin will be equal to the time taken by the sound to travel the distance SP. (See the diagram below.)
Diagram for Q-86
Hence OS = ut, {where u = speed of the source}
and OP = Vt, {where V = speed of sound}
Hence for the angle ∠PSO,
cos α = OS/OP =ut/Vt =u/v
Hence the instance speed of the source towards the listener u' = u.cosα =u.u/V = u²/V
Hence the apparent frequency heard by the listener
𝜈' = V𝜈/(V-u')
→𝜈' = V𝜈/{V-u²/V}= V²𝜈/(V²-u²)
V = 330 m/s, 𝜈 = 4000 Hz, u = 22 m/s
→𝜈' = 330²*4000/(330²-22²)
= 4018 Hz
87. A source of sound emitting a 1200 Hz note travels along a straight line at a speed of 170 m/s. A detector is placed at a distance of 200 m from the line of motion of the source. (a) Find the frequency of the sound received by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant it detects the frequency 1200 Hz. The velocity of sound in air = 340 m/s.
ANSWER: The frequency of the sound, 𝜈 = 1200 Hz.
u = 170 m/s, the speed of sound, V = 340 m/s.
(a) The source is closest at O. {see the diagram below}.
Diagram for Q-87
But at this instant, the listener hears the sound produced at S. The time taken t by the source in coming from S to O is the same as the time taken by the sound in coming from S to P. Hence SP = Vt and SO = ut.
Hence cos α = SO/SP = ut/Vt = u/V. The sound is approaching the listener at S with a speed
u' = u.cos α =u²/V.
Hence the frequency of sound heard by the listener
𝜈' = V𝜈/(V-u') = V𝜈/(V-u²/V) =V²𝜈/(V²-u²)
→𝜈' = 340²*1200/(340²-170²)
=340²*1200/(340+170)(340-170) Hz
=340²*1200/(510*170) Hz
=2*340*1200/510 Hz
=2*2*1200/3 Hz
=4*400 Hz
=1600 Hz
(b) The detector will detect the original frequency of 1200 Hz when the sound produced at O reaches P, because when at P, the relative motion between O and P is zero. Time taken by the sound in traveling from O to P,
t' = OP/V =200/340 s =10/17 s
In this time the source reaches Q. Hence,
OQ = u*t' = 170*10/17 m =100 m.
The distance between the source and the detector when it detects the frequency of 1200 Hz = PQ.
→PQ² = PO²+OQ²
→PQ² =200²+100² =100²(2²+1)=100²*5
→PQ = 100√(5) ≈224 m
88. A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point
(a) An observer is located in the same vertical plane at a large distance and at the same height as the center of the circle (Figure 16-E13). The speed of sound in air = 330 m/s and g =10 m/s². Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the center of the circle. Find the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the same height as the center.
The figure for Q-88
ANSWER: (a) The radius of the circle, r = 1.6 m. If the speed of the source be u, and its mass =m then at the highest point the weight mg is just equal to the centrifugal force mu²/r.
→mu²/r = mg
→u² = gr
→u =√gr
The maximum frequency is heard by the listener when the source is at the lowest point of the circle because it is then approaching with the maximum speed u'.
½mu'² - ½mu² = mg*2r
→u'² = 4gr+u² =4gr+gr =5gr
→u' = √(5gr) =√(5*10*1.6) =√80 =4√5 m/s
The frequency heard = V𝜈/(V-u') =V𝜈/(V-√5gr)
=330*500/{330-4√5} Hz
=514 Hz
Note: If we take the source moving clockwise, then at the top point it is approaching the listener at the maximum speed u = √gr =√(10*1.6) =√16 =4 m/s.
Hence the maximum frequency heard
=V𝜈/(V-u)
=330*500/(330-4) Hz
=330*500/326 Hz
=506 Hz
(b) When the source is at the same height as the center, it has two positions- one at the left side and another on the right side. At one place it is going down while at the other place it is coming up. Let the speed at this level = U. Now
the difference of the K.E. = change in P.E.
→½mU² - ½mu² = mgr
→U² = u² + 2gr =gr + 2gr = 3gr
→U = √(3gr) =√(3*10*1.6) =4√3 m/s
Now the maximum frequency heard when the source is going up with U.
Maximum frequency heard = V𝜈/(V-U)
=330*500/(330-4√3)
=511 Hz
When the source is going down with speed U, the minimum frequency is heard,
=V𝜈/(V+U)
=330*500/(330+4√3)
=490 Hz
89. A source emitting a sound of frequency 𝜈 is placed at a large distance from the observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is V.
