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SOUND WAVES
EXERCISES- Q-61 to Q-70
61. A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?
ANSWER: The fundamental frequency of vibration of wire fixed at both ends, ν = (1/2L)√(F/µ). When the length is shortened this frequency increases. The beat frequency is the difference between the two frequencies. Since the beat-frequency decreases with the increase of wire frequency, hence the tuning fork frequency is greater than the present wire frequency. Thus the present wire frequency ν = 256 - 4 =252 Hz.
There will be no beat if the wire-frequency is the same as the frequency of the tuning fork i.e. ν' =256 Hz. Suppose for this frequency the shortened length of the wire = L' and it is still in the fundamental mode.
The ratio of the two frequencies can be written as
ν/ν' = (1/2L)/(1/2L')
→ν/ν' = L'/L
→L' = νL/ν' = 252*25/256 = 24.61 cm.
Hence for no beat, the length to be shortened by is 25 -24.61 cm = 0.39 cm
62. A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the apparent frequency heard by a scooter driver approaching the policeman at a speed of 36.00 km/h? The speed of sound in air = 340 m/s.
ANSWER: The frequency of sound, ν = 2.00 kHz.
The source is stationary. The speed of the observer, u= 36 km/h = 36000/3600 m/s = 10 m/s. The speed of sound, V = 340 m/s.
Hence the apparent frequency heard, ν' = (V+u)ν/V
=(340+10)*2.00/340 kHz
=2.06 Hz
63. The horn of a car emits sound with a dominant frequency of 2400 Hz. What will be the apparent dominant frequency heard by a person standing on the road in front of the car if the car is going at 18.0 km/h? The speed of sound in air = 340 m/s.
ANSWER: Frequency of the source, ν = 2400 Hz.
Speed of the source, u = 18.0 km/h =18000/3600 m/s = 5 m/s
Speed of sound in air, V = 340 m/s.
Here the source is approaching the observer and the observer is stationary. Hence the apparent frequency for the observer, ν' = Vν/(V-u)
=340*2400/(340-5) Hz
=2436 Hz
64. A person riding a car moving at 72 km/h sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car? The speed of sound in air = 340 m/s.
ANSWER: The frequency of the source of the sound, ν = 1250 Hz.
The speed of the source, u = 72 km/h
=72000/3600 m/s
=20 m/s
The speed of sound in air, V =340 m/s.
(a) For the person standing in front of the car, the source is approaching. Hence the apparent frequency, ν' =Vν/(V-u)
=340*1250/(340-20) Hz
=1328 Hz
(b) For the person standing behind the car, the source is receding. Hence the apparent frequency,
ν' = Vν/(V+u)
=340*1250/(340+20) Hz
=1181 Hz
65. A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m/s.
ANSWER: The speed of sound in air, V =332 m/s.
The speed of the source, u = 54 km/h = 54000/3600 m/s = 15 m/s. Apparent frequency heard by the observer when the source is approaching, ν' = 1620 Hz. Let the actual frequency of the source = ν, then
ν' = Vν/(V-u)
→1620 = 332ν/(332-15) =332ν/317
→ν = 1620*317/332 Hz
When the train has crossed the platform the source is receding, hence the apparent frequency now,
ν" = Vν/(332+15)=332ν/347
=(332/347)*(1620*317/332)
{Putting the value of ν}
=1620*317/347 Hz
=1480 Hz
66. A bat emitting an ultrasonic wave of frequency 4.5x10⁴Hz flies at a speed of 6 m/s between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m/s.
ANSWER: Let W and W' be the two parallel walls and the bat is flying towards the wall W. The bat will hear two sounds, one from the reflection of the sound produced by it from the front wall W and other from the reflection from the hind wall W'.
Diagram for Q-66
The frequency of sound produced by the bat, ν= 4.5x10⁴ Hz. For the front wall W, the apparent frequency coming to it
ν'=330ν/(330-6) Hz,
{Since source approaching the observer}
=330ν/324 Hz
This apparent frequency is reflected and for the bat now the wall W is source and observer (bat) moving.
