My Channel on YouTube → SimplePhysics with KK
For links to
other chapters - See bottom of the page
SOUND WAVES
EXERCISES- Q-71 to Q-80
71. Figure (16-E11) shows a person standing somewhere in between two identical tuning forks, each vibrating at 512 Hz. If both the tuning forks move towards the right at a speed of 5.5 m/s, find the number of beats heard by the listener. The speed of sound in air = 330 m/s.
The figure for Q-71
ANSWER: The observer is standing and the sources are moving. V = 330 m/s. u = 5.5 m/s. ν = 512 Hz.
The left tuning fork is approaching the observer, hence the apparent frequency ν' =Vν/(V-u)
=330*512/(330-5.5) Hz
=520.68 Hz
The right tuning fork is leaving the observer, hence the apparent frequency ν" =Vν/(V+u)
=330*512/(330+5.5) Hz
=503.61 Hz
So the number of beats produced for the listener = ν'-ν"
=520.68 - 503.61 Hz
=17.07 Hz ≈ 17.1 Hz > 16 Hz.
Since an average human ear can not distinguish the beats of more than 16 Hz frequency, hence the listener, in this case, may not be able to distinguish the beats.
72. A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius of 100/π cm at a constant angular speed of 5.0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and maximum frequency of the sound observed. The speed of sound in air =332 m/s.
ANSWER: The frequency of the sound, ν = 500 Hz.
The radius of the circle, r = 100/π cm =1/π m.
Angular speed ⍵ = 5*2π rad/s =10π rad/s.
Hence the magnitude of the instantaneous velocity of the source,
u = ⍵r = 10π*1/π m/s =10 m/s.
As it is clear from the diagram below, the maximum apparent frequency will be observed by the listener when the direction of the velocity is towards him, i.e. at A.
Diagram for Q-72
Similarly, the minimum frequency will be observed at B when the direction of the source is away from him.
In both the cases, the observer is stationary and the source is moving. Hence the apparent frequency heard when the source is at A
ν' =Vν/(V-u) =332*500/(332-10) Hz
=332*500/322 Hz
≈515 Hz
The apparent frequency heard when the source is at B
ν" =Vν/(V+u)
=332*500/(332+10) Hz
=332*500/342 Hz
≈485 Hz
73. Two trains are traveling towards each other both at a speed of 90 km/h. If one of the trains sounds a whistle at 500 Hz, what will be the apparent frequency heard in the other train? The speed of sound in air = 350 m/s.
ANSWER: The frequency of the source, ν = 500 Hz.
The speed of the source uₛ = 90 km/h
=90000/3600 m/s =25 m/s
The speed of the observer, uₒ =90 km/h =25 m/s.
The speed of the sound V = 350 m/s.
Hence the apparent frequency heard in the other train ν' = (V+uₒ)ν/(V-uₛ)
=(350+25)*500/(350-25)
=375*500/325 Hz
=577 Hz
74. A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car driver does not stop and takes the plea in court that because of the Doppler shift the frequency of the whistle reaching him might have gone beyond the audible limit of 20 kHz and he did not hear it. Experiments showed that the whistle emits a sound with a frequency close to 16 kHz. Assuming that the claim of the driver is true, how fast was he driving the car? Take the speed of sound in air to be 330 m/s. Is this speed practical with today's technology?
ANSWER: The frequency of sound, ν = 16 kHz.
V = 330 m/s. The apparent frequency (minimum for beyond the hearing range) ν' = 20 kHz. The speed of observer =?
ν' = (V+u)ν/V
→20 = (330+u)*16/330
→330+u = 20*330/16 =412.5
→u = 82.5 m/s = 82.5*3600/1000 km/h
→u = 297 km/h
Hence if the claim of the driver is assumed to be true, then he must be driving the car more than 297 km/h ≈300 km/h
In today's technology, general make of cars and road conditions, it is not practical.
75. A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has a dominant frequency of 800 Hz. What will be the apparent frequency heard by the driver in the front car? The speed of sound in air = 330 m/s.
ANSWER: The frequency of the sound, ν = 800 Hz.
The speed of the source, uₛ = 108 km/h =108000/3600 m/s =30 m/s. The speed of the observer, uₒ = 72 km/h =72000/3600 m/s = 20 m/s. V = 330 m/s.