ANSWER: The first pulse emitted (at t = 0 s) by the source at A reaches the observer at O in time AO/V =d/V, where AO = d.
Let the second pulse be emitted by the source at time t = T. In this time the source moves a distance
s = 0*T+½aT² = ½aT²
Now the source is at a distance = d-s =d-½aT²
The pulse produced here will reach the listener in time = (d-½aT²)/V
Diagram for Q-89
Hence the time between two consecutive pulses heard
={T+ (d-½aT²)/V} -d/V
=T-aT²/2V
=Time period of the pulse heard =T'
Hence the frequency of the sound heard
𝜈' =1/T' = 1/{T-aT²/2V}
But T =1/𝜈
→𝜈' = 1/{1/𝜈 - a/2V𝜈²} =2V𝜈²/{2V𝜈-a}
81. A boy riding on a bicycle going at 12 km/h towards a vertical wall, whistles at his dog on the ground. If the frequency of the whistle is 1600 Hz and the speed of sound in air is 330 m/s, find (a) the frequency of the whistle as received by the wall (b) the frequency of the reflected whistle as received by the boy.
ANSWER: (a) The speed of the boy u = 12 km/h
=12000/3600 m/s
=10/3 m/s
The frequency of the whistle ν = 1600 Hz.
The speed of sound in air V = 330 m/s
The apparent frequency received by the wall
ν' = Vν/(V-u)
= 330*1600/(330-10/3) Hz
= 330*1600*3/(990-10) Hz
= 330*1600*3/980 Hz
= 1616 Hz
(b) The apparent frequency is reflected by the wall which acts as a source now. The boy as an observer is moving. Hence the apparent frequency received by the boy = (V+u)𝜈'/V
= (330+10/3)1616/330 Hz
= 1000*1616/330*3 Hz
= 1632 Hz
82. A person standing on the road sends a sound signal to the driver of a car going away from him at a speed of 72 km/h. The signal traveling at 330 m/s in the air and having a frequency of 1600 Hz gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person.
ANSWER: The speed of the car u = 72 km/h
=72000/3600 m/s
=20 m/s
The frequency of the sound 𝜈 = 1600 Hz
The speed of the sound V = 330 m/s
Apparent frequency received by the car
𝜈' = (V-u)𝜈/V
= (330-20)1600/330 Hz
= 310*1600/330 Hz
= 1503 Hz
This is the frequency reflected by the car. The apparent frequency received by the person
= V𝜈/(V+u)
= 330*1503/(330+20) Hz
= 330*1503/350 Hz
= 1417 Hz
83. A car moves with a speed of 54 km/h towards a cliff. The horn of the car emits a sound of frequency 400 Hz at a speed of 335 m/s. (a) Find the wavelength of the sound emitted by the horn in front of the car. (b) Find the wavelength of the wave reflected from the cliff. (c) What frequency does a person sitting in the car hear for the reflected sound wave? (d) How many beats does he hear in 10 seconds between the sound coming directly from the horn and that coming after the reflection?
ANSWER: The frequency of the sound 𝜈 = 400 Hz
The speed of sound in air V = 335 m/s
The speed of car u = 54 km/h = 54000/3600 m/s
=15 m/s
(a) The apparent frequency in front of the car
𝜈' = V𝜈/(V-u) =335*400/(335-15) Hz
=418.75 Hz
Hence the wavelength = V/𝜈' = 335/418.75 m
= 0.80 m = 80 cm
(b) The same apparent frequency will be reflected by the cliff, hence the wavelength of the reflected frequency will be the same i.e. = 80 cm
(c) For the person sitting in the car, the apparent frequency reflected by the cliff is the frequency from the stationary source while the observer is moving at a speed of 15 m/s. Hence the frequency heard by him = (V+u)𝜈'/V
= (335+15)418.75/335 Hz
= 350*418.75/335 Hz
≈ 437 Hz
(d) The beat frequency reached the person in the car
= 437 -400 Hz
= 37 Hz
Since the beat frequency is more than 16 Hz which is beyond the normal person's perception, hence the person may not hear any beat.
84. An operator sitting in his base camp sends a sound signal of frequency 400 Hz. The signal is reflected back from a car moving towards him. The frequency of the reflected sound is found to be 410 Hz. Find the speed of the car. The speed of sound in air = 324 m/s.
ANSWER: The speed of sound in air V = 324 m/s.