The apparent frequency for the bat is now
ν₁ =(V+u)ν'/V
→ν₁ ={(330+6)/330}ν'
→ν₁ = 336ν'/330 =(336/330)*(330/324)ν
→ν₁ = 336ν/324
→ν₁ = 336*4.5x10⁴/324 Hz
→ν₁ = 1.037*4.5x10⁴ Hz
→ν₁ = 4.67x10⁴ Hz
At the back wall W', the coming sound has an apparent frequency
ν" = 330ν/(330+6)
{The source is receding, observer stationary}
ν" = 330ν/336
When it is reflected, it is the source for the bat. This source is stationary and the observer receding.
The apparent frequency
ν₂ =(V-u)ν"/V
→ν₂ = (330-6)ν"/330 Hz
→ν₂ =324ν"/330 Hz
→ν₂ =(324/330)(330/336)ν Hz
→ν₂ =(324/336)*4.5x10⁴ Hz
→ν₂ =0.964*4.5x10⁴ Hz
→ν₂ =4.34x10⁴ Hz
The beat frequency = Difference between the two frequencies =ν₁-ν₂ = 336ν/324 - 324ν/336
=(336²-324²)ν/(336*324)
=(336+324)(336-324)ν/(336*324)
=660*12ν/(336*324)
=110*2ν/(56*54)
=55ν/(28*27) =0.0727*4.5x10⁴ Hz
=0.327x10⁴ Hz
=3270 Hz
67. A bullet passes past a person at a speed of 220 m/s. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. The speed of sound in air = 330 m/s.
ANSWER: Speed of the source u = 220 m/s. The speed of sound in air = 330 m/s. Let the original frequency =ν.
When the bullet is approaching, the source is approaching and the observer is stationary, the apparent frequency
ν₁ =Vν/(V-u) =330ν/(330-220)
→ν₁ = 330ν/110 =3ν
When the bullet has crossed, the source is receding and the observer is stationary. The apparent frequency now,
ν₂ = Vν/(V+u) =330ν/(330+220)
→ν₂ = 330ν/550 =3ν/5
The change in apparent frequency = ν₁-ν₂
=3ν-3ν/5 = 12ν/5
Hence the fractional change in apparent frequency with respect to initial frequency = (Change in apparent frequency)÷(Initial apparent frequency)
=(12ν/5)÷(3ν)
=4/5
=0.8
68. Two electric trains run at the same speed of 72 km/h along the same track and in the same direction with a separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340m/s, find the frequencies heard by the person.
ANSWER: The speed of train u = 72 km/h
→u = 72000/3600 m/s =20 m/s.
Diagram for Q-68
The distance of the person from each of the trains
=√(500²+1200²) =√(1690000) =√(1300²) =1300 m. {From the right-angled triangle}
If α be the angle between the track and the line joining the person with the train, the instantaneous speed of train towards the person, u'=u cosα.
=20*1200/1300 m/s
=240/13 m/s
=18.46 m/s
The frequency of the whistle = ν = 500 Hz
Speed of sound in air = V = 340 m/s
Hence the frequency heard by the person from the approaching train = Vν/(V-u')
=340*500/(340-18.46)
=529 Hz
And the frequency heard by the person from the leaving train = Vν/(V+u')
=340*500/(340+18.46)
=474 Hz
69. A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4 beats per second. The speed of sound in air = 340 m/s.
(a) Calculate the speed of the train.
(b) What beat frequency is heard by the player in the train?
ANSWER: (a) The apparent frequency heard by the standing person will be more than 440 Hz due to Doppler's effect. Since he hears 4 beats per second, the apparent frequency of the violin in the train, ν' = 440+4 Hz =444 Hz.
Here ν =440 Hz
V = 340 m/s
u =?
ν' = Vν/(V-u)
→444 =340*440/(340-u)
→340-u = 340*440/444 =336.94
→u = 340 - 336.94 m/s = 3.06 m/s
=3.06*3600/1000 km/h ≈11 km/h
(b) For the person in the train, the source on the ground is stationary and observer approaches.