Apparent frequency heard by the driver in the front car
ν' = (V+uₒ)ν/(V-uₛ)
But the source is approaching and the observer is leaving, hence
ν' = (V-uₒ)ν/(V-uₛ)
= (330-20)*800/(330-30) Hz
= 310*800/300 Hz
= 827 Hz
76. Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/h and the other at 54 km/h relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz.
(a) At what frequency is the signal received by the second submarine?
(b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be 1500 m/s.
ANSWER: (a)The frequency of the sound, ν = 2000 Hz
Speed of sound V = 1500 m/s.
The speed of the source uₛ = 36 km/h
=36000/3600 m/s =10 m/s
The speed of the observer uₒ = 54 km/h
=54000/3600 m/s
=30/2 m/s =15 m/s
Hence the apparent frequency received by the second submarine
ν' = (V+uₒ)ν/(V-uₛ)
=(1500+15)2000/(1500-10)
=1515*2000/1490 Hz
=2034 Hz
(b) This apparent frequency is now a source when the signal is reflected. The frequency of the sound received by the first submarine
ν" =(V+uₒ)ν/(V-uₛ)
But now uₒ = 10 m/s and uₛ = 15 m/s and ν = ν' =2034 Hz
Hence, ν"=(1500+10)*2034/(1500-15)
=1510*2034/1485
=2068 Hz
77. 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 m/s. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.
ANSWER: The frequency of the sound ν =800 Hz
V =340 m/s. Since the detector is placed along the line of motion the frequency detected is less than 800 Hz when the source is leaving and the frequency detected is more than 800 Hz when the source is approaching. The extreme frequencies detected will be when the source is at its maximum speed u at the mean position.
ν' = frequency detected when the source speed is u and approaching.
ν' = Vν/(V-u)
→ν' = 340*800/(340-u)
ν'' = frequency detected when the source speed is u and leaving.
ν" = Vν/(V+u)
→ν'' = 340*800/(340+u)
Given that width of frequency band detected is 8 Hz.
Hence, ν' - ν" = 8
340*800/(340-u) - 340*800/(340+u) = 8
→340*800*[340+u-340+u]/(340²-u²) = 8
→340*800*2u = 8(340²-u²)
→68000 u = 340² - u²
→u² +68000 u -115600 = 0
→u = -68000士√{68000²+4*1*115600}]/2
=-34000士½*68003.4
=-34000士34001.7
=1.7 m/s {since we have taken only numerical value of u, negative value is discarded}
This is the maximum speed of the SHM. The amplitude of the SHM = 17 cm =0.17.
Hence u =A⍵
→⍵ = u/A =1.7/0.17
→2π/T = 10 {where T =time period}
→T = 2π/10 s =2*3.14/10 s
=6.28/10 s
≈0.63 s
78. A boy riding on his bike is going towards the east at a speed of 4√2 m/s. At a certain point, he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of 334 m/s. A second boy stands on the ground 45° south of east from him. Find the frequency of the pulse received by the second boy.
ANSWER: The speed of source u = 4√2 m/s. The speed of sound V = 334 m/s. The frequency of sound ν = 1650 Hz.
Since the observer is standing at 45° from the direction of motion, the speed of the source for the observer = u cos 45° =u/√2
Diagram for Q - 78
Hence the apparent frequency received by the second boy ν' = Vν/(V-u/√2)
=334*1650/(334-4)
=334*1650/330 Hz
=1670 Hz
79. A sound source fixed at the origin is continuously emitting sound at a frequency 660 Hz. The sound travels in air at a speed of 330 m/s. A listener is moving along the line x = 336 m at a constant speed of 26 m/s. Find the frequency of sound as observed by the listener when he is (a) at y = -140 m, (b) at y = 0 and (c) at y = 140 m.
ANSWER: The frequency of source ν = 660 Hz.
V = 330 m/s. The speed of the observer u = 26 m/s.
Let the listener be at a distance y meter from the X-axis at any instant t seconds. The distance between the source and the listener z = OA =√(y²+336²)
Diagram for Q-79
Hence the instantaneous speed of the listener with respect to the source u' = dz/dt =d√(y²+336²)/dt
u' = {1/2√(y²+336²)}*2y*dy/dt
But dy/dt = u
hence u' = uy/√(y²+336²)
(a) The frequency of sound observed by the listener when at y = -140 m
ν' = (V+u')ν/V
But u' = -140*26/√(140²+336²) m/s
=-140*26/364 =-10 m/s
The negative sign says that the listener is approaching the origin. Hence
ν' = (330+10)*660/330=340*2 m/s =680 m/s
(b) When y = 0, u' = 0
hence ν' = (V+0)ν/V = ν = 660 Hz
(c) When y = 140 m
u' = 140*26/√(140²+336²) m/s = 10 m/s
The positive u' says that the listener is going away from the source. Hence
ν' = (V-u')ν/V
=(330-10)*660/330
=320*2 Hz =640 Hz
80. A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. The speed of sound in air is 340 m/s.
(a) What frequency will a passenger sitting near the open window hear?