Original frequency produced 𝜈 = 400 Hz. Let the speed of the car = u.
The frequency received by the car 𝜈' = (V+u)𝜈/V
The frequency 𝜈' is reflected by the car. Now for the reflected frequency, the source is approaching and the observer is stationary. The frequency of reflected sound received by the operator
𝜈" = V𝜈'/(V-u) = (V+u)𝜈/(V-u)
Putting the values,
410 = (324+u)*400/(324-u)
→410*324-410 u = 324*400 + 400 u
→810 u = 324(410-400) =324*10
→u = 324/81 m/s =4 m/s
85. Figure (16-E12) shows a source of sound moving along the X-axis at a speed of 22 m/s continuously emitting a sound of frequency 2.0 kHz which travels in air at a speed of 330 m/s. A listener Q stands on the Y-axis at a distance of 330 m from the origin. At t = 0, the source crosses the origin P. (a) When does the sound emitted from the source at P reach the listener at Q? (b) What will be the frequency heard by the listener at this instant? (c) Where will the source be at this instant?
ANSWER: (a) The distance of the listener from the origin, d = 330 m. The speed of sound in air, V = 330 m/s. The time taken by the sound to reach the listener,
t = d/V =330/330 s = 1 second.
(b) Since the direction of the movement of the source is perpendicular to the line joining the source and the person, hence the speed of the source towards the listener will be zero. The same frequency of 2.0 kHz will be heard by the person.
(c) Since the time to reach the listener = 1 s and the speed of the source = 22 m/s. Hence at this instant, the source will move 22 m. Its position will be
at x = 22 m.
86. A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22m/s. A listener is situated on the ground at the position (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. The speed of sound in air = 330 m/s.
ANSWER: Assume that when the source crosses the origin, the sound received by the listener was produced at S on the Y-axis. Hence the time taken t by the source to reach the origin will be equal to the time taken by the sound to travel the distance SP. (See the diagram below.)
Hence OS = ut, {where u = speed of the source}
and OP = Vt, {where V = speed of sound}
Hence for the angle ∠PSO,
cos α = OS/OP =ut/Vt =u/v
Hence the instance speed of the source towards the listener u' = u.cosα =u.u/V = u²/V
Hence the apparent frequency heard by the listener
𝜈' = V𝜈/(V-u')
→𝜈' = V𝜈/{V-u²/V}= V²𝜈/(V²-u²)
V = 330 m/s, 𝜈 = 4000 Hz, u = 22 m/s
→𝜈' = 330²*4000/(330²-22²)
= 4018 Hz
87. A source of sound emitting a 1200 Hz note travels along a straight line at a speed of 170 m/s. A detector is placed at a distance of 200 m from the line of motion of the source. (a) Find the frequency of the sound received by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant it detects the frequency 1200 Hz. The velocity of sound in air = 340 m/s.
ANSWER: The frequency of the sound, 𝜈 = 1200 Hz.
u = 170 m/s, the speed of sound, V = 340 m/s.
(a) The source is closest at O. {see the diagram below}.
But at this instant, the listener hears the sound produced at S. The time taken t by the source in coming from S to O is the same as the time taken by the sound in coming from S to P. Hence SP = Vt and SO = ut.
Hence cos α = SO/SP = ut/Vt = u/V. The sound is approaching the listener at S with a speed
u' = u.cos α =u²/V.
Hence the frequency of sound heard by the listener
𝜈' = V𝜈/(V-u') = V𝜈/(V-u²/V) =V²𝜈/(V²-u²)
→𝜈' = 340²*1200/(340²-170²)
=340²*1200/(340+170)(340-170) Hz
=340²*1200/(510*170) Hz
=2*340*1200/510 Hz
=2*2*1200/3 Hz
=4*400 Hz
=1600 Hz
(b) The detector will detect the original frequency of 1200 Hz when the sound produced at O reaches P, because when at P, the relative motion between O and P is zero. Time taken by the sound in traveling from O to P,
t' = OP/V =200/340 s =10/17 s
In this time the source reaches Q. Hence,
OQ = u*t' = 170*10/17 m =100 m.
The distance between the source and the detector when it detects the frequency of 1200 Hz = PQ.
→PQ² = PO²+OQ²
→PQ² =200²+100² =100²(2²+1)=100²*5
→PQ = 100√(5) ≈224 m
88. A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point
(a) An observer is located in the same vertical plane at a large distance and at the same height as the center of the circle (Figure 16-E13). The speed of sound in air = 330 m/s and g =10 m/s². Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the center of the circle. Find the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the same height as the center.