The apparent frequency heard by the person in the train ν'' =ν(V+u)/V
= 440*(340+3.06)/340 Hz
=440*1.009 Hz
=443.96 Hz
The beat frequency heard when the train approaches =443.96 - 440 = 3.96 beats/s.
i.e. a little less than 4 beats/s
70. Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0 m/s so that he approaches one tuning fork and recedes from the other (figure 16-E10). Find the beat frequency observed by the listener. The speed of sound in air = 332 m/s.
Figure for Q-70
ANSWER: Speed of sound in air V = 332 m/s,
The speed of the observer, u = 3.0 m/s.
The frequency of the sources ν = 256 Hz.
The sources are stationary and observer moving, hence the apparent frequency heard by the observer from the front tuning fork (approaching)
ν' = (V+u)ν/V
and the apparent frequency heard by the observer from the receding source
ν" = (V-u)ν/V
Hence the beat frequency observed by the listener
=ν' - ν"
=(V+u)ν/V - (V-u)ν/V
=2uν/V
=2*3.0*256/332 Hz
=4.6 Hz
61. A tuning fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork?
ANSWER: The fundamental frequency of vibration of wire fixed at both ends, ν = (1/2L)√(F/µ). When the length is shortened this frequency increases. The beat frequency is the difference between the two frequencies. Since the beat-frequency decreases with the increase of wire frequency, hence the tuning fork frequency is greater than the present wire frequency. Thus the present wire frequency ν = 256 - 4 =252 Hz.
There will be no beat if the wire-frequency is the same as the frequency of the tuning fork i.e. ν' =256 Hz. Suppose for this frequency the shortened length of the wire = L' and it is still in the fundamental mode.
The ratio of the two frequencies can be written as
ν/ν' = (1/2L)/(1/2L')
→ν/ν' = L'/L
→L' = νL/ν' = 252*25/256 = 24.61 cm.
Hence for no beat, the length to be shortened by is 25 -24.61 cm = 0.39 cm
62. A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the apparent frequency heard by a scooter driver approaching the policeman at a speed of 36.00 km/h? The speed of sound in air = 340 m/s.
ANSWER: The frequency of sound, ν = 2.00 kHz.
The source is stationary. The speed of the observer, u= 36 km/h = 36000/3600 m/s = 10 m/s. The speed of sound, V = 340 m/s.
Hence the apparent frequency heard, ν' = (V+u)ν/V
=(340+10)*2.00/340 kHz
=2.06 Hz
63. The horn of a car emits sound with a dominant frequency of 2400 Hz. What will be the apparent dominant frequency heard by a person standing on the road in front of the car if the car is going at 18.0 km/h? The speed of sound in air = 340 m/s.
ANSWER: Frequency of the source, ν = 2400 Hz.
Speed of the source, u = 18.0 km/h =18000/3600 m/s = 5 m/s
Speed of sound in air, V = 340 m/s.
Here the source is approaching the observer and the observer is stationary. Hence the apparent frequency for the observer, ν' = Vν/(V-u)
=340*2400/(340-5) Hz
=2436 Hz
64. A person riding a car moving at 72 km/h sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car? The speed of sound in air = 340 m/s.
ANSWER: The frequency of the source of the sound, ν = 1250 Hz.
The speed of the source, u = 72 km/h
=72000/3600 m/s
=20 m/s
The speed of sound in air, V =340 m/s.
(a) For the person standing in front of the car, the source is approaching. Hence the apparent frequency, ν' =Vν/(V-u)
=340*1250/(340-20) Hz
=1328 Hz
(b) For the person standing behind the car, the source is receding. Hence the apparent frequency,
ν' = Vν/(V+u)
=340*1250/(340+20) Hz
=1181 Hz
65. A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m/s.
ANSWER: The speed of sound in air, V =332 m/s.
The speed of the source, u = 54 km/h = 54000/3600 m/s = 15 m/s. Apparent frequency heard by the observer when the source is approaching, ν' = 1620 Hz. Let the actual frequency of the source = ν, then
ν' = Vν/(V-u)
→1620 = 332ν/(332-15) =332ν/317
→ν = 1620*317/332 Hz
When the train has crossed the platform the source is receding, hence the apparent frequency now,
ν" = Vν/(332+15)=332ν/347
=(332/347)*(1620*317/332)
{Putting the value of ν}
=1620*317/347 Hz
=1480 Hz
66. A bat emitting an ultrasonic wave of frequency 4.5x10⁴Hz flies at a speed of 6 m/s between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m/s.