(b) What frequency will a person standing near the track hear whom the train has just passed?
(c) A wind starts blowing towards the east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.
ANSWER: (a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s.
The speed of the observer uₒ = 108 km/k =30 m/s.
In such case apparent frequency
ν' = (V+uₒ)ν/(V-uₛ)
But here the source is leaving hence
ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz
{since uₒ = uₛ}
It can also be concluded by the fact that the relative motion between the source and the observer is zero.
(b) For the person standing near the track uₒ = 0 and the source is leaving.
ν' = Vν/(V+uₛ) =340*500/(340+30) Hz
=340*500/370 Hz =459 Hz
(c) The speed of the medium uₘ =36 km/h
=36000/3600 m/s = 10 m/s towards east.
Hence the effective speed of source = uₛ+uₘ
and the effective speed of the passenger = uₒ+uₘ
Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz.
The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual
ν' = Vν/(V+uₛ)
=330*500/(330+30) Hz
=330*500/360 Hz
=458 Hz
71. Figure (16-E11) shows a person standing somewhere in between two identical tuning forks, each vibrating at 512 Hz. If both the tuning forks move towards the right at a speed of 5.5 m/s, find the number of beats heard by the listener. The speed of sound in air = 330 m/s.
ANSWER: The observer is standing and the sources are moving. V = 330 m/s. u = 5.5 m/s. ν = 512 Hz.
The left tuning fork is approaching the observer, hence the apparent frequency ν' =Vν/(V-u)
=330*512/(330-5.5) Hz
=520.68 Hz
The right tuning fork is leaving the observer, hence the apparent frequency ν" =Vν/(V+u)
=330*512/(330+5.5) Hz
=503.61 Hz
So the number of beats produced for the listener = ν'-ν"
=520.68 - 503.61 Hz
=17.07 Hz ≈ 17.1 Hz > 16 Hz.
Since an average human ear can not distinguish the beats of more than 16 Hz frequency, hence the listener, in this case, may not be able to distinguish the beats.
72. A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius of 100/π cm at a constant angular speed of 5.0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and maximum frequency of the sound observed. The speed of sound in air =332 m/s.
ANSWER: The frequency of the sound, ν = 500 Hz.
The radius of the circle, r = 100/π cm =1/π m.
Angular speed ⍵ = 5*2π rad/s =10π rad/s.
Hence the magnitude of the instantaneous velocity of the source,
u = ⍵r = 10π*1/π m/s =10 m/s.
As it is clear from the diagram below, the maximum apparent frequency will be observed by the listener when the direction of the velocity is towards him, i.e. at A.
Similarly, the minimum frequency will be observed at B when the direction of the source is away from him.
In both the cases, the observer is stationary and the source is moving. Hence the apparent frequency heard when the source is at A
ν' =Vν/(V-u) =332*500/(332-10) Hz
=332*500/322 Hz
≈515 Hz
The apparent frequency heard when the source is at B
ν" =Vν/(V+u)
=332*500/(332+10) Hz
=332*500/342 Hz
≈485 Hz
73. Two trains are traveling towards each other both at a speed of 90 km/h. If one of the trains sounds a whistle at 500 Hz, what will be the apparent frequency heard in the other train? The speed of sound in air = 350 m/s.
ANSWER: The frequency of the source, ν = 500 Hz.
The speed of the source uₛ = 90 km/h
=90000/3600 m/s =25 m/s
The speed of the observer, uₒ =90 km/h =25 m/s.
The speed of the sound V = 350 m/s.
Hence the apparent frequency heard in the other train ν' = (V+uₒ)ν/(V-uₛ)
=(350+25)*500/(350-25)
=375*500/325 Hz
=577 Hz
74. A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car driver does not stop and takes the plea in court that because of the Doppler shift the frequency of the whistle reaching him might have gone beyond the audible limit of 20 kHz and he did not hear it. Experiments showed that the whistle emits a sound with a frequency close to 16 kHz. Assuming that the claim of the driver is true, how fast was he driving the car? Take the speed of sound in air to be 330 m/s. Is this speed practical with today's technology?