ANSWER: (a) The radius of the circle, r = 1.6 m. If the speed of the source be u, and its mass =m then at the highest point the weight mg is just equal to the centrifugal force mu²/r.
→mu²/r = mg
→u² = gr
→u =√gr
The maximum frequency is heard by the listener when the source is at the lowest point of the circle because it is then approaching with the maximum speed u'.
½mu'² - ½mu² = mg*2r
→u'² = 4gr+u² =4gr+gr =5gr
→u' = √(5gr) =√(5*10*1.6) =√80 =4√5 m/s
The frequency heard = V𝜈/(V-u') =V𝜈/(V-√5gr)
=330*500/{330-4√5} Hz
=514 Hz
Note: If we take the source moving clockwise, then at the top point it is approaching the listener at the maximum speed u = √gr =√(10*1.6) =√16 =4 m/s.
Hence the maximum frequency heard
=V𝜈/(V-u)
=330*500/(330-4) Hz
=330*500/326 Hz
=506 Hz
(b) When the source is at the same height as the center, it has two positions- one at the left side and another on the right side. At one place it is going down while at the other place it is coming up. Let the speed at this level = U. Now
the difference of the K.E. = change in P.E.
→½mU² - ½mu² = mgr
→U² = u² + 2gr =gr + 2gr = 3gr
→U = √(3gr) =√(3*10*1.6) =4√3 m/s
Now the maximum frequency heard when the source is going up with U.
Maximum frequency heard = V𝜈/(V-U)
=330*500/(330-4√3)
=511 Hz
When the source is going down with speed U, the minimum frequency is heard,
=V𝜈/(V+U)
=330*500/(330+4√3)
=490 Hz
89. A source emitting a sound of frequency 𝜈 is placed at a large distance from the observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is V.
ANSWER: The first pulse emitted (at t = 0 s) by the source at A reaches the observer at O in time AO/V =d/V, where AO = d.
Let the second pulse be emitted by the source at time t = T. In this time the source moves a distance
s = 0*T+½aT² = ½aT²
Now the source is at a distance = d-s =d-½aT²
The pulse produced here will reach the listener in time = (d-½aT²)/V
Hence the time between two consecutive pulses heard
={T+ (d-½aT²)/V} -d/V
=T-aT²/2V
=Time period of the pulse heard =T'
Hence the frequency of the sound heard
𝜈' =1/T' = 1/{T-aT²/2V}
But T =1/𝜈
→𝜈' = 1/{1/𝜈 - a/2V𝜈²} =2V𝜈²/{2V𝜈-a}
ANSWER: (a) The speed of the boy u = 12 km/h
=12000/3600 m/s
=10/3 m/s
The frequency of the whistle ν = 1600 Hz.
The speed of sound in air V = 330 m/s
The apparent frequency received by the wall
ν' = Vν/(V-u)
= 330*1600/(330-10/3) Hz
= 330*1600*3/(990-10) Hz
= 330*1600*3/980 Hz
= 1616 Hz
(b) The apparent frequency is reflected by the wall which acts as a source now. The boy as an observer is moving. Hence the apparent frequency received by the boy = (V+u)𝜈'/V
= (330+10/3)1616/330 Hz
= 1000*1616/330*3 Hz
= 1632 Hz
82. A person standing on the road sends a sound signal to the driver of a car going away from him at a speed of 72 km/h. The signal traveling at 330 m/s in the air and having a frequency of 1600 Hz gets reflected from the body of the car and returns. Find the frequency of the reflected signal as heard by the person.
ANSWER: The speed of the car u = 72 km/h
=72000/3600 m/s
=20 m/s
The frequency of the sound 𝜈 = 1600 Hz
The speed of the sound V = 330 m/s
Apparent frequency received by the car
𝜈' = (V-u)𝜈/V
= (330-20)1600/330 Hz
= 310*1600/330 Hz
= 1503 Hz
This is the frequency reflected by the car. The apparent frequency received by the person
= V𝜈/(V+u)
= 330*1503/(330+20) Hz
= 330*1503/350 Hz
= 1417 Hz
83. A car moves with a speed of 54 km/h towards a cliff. The horn of the car emits a sound of frequency 400 Hz at a speed of 335 m/s. (a) Find the wavelength of the sound emitted by the horn in front of the car. (b) Find the wavelength of the wave reflected from the cliff. (c) What frequency does a person sitting in the car hear for the reflected sound wave? (d) How many beats does he hear in 10 seconds between the sound coming directly from the horn and that coming after the reflection?