ANSWER: Let W and W' be the two parallel walls and the bat is flying towards the wall W. The bat will hear two sounds, one from the reflection of the sound produced by it from the front wall W and other from the reflection from the hind wall W'.
The frequency of sound produced by the bat, ν= 4.5x10⁴ Hz. For the front wall W, the apparent frequency coming to it
ν'=330ν/(330-6) Hz,
{Since source approaching the observer}
=330ν/324 Hz
This apparent frequency is reflected and for the bat now the wall W is source and observer (bat) moving.
The apparent frequency for the bat is now
ν₁ =(V+u)ν'/V
→ν₁ ={(330+6)/330}ν'
→ν₁ = 336ν'/330 =(336/330)*(330/324)ν
→ν₁ = 336ν/324
→ν₁ = 336*4.5x10⁴/324 Hz
→ν₁ = 1.037*4.5x10⁴ Hz
→ν₁ = 4.67x10⁴ Hz
At the back wall W', the coming sound has an apparent frequency
ν" = 330ν/(330+6)
{The source is receding, observer stationary}
ν" = 330ν/336
When it is reflected, it is the source for the bat. This source is stationary and the observer receding.
The apparent frequency
ν₂ =(V-u)ν"/V
→ν₂ = (330-6)ν"/330 Hz
→ν₂ =324ν"/330 Hz
→ν₂ =(324/330)(330/336)ν Hz
→ν₂ =(324/336)*4.5x10⁴ Hz
→ν₂ =0.964*4.5x10⁴ Hz
→ν₂ =4.34x10⁴ Hz
The beat frequency = Difference between the two frequencies =ν₁-ν₂ = 336ν/324 - 324ν/336
=(336²-324²)ν/(336*324)
=(336+324)(336-324)ν/(336*324)
=660*12ν/(336*324)
=110*2ν/(56*54)
=55ν/(28*27) =0.0727*4.5x10⁴ Hz
=0.327x10⁴ Hz
=3270 Hz
67. A bullet passes past a person at a speed of 220 m/s. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. The speed of sound in air = 330 m/s.
ANSWER: Speed of the source u = 220 m/s. The speed of sound in air = 330 m/s. Let the original frequency =ν.
When the bullet is approaching, the source is approaching and the observer is stationary, the apparent frequency
ν₁ =Vν/(V-u) =330ν/(330-220)
→ν₁ = 330ν/110 =3ν
When the bullet has crossed, the source is receding and the observer is stationary. The apparent frequency now,
ν₂ = Vν/(V+u) =330ν/(330+220)
→ν₂ = 330ν/550 =3ν/5
The change in apparent frequency = ν₁-ν₂
=3ν-3ν/5 = 12ν/5
Hence the fractional change in apparent frequency with respect to initial frequency = (Change in apparent frequency)÷(Initial apparent frequency)
=(12ν/5)÷(3ν)
=4/5
=0.8
68. Two electric trains run at the same speed of 72 km/h along the same track and in the same direction with a separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340m/s, find the frequencies heard by the person.
ANSWER: The speed of train u = 72 km/h
→u = 72000/3600 m/s =20 m/s.
The distance of the person from each of the trains
=√(500²+1200²) =√(1690000) =√(1300²) =1300 m. {From the right-angled triangle}
If α be the angle between the track and the line joining the person with the train, the instantaneous speed of train towards the person, u'=u cosα.
=20*1200/1300 m/s
=240/13 m/s
=18.46 m/s
The frequency of the whistle = ν = 500 Hz
Speed of sound in air = V = 340 m/s
Hence the frequency heard by the person from the approaching train = Vν/(V-u')
=340*500/(340-18.46)
=529 Hz
And the frequency heard by the person from the leaving train = Vν/(V+u')
=340*500/(340+18.46)
=474 Hz
69. A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4 beats per second. The speed of sound in air = 340 m/s.
(a) Calculate the speed of the train.
(b) What beat frequency is heard by the player in the train?