ANSWER: The frequency of sound, ν = 16 kHz.
V = 330 m/s. The apparent frequency (minimum for beyond the hearing range) ν' = 20 kHz. The speed of observer =?
ν' = (V+u)ν/V
→20 = (330+u)*16/330
→330+u = 20*330/16 =412.5
→u = 82.5 m/s = 82.5*3600/1000 km/h
→u = 297 km/h
Hence if the claim of the driver is assumed to be true, then he must be driving the car more than 297 km/h ≈300 km/h
In today's technology, general make of cars and road conditions, it is not practical.
75. A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has a dominant frequency of 800 Hz. What will be the apparent frequency heard by the driver in the front car? The speed of sound in air = 330 m/s.
ANSWER: The frequency of the sound, ν = 800 Hz.
The speed of the source, uₛ = 108 km/h =108000/3600 m/s =30 m/s. The speed of the observer, uₒ = 72 km/h =72000/3600 m/s = 20 m/s. V = 330 m/s.
Apparent frequency heard by the driver in the front car
ν' = (V+uₒ)ν/(V-uₛ)
But the source is approaching and the observer is leaving, hence
ν' = (V-uₒ)ν/(V-uₛ)
= (330-20)*800/(330-30) Hz
= 310*800/300 Hz
= 827 Hz
76. Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/h and the other at 54 km/h relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz.
(a) At what frequency is the signal received by the second submarine?
(b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be 1500 m/s.
ANSWER: (a)The frequency of the sound, ν = 2000 Hz
Speed of sound V = 1500 m/s.
The speed of the source uₛ = 36 km/h
=36000/3600 m/s =10 m/s
The speed of the observer uₒ = 54 km/h
=54000/3600 m/s
=30/2 m/s =15 m/s
Hence the apparent frequency received by the second submarine
ν' = (V+uₒ)ν/(V-uₛ)
=(1500+15)2000/(1500-10)
=1515*2000/1490 Hz
=2034 Hz
(b) This apparent frequency is now a source when the signal is reflected. The frequency of the sound received by the first submarine
ν" =(V+uₒ)ν/(V-uₛ)
But now uₒ = 10 m/s and uₛ = 15 m/s and ν = ν' =2034 Hz
Hence, ν"=(1500+10)*2034/(1500-15)
=1510*2034/1485
=2068 Hz
77. 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 m/s. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.
ANSWER: The frequency of the sound ν =800 Hz
V =340 m/s. Since the detector is placed along the line of motion the frequency detected is less than 800 Hz when the source is leaving and the frequency detected is more than 800 Hz when the source is approaching. The extreme frequencies detected will be when the source is at its maximum speed u at the mean position.
ν' = frequency detected when the source speed is u and approaching.
ν' = Vν/(V-u)
→ν' = 340*800/(340-u)
ν'' = frequency detected when the source speed is u and leaving.
This is the maximum speed of the SHM. The amplitude of the SHM = 17 cm =0.17.
Hence u =A⍵
→⍵ = u/A =1.7/0.17
→2π/T = 10 {where T =time period}
→T = 2π/10 s =2*3.14/10 s
=6.28/10 s
≈0.63 s
78. A boy riding on his bike is going towards the east at a speed of 4√2 m/s. At a certain point, he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of 334 m/s. A second boy stands on the ground 45° south of east from him. Find the frequency of the pulse received by the second boy.
ANSWER: The speed of source u = 4√2 m/s. The speed of sound V = 334 m/s. The frequency of sound ν = 1650 Hz.
Since the observer is standing at 45° from the direction of motion, the speed of the source for the observer = u cos 45° =u/√2
Hence the apparent frequency received by the second boy ν' = Vν/(V-u/√2)
=334*1650/(334-4)
=334*1650/330 Hz
=1670 Hz
79. A sound source fixed at the origin is continuously emitting sound at a frequency 660 Hz. The sound travels in air at a speed of 330 m/s. A listener is moving along the line x = 336 m at a constant speed of 26 m/s. Find the frequency of sound as observed by the listener when he is (a) at y = -140 m, (b) at y = 0 and (c) at y = 140 m.
ANSWER: The frequency of source ν = 660 Hz.
V = 330 m/s. The speed of the observer u = 26 m/s.