ANSWER: The frequency of the sound 𝜈 = 400 Hz
The speed of sound in air V = 335 m/s
The speed of car u = 54 km/h = 54000/3600 m/s
=15 m/s
(a) The apparent frequency in front of the car
𝜈' = V𝜈/(V-u) =335*400/(335-15) Hz
=418.75 Hz
Hence the wavelength = V/𝜈' = 335/418.75 m
= 0.80 m = 80 cm
(b) The same apparent frequency will be reflected by the cliff, hence the wavelength of the reflected frequency will be the same i.e. = 80 cm
(c) For the person sitting in the car, the apparent frequency reflected by the cliff is the frequency from the stationary source while the observer is moving at a speed of 15 m/s. Hence the frequency heard by him = (V+u)𝜈'/V
= (335+15)418.75/335 Hz
= 350*418.75/335 Hz
≈ 437 Hz
(d) The beat frequency reached the person in the car
= 437 -400 Hz
= 37 Hz
Since the beat frequency is more than 16 Hz which is beyond the normal person's perception, hence the person may not hear any beat.
84. An operator sitting in his base camp sends a sound signal of frequency 400 Hz. The signal is reflected back from a car moving towards him. The frequency of the reflected sound is found to be 410 Hz. Find the speed of the car. The speed of sound in air = 324 m/s.
ANSWER: The speed of sound in air V = 324 m/s.
Original frequency produced 𝜈 = 400 Hz. Let the speed of the car = u.
The frequency received by the car 𝜈' = (V+u)𝜈/V
The frequency 𝜈' is reflected by the car. Now for the reflected frequency, the source is approaching and the observer is stationary. The frequency of reflected sound received by the operator
𝜈" = V𝜈'/(V-u) = (V+u)𝜈/(V-u)
Putting the values,
410 = (324+u)*400/(324-u)
→410*324-410 u = 324*400 + 400 u
→810 u = 324(410-400) =324*10
→u = 324/81 m/s =4 m/s
85. Figure (16-E12) shows a source of sound moving along the X-axis at a speed of 22 m/s continuously emitting a sound of frequency 2.0 kHz which travels in air at a speed of 330 m/s. A listener Q stands on the Y-axis at a distance of 330 m from the origin. At t = 0, the source crosses the origin P. (a) When does the sound emitted from the source at P reach the listener at Q? (b) What will be the frequency heard by the listener at this instant? (c) Where will the source be at this instant?
 |
| The figure for Q-85 |
ANSWER: (a) The distance of the listener from the origin, d = 330 m. The speed of sound in air, V = 330 m/s. The time taken by the sound to reach the listener,
t = d/V =330/330 s = 1 second.
(b) Since the direction of the movement of the source is perpendicular to the line joining the source and the person, hence the speed of the source towards the listener will be zero. The same frequency of 2.0 kHz will be heard by the person.
(c) Since the time to reach the listener = 1 s and the speed of the source = 22 m/s. Hence at this instant, the source will move 22 m. Its position will be
at x = 22 m.
86. A source emitting sound at frequency 4000 Hz, is moving along the Y-axis with a speed of 22m/s. A listener is situated on the ground at the position (660 m, 0). Find the frequency of the sound received by the listener at the instant the source crosses the origin. The speed of sound in air = 330 m/s.
ANSWER: Assume that when the source crosses the origin, the sound received by the listener was produced at S on the Y-axis. Hence the time taken t by the source to reach the origin will be equal to the time taken by the sound to travel the distance SP. (See the diagram below.)
 |
| Diagram for Q-86 |
Hence OS = ut, {where u = speed of the source}
and OP = Vt, {where V = speed of sound}
Hence for the angle ∠PSO,
cos α = OS/OP =ut/Vt =u/v
Hence the instance speed of the source towards the listener u' = u.cosα =u.u/V = u²/V
Hence the apparent frequency heard by the listener
𝜈' = V𝜈/(V-u')
→𝜈' = V𝜈/{V-u²/V}= V²𝜈/(V²-u²)
V = 330 m/s, 𝜈 = 4000 Hz, u = 22 m/s
→𝜈' = 330²*4000/(330²-22²)
= 4018 Hz
87. A source of sound emitting a 1200 Hz note travels along a straight line at a speed of 170 m/s. A detector is placed at a distance of 200 m from the line of motion of the source. (a) Find the frequency of the sound received by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant it detects the frequency 1200 Hz. The velocity of sound in air = 340 m/s.