ANSWER: (a) The apparent frequency heard by the standing person will be more than 440 Hz due to Doppler's effect. Since he hears 4 beats per second, the apparent frequency of the violin in the train, ν' = 440+4 Hz =444 Hz.
Here ν =440 Hz
V = 340 m/s
u =?
ν' = Vν/(V-u)
→444 =340*440/(340-u)
→340-u = 340*440/444 =336.94
→u = 340 - 336.94 m/s = 3.06 m/s
=3.06*3600/1000 km/h ≈11 km/h
(b) For the person in the train, the source on the ground is stationary and observer approaches.
The apparent frequency heard by the person in the train ν'' =ν(V+u)/V
= 440*(340+3.06)/340 Hz
=440*1.009 Hz
=443.96 Hz
The beat frequency heard when the train approaches =443.96 - 440 = 3.96 beats/s.
i.e. a little less than 4 beats/s
70. Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0 m/s so that he approaches one tuning fork and recedes from the other (figure 16-E10). Find the beat frequency observed by the listener. The speed of sound in air = 332 m/s.
ANSWER: Speed of sound in air V = 332 m/s,
The speed of the observer, u = 3.0 m/s.
The frequency of the sources ν = 256 Hz.
The sources are stationary and observer moving, hence the apparent frequency heard by the observer from the front tuning fork (approaching)
ν' = (V+u)ν/V
and the apparent frequency heard by the observer from the receding source
ν" = (V-u)ν/V
Hence the beat frequency observed by the listener
=ν' - ν"
=(V+u)ν/V - (V-u)ν/V
=2uν/V
=2*3.0*256/332 Hz
=4.6 Hz
ANSWER: The fundamental frequency of vibration of wire fixed at both ends, ν = (1/2L)√(F/µ). When the length is shortened this frequency increases. The beat frequency is the difference between the two frequencies. Since the beat-frequency decreases with the increase of wire frequency, hence the tuning fork frequency is greater than the present wire frequency. Thus the present wire frequency ν = 256 - 4 =252 Hz.
There will be no beat if the wire-frequency is the same as the frequency of the tuning fork i.e. ν' =256 Hz. Suppose for this frequency the shortened length of the wire = L' and it is still in the fundamental mode.
The ratio of the two frequencies can be written as
ν/ν' = (1/2L)/(1/2L')
→ν/ν' = L'/L
→L' = νL/ν' = 252*25/256 = 24.61 cm.
Hence for no beat, the length to be shortened by is 25 -24.61 cm = 0.39 cm
62. A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the apparent frequency heard by a scooter driver approaching the policeman at a speed of 36.00 km/h? The speed of sound in air = 340 m/s.
ANSWER: The frequency of sound, ν = 2.00 kHz.
The source is stationary. The speed of the observer, u= 36 km/h = 36000/3600 m/s = 10 m/s. The speed of sound, V = 340 m/s.
Hence the apparent frequency heard, ν' = (V+u)ν/V
=(340+10)*2.00/340 kHz
=2.06 Hz
63. The horn of a car emits sound with a dominant frequency of 2400 Hz. What will be the apparent dominant frequency heard by a person standing on the road in front of the car if the car is going at 18.0 km/h? The speed of sound in air = 340 m/s.
ANSWER: Frequency of the source, ν = 2400 Hz.
Speed of the source, u = 18.0 km/h =18000/3600 m/s = 5 m/s
Speed of sound in air, V = 340 m/s.
Here the source is approaching the observer and the observer is stationary. Hence the apparent frequency for the observer, ν' = Vν/(V-u)
=340*2400/(340-5) Hz
=2436 Hz
64. A person riding a car moving at 72 km/h sounds a whistle emitting a wave of frequency 1250 Hz. What frequency will be heard by another person standing on the road (a) in front of the car (b) behind the car? The speed of sound in air = 340 m/s.
ANSWER: The frequency of the source of the sound, ν = 1250 Hz.
The speed of the source, u = 72 km/h
=72000/3600 m/s
=20 m/s
The speed of sound in air, V =340 m/s.