Let the listener be at a distance y meter from the X-axis at any instant t seconds. The distance between the source and the listener z = OA =√(y²+336²)
Hence the instantaneous speed of the listener with respect to the source u' = dz/dt =d√(y²+336²)/dt
u' = {1/2√(y²+336²)}*2y*dy/dt
But dy/dt = u
hence u' = uy/√(y²+336²)
(a) The frequency of sound observed by the listener when at y = -140 m
ν' = (V+u')ν/V
But u' = -140*26/√(140²+336²) m/s
=-140*26/364 =-10 m/s
The negative sign says that the listener is approaching the origin. Hence
ν' = (330+10)*660/330=340*2 m/s =680 m/s
(b) When y = 0, u' = 0
hence ν' = (V+0)ν/V = ν = 660 Hz
(c) When y = 140 m
u' = 140*26/√(140²+336²) m/s = 10 m/s
The positive u' says that the listener is going away from the source. Hence
ν' = (V-u')ν/V
=(330-10)*660/330
=320*2 Hz =640 Hz
80. A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. The speed of sound in air is 340 m/s.
(a) What frequency will a passenger sitting near the open window hear?
(b) What frequency will a person standing near the track hear whom the train has just passed?
(c) A wind starts blowing towards the east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.
ANSWER: (a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s.
The speed of the observer uₒ = 108 km/k =30 m/s.
In such case apparent frequency
ν' = (V+uₒ)ν/(V-uₛ)
But here the source is leaving hence
ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz
{since uₒ = uₛ}
It can also be concluded by the fact that the relative motion between the source and the observer is zero.
(b) For the person standing near the track uₒ = 0 and the source is leaving.
ν' = Vν/(V+uₛ) =340*500/(340+30) Hz
=340*500/370 Hz =459 Hz
(c) The speed of the medium uₘ =36 km/h
=36000/3600 m/s = 10 m/s towards east.
Hence the effective speed of source = uₛ+uₘ
and the effective speed of the passenger = uₒ+uₘ
Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz.
The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual
ν' = Vν/(V+uₛ)
=330*500/(330+30) Hz
=330*500/360 Hz
=458 Hz
 |
| The figure for Q-71 |
ANSWER: The observer is standing and the sources are moving. V = 330 m/s. u = 5.5 m/s. ν = 512 Hz.
The left tuning fork is approaching the observer, hence the apparent frequency ν' =Vν/(V-u)
=330*512/(330-5.5) Hz
=520.68 Hz
The right tuning fork is leaving the observer, hence the apparent frequency ν" =Vν/(V+u)
=330*512/(330+5.5) Hz
=503.61 Hz
So the number of beats produced for the listener = ν'-ν"
=520.68 - 503.61 Hz
=17.07 Hz ≈ 17.1 Hz > 16 Hz.
Since an average human ear can not distinguish the beats of more than 16 Hz frequency, hence the listener, in this case, may not be able to distinguish the beats.
72. A small source of sound vibrating at frequency 500 Hz is rotated in a circle of radius of 100/π cm at a constant angular speed of 5.0 revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and maximum frequency of the sound observed. The speed of sound in air =332 m/s.
ANSWER: The frequency of the sound, ν = 500 Hz.
The radius of the circle, r = 100/π cm =1/π m.
Angular speed ⍵ = 5*2π rad/s =10π rad/s.
Hence the magnitude of the instantaneous velocity of the source,
u = ⍵r = 10π*1/π m/s =10 m/s.
As it is clear from the diagram below, the maximum apparent frequency will be observed by the listener when the direction of the velocity is towards him, i.e. at A.
 |
| Diagram for Q-72 |
Similarly, the minimum frequency will be observed at B when the direction of the source is away from him.
In both the cases, the observer is stationary and the source is moving. Hence the apparent frequency heard when the source is at A
ν' =Vν/(V-u) =332*500/(332-10) Hz
=332*500/322 Hz
≈515 Hz
The apparent frequency heard when the source is at B
ν" =Vν/(V+u)
=332*500/(332+10) Hz
=332*500/342 Hz
≈485 Hz
73. Two trains are traveling towards each other both at a speed of 90 km/h. If one of the trains sounds a whistle at 500 Hz, what will be the apparent frequency heard in the other train? The speed of sound in air = 350 m/s.
ANSWER: The frequency of the source, ν = 500 Hz.