ANSWER: The frequency of the sound, 𝜈 = 1200 Hz.
u = 170 m/s, the speed of sound, V = 340 m/s.
(a) The source is closest at O. {see the diagram below}.
 |
| Diagram for Q-87 |
But at this instant, the listener hears the sound produced at S. The time taken t by the source in coming from S to O is the same as the time taken by the sound in coming from S to P. Hence SP = Vt and SO = ut.
Hence cos α = SO/SP = ut/Vt = u/V. The sound is approaching the listener at S with a speed
u' = u.cos α =u²/V.
Hence the frequency of sound heard by the listener
𝜈' = V𝜈/(V-u') = V𝜈/(V-u²/V) =V²𝜈/(V²-u²)
→𝜈' = 340²*1200/(340²-170²)
=340²*1200/(340+170)(340-170) Hz
=340²*1200/(510*170) Hz
=2*340*1200/510 Hz
=2*2*1200/3 Hz
=4*400 Hz
=1600 Hz
(b) The detector will detect the original frequency of 1200 Hz when the sound produced at O reaches P, because when at P, the relative motion between O and P is zero. Time taken by the sound in traveling from O to P,
t' = OP/V =200/340 s =10/17 s
In this time the source reaches Q. Hence,
OQ = u*t' = 170*10/17 m =100 m.
The distance between the source and the detector when it detects the frequency of 1200 Hz = PQ.
→PQ² = PO²+OQ²
→PQ² =200²+100² =100²(2²+1)=100²*5
→PQ = 100√(5) ≈224 m
88. A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point
(a) An observer is located in the same vertical plane at a large distance and at the same height as the center of the circle (Figure 16-E13). The speed of sound in air = 330 m/s and g =10 m/s². Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the center of the circle. Find the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the same height as the center.
 |
| The figure for Q-88 |
ANSWER: (a) The radius of the circle, r = 1.6 m. If the speed of the source be u, and its mass =m then at the highest point the weight mg is just equal to the centrifugal force mu²/r.
→mu²/r = mg
→u² = gr
→u =√gr
The maximum frequency is heard by the listener when the source is at the lowest point of the circle because it is then approaching with the maximum speed u'.
½mu'² - ½mu² = mg*2r
→u'² = 4gr+u² =4gr+gr =5gr
→u' = √(5gr) =√(5*10*1.6) =√80 =4√5 m/s
The frequency heard = V𝜈/(V-u') =V𝜈/(V-√5gr)
=330*500/{330-4√5} Hz
=514 Hz
Note: If we take the source moving clockwise, then at the top point it is approaching the listener at the maximum speed u = √gr =√(10*1.6) =√16 =4 m/s.
Hence the maximum frequency heard
=V𝜈/(V-u)
=330*500/(330-4) Hz
=330*500/326 Hz
=506 Hz
(b) When the source is at the same height as the center, it has two positions- one at the left side and another on the right side. At one place it is going down while at the other place it is coming up. Let the speed at this level = U. Now
the difference of the K.E. = change in P.E.
→½mU² - ½mu² = mgr
→U² = u² + 2gr =gr + 2gr = 3gr
→U = √(3gr) =√(3*10*1.6) =4√3 m/s
Now the maximum frequency heard when the source is going up with U.
Maximum frequency heard = V𝜈/(V-U)
=330*500/(330-4√3)
=511 Hz
When the source is going down with speed U, the minimum frequency is heard,
=V𝜈/(V+U)
=330*500/(330+4√3)
=490 Hz
89. A source emitting a sound of frequency 𝜈 is placed at a large distance from the observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is V.
ANSWER: The first pulse emitted (at t = 0 s) by the source at A reaches the observer at O in time AO/V =d/V, where AO = d.
Let the second pulse be emitted by the source at time t = T. In this time the source moves a distance
s = 0*T+½aT² = ½aT²
Now the source is at a distance = d-s =d-½aT²
The pulse produced here will reach the listener in time = (d-½aT²)/V
 |
| Diagram for Q-89 |
Hence the time between two consecutive pulses heard
={T+ (d-½aT²)/V} -d/V
=T-aT²/2V
=Time period of the pulse heard =T'
Hence the frequency of the sound heard
𝜈' =1/T' = 1/{T-aT²/2V}
But T =1/𝜈
→𝜈' = 1/{1/𝜈 - a/2V𝜈²} =2V𝜈²/{2V𝜈-a}
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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