The speed of the source uₛ = 90 km/h
=90000/3600 m/s =25 m/s
The speed of the observer, uₒ =90 km/h =25 m/s.
The speed of the sound V = 350 m/s.
Hence the apparent frequency heard in the other train ν' = (V+uₒ)ν/(V-uₛ)
=(350+25)*500/(350-25)
=375*500/325 Hz
=577 Hz
74. A traffic policeman sounds a whistle to stop a car-driver approaching towards him. The car driver does not stop and takes the plea in court that because of the Doppler shift the frequency of the whistle reaching him might have gone beyond the audible limit of 20 kHz and he did not hear it. Experiments showed that the whistle emits a sound with a frequency close to 16 kHz. Assuming that the claim of the driver is true, how fast was he driving the car? Take the speed of sound in air to be 330 m/s. Is this speed practical with today's technology?
ANSWER: The frequency of sound, ν = 16 kHz.
V = 330 m/s. The apparent frequency (minimum for beyond the hearing range) ν' = 20 kHz. The speed of observer =?
ν' = (V+u)ν/V
→20 = (330+u)*16/330
→330+u = 20*330/16 =412.5
→u = 82.5 m/s = 82.5*3600/1000 km/h
→u = 297 km/h
Hence if the claim of the driver is assumed to be true, then he must be driving the car more than 297 km/h ≈300 km/h
In today's technology, general make of cars and road conditions, it is not practical.
75. A car moving at 108 km/h finds another car in front of it going in the same direction at 72 km/h. The first car sounds a horn that has a dominant frequency of 800 Hz. What will be the apparent frequency heard by the driver in the front car? The speed of sound in air = 330 m/s.
ANSWER: The frequency of the sound, ν = 800 Hz.
The speed of the source, uₛ = 108 km/h =108000/3600 m/s =30 m/s. The speed of the observer, uₒ = 72 km/h =72000/3600 m/s = 20 m/s. V = 330 m/s.
Apparent frequency heard by the driver in the front car
ν' = (V+uₒ)ν/(V-uₛ)
But the source is approaching and the observer is leaving, hence
ν' = (V-uₒ)ν/(V-uₛ)
= (330-20)*800/(330-30) Hz
= 310*800/300 Hz
= 827 Hz
76. Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/h and the other at 54 km/h relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz.
(a) At what frequency is the signal received by the second submarine?
(b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be 1500 m/s.
ANSWER: (a)The frequency of the sound, ν = 2000 Hz
Speed of sound V = 1500 m/s.
The speed of the source uₛ = 36 km/h
=36000/3600 m/s =10 m/s
The speed of the observer uₒ = 54 km/h
=54000/3600 m/s
=30/2 m/s =15 m/s
Hence the apparent frequency received by the second submarine
ν' = (V+uₒ)ν/(V-uₛ)
=(1500+15)2000/(1500-10)
=1515*2000/1490 Hz
=2034 Hz
(b) This apparent frequency is now a source when the signal is reflected. The frequency of the sound received by the first submarine
ν" =(V+uₒ)ν/(V-uₛ)
But now uₒ = 10 m/s and uₛ = 15 m/s and ν = ν' =2034 Hz
Hence, ν"=(1500+10)*2034/(1500-15)
=1510*2034/1485
=2068 Hz
77. 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 m/s. If the width of the frequency band detected by the detector is 8 Hz, find the time period of the source.
ANSWER: The frequency of the sound ν =800 Hz
V =340 m/s. Since the detector is placed along the line of motion the frequency detected is less than 800 Hz when the source is leaving and the frequency detected is more than 800 Hz when the source is approaching. The extreme frequencies detected will be when the source is at its maximum speed u at the mean position.
ν' = frequency detected when the source speed is u and approaching.
ν' = Vν/(V-u)
→ν' = 340*800/(340-u)
ν'' = frequency detected when the source speed is u and leaving.
ν" = Vν/(V+u)
→ν'' = 340*800/(340+u)
Given that width of frequency band detected is 8 Hz.
Hence, ν' - ν" = 8
340*800/(340-u) - 340*800/(340+u) = 8
→340*800*[340+u-340+u]/(340²-u²) = 8
→340*800*2u = 8(340²-u²)
→68000 u = 340² - u²
→u² +68000 u -115600 = 0
→u = -68000士√{68000²+4*1*115600}]/2
=-34000士½*68003.4
=-34000士34001.7
=1.7 m/s {since we have taken only numerical value of u, negative value is discarded}
This is the maximum speed of the SHM. The amplitude of the SHM = 17 cm =0.17.
Hence u =A⍵
→⍵ = u/A =1.7/0.17
→2π/T = 10 {where T =time period}
→T = 2π/10 s =2*3.14/10 s
=6.28/10 s
≈0.63 s
78. A boy riding on his bike is going towards the east at a speed of 4√2 m/s. At a certain point, he produces a sound pulse of frequency 1650 Hz that travels in air at a speed of 334 m/s. A second boy stands on the ground 45° south of east from him. Find the frequency of the pulse received by the second boy.
ANSWER: The speed of source u = 4√2 m/s. The speed of sound V = 334 m/s. The frequency of sound ν = 1650 Hz.
Since the observer is standing at 45° from the direction of motion, the speed of the source for the observer = u cos 45° =u/√2
 |
| Diagram for Q - 78 |
Hence the apparent frequency received by the second boy ν' = Vν/(V-u/√2)
=334*1650/(334-4)
=334*1650/330 Hz
=1670 Hz
79. A sound source fixed at the origin is continuously emitting sound at a frequency 660 Hz. The sound travels in air at a speed of 330 m/s. A listener is moving along the line x = 336 m at a constant speed of 26 m/s. Find the frequency of sound as observed by the listener when he is (a) at y = -140 m, (b) at y = 0 and (c) at y = 140 m.
ANSWER: The frequency of source ν = 660 Hz.
V = 330 m/s. The speed of the observer u = 26 m/s.
Let the listener be at a distance y meter from the X-axis at any instant t seconds. The distance between the source and the listener z = OA =√(y²+336²)
 |
| Diagram for Q-79 |
Hence the instantaneous speed of the listener with respect to the source u' = dz/dt =d√(y²+336²)/dt
u' = {1/2√(y²+336²)}*2y*dy/dt
But dy/dt = u
hence u' = uy/√(y²+336²)
(a) The frequency of sound observed by the listener when at y = -140 m
ν' = (V+u')ν/V
But u' = -140*26/√(140²+336²) m/s
=-140*26/364 =-10 m/s
The negative sign says that the listener is approaching the origin. Hence
ν' = (330+10)*660/330=340*2 m/s =680 m/s
(b) When y = 0, u' = 0
hence ν' = (V+0)ν/V = ν = 660 Hz
(c) When y = 140 m
u' = 140*26/√(140²+336²) m/s = 10 m/s
The positive u' says that the listener is going away from the source. Hence
ν' = (V-u')ν/V
=(330-10)*660/330
=320*2 Hz =640 Hz
80. A train running at 108 km/h towards east whistles at a dominant frequency of 500 Hz. The speed of sound in air is 340 m/s.
(a) What frequency will a passenger sitting near the open window hear?
(b) What frequency will a person standing near the track hear whom the train has just passed?
(c) A wind starts blowing towards the east at a speed of 36 km/h. Calculate the frequencies heard by the passenger in the train and by the person standing near the track.
ANSWER: (a) The frequency of the sound ν = 500 Hz. The speed of sound in air V = 340 m/s. Speed of the source uₛ = 108 km/h =108000/3600 m/s =30 m/s.
The speed of the observer uₒ = 108 km/k =30 m/s.
In such case apparent frequency
ν' = (V+uₒ)ν/(V-uₛ)
But here the source is leaving hence
ν' = (V+uₒ)ν/(V+uₛ) = ν = 500 Hz
{since uₒ = uₛ}
It can also be concluded by the fact that the relative motion between the source and the observer is zero.
(b) For the person standing near the track uₒ = 0 and the source is leaving.
ν' = Vν/(V+uₛ) =340*500/(340+30) Hz
=340*500/370 Hz =459 Hz
(c) The speed of the medium uₘ =36 km/h
=36000/3600 m/s = 10 m/s towards east.
Hence the effective speed of source = uₛ+uₘ
and the effective speed of the passenger = uₒ+uₘ
Both are in the same direction hence no relative motion. Thus the frequency observed by the passenger = ν = 500 Hz.
The effective speed of the sound in the air for the person standing due to the wind towards the east = V-uₘ =340-10 =330 m/s. Now the apparent frequency for the observer can be calculated as usual
ν' = Vν/(V+uₛ)
=330*500/(330+30) Hz
=330*500/360 Hz
=458 Hz
===<<<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
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
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
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
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)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
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
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
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"
No comments:
Post a Comment