(a) For the person standing in front of the car, the source is approaching. Hence the apparent frequency, ν' =Vν/(V-u)
=340*1250/(340-20) Hz
=1328 Hz
(b) For the person standing behind the car, the source is receding. Hence the apparent frequency,
ν' = Vν/(V+u)
=340*1250/(340+20) Hz
=1181 Hz
65. A train approaching a platform at a speed of 54 km/h sounds a whistle. An observer on the platform finds its frequency to be 1620 Hz. The train passes the platform keeping the whistle on and without slowing down. What frequency will the observer hear after the train has crossed the platform? The speed of sound in air = 332 m/s.
ANSWER: The speed of sound in air, V =332 m/s.
The speed of the source, u = 54 km/h = 54000/3600 m/s = 15 m/s. Apparent frequency heard by the observer when the source is approaching, ν' = 1620 Hz. Let the actual frequency of the source = ν, then
ν' = Vν/(V-u)
→1620 = 332ν/(332-15) =332ν/317
→ν = 1620*317/332 Hz
When the train has crossed the platform the source is receding, hence the apparent frequency now,
ν" = Vν/(332+15)=332ν/347
=(332/347)*(1620*317/332)
{Putting the value of ν}
=1620*317/347 Hz
=1480 Hz
66. A bat emitting an ultrasonic wave of frequency 4.5x10⁴Hz flies at a speed of 6 m/s between two parallel walls. Find the two frequencies heard by the bat and the beat frequency between the two. The speed of sound is 330 m/s.
ANSWER: Let W and W' be the two parallel walls and the bat is flying towards the wall W. The bat will hear two sounds, one from the reflection of the sound produced by it from the front wall W and other from the reflection from the hind wall W'.
 |
| Diagram for Q-66 |
The frequency of sound produced by the bat, ν= 4.5x10⁴ Hz. For the front wall W, the apparent frequency coming to it
ν'=330ν/(330-6) Hz,
{Since source approaching the observer}
=330ν/324 Hz
This apparent frequency is reflected and for the bat now the wall W is source and observer (bat) moving.
The apparent frequency for the bat is now
ν₁ =(V+u)ν'/V
→ν₁ ={(330+6)/330}ν'
→ν₁ = 336ν'/330 =(336/330)*(330/324)ν
→ν₁ = 336ν/324
→ν₁ = 336*4.5x10⁴/324 Hz
→ν₁ = 1.037*4.5x10⁴ Hz
→ν₁ = 4.67x10⁴ Hz
At the back wall W', the coming sound has an apparent frequency
ν" = 330ν/(330+6)
{The source is receding, observer stationary}
ν" = 330ν/336
When it is reflected, it is the source for the bat. This source is stationary and the observer receding.
The apparent frequency
ν₂ =(V-u)ν"/V
→ν₂ = (330-6)ν"/330 Hz
→ν₂ =324ν"/330 Hz
→ν₂ =(324/330)(330/336)ν Hz
→ν₂ =(324/336)*4.5x10⁴ Hz
→ν₂ =0.964*4.5x10⁴ Hz
→ν₂ =4.34x10⁴ Hz
The beat frequency = Difference between the two frequencies =ν₁-ν₂ = 336ν/324 - 324ν/336
=(336²-324²)ν/(336*324)
=(336+324)(336-324)ν/(336*324)
=660*12ν/(336*324)
=110*2ν/(56*54)
=55ν/(28*27) =0.0727*4.5x10⁴ Hz
=0.327x10⁴ Hz
=3270 Hz
67. A bullet passes past a person at a speed of 220 m/s. Find the fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person. The speed of sound in air = 330 m/s.
ANSWER: Speed of the source u = 220 m/s. The speed of sound in air = 330 m/s. Let the original frequency =ν.
When the bullet is approaching, the source is approaching and the observer is stationary, the apparent frequency
ν₁ =Vν/(V-u) =330ν/(330-220)
→ν₁ = 330ν/110 =3ν
When the bullet has crossed, the source is receding and the observer is stationary. The apparent frequency now,
ν₂ = Vν/(V+u) =330ν/(330+220)
→ν₂ = 330ν/550 =3ν/5
The change in apparent frequency = ν₁-ν₂
=3ν-3ν/5 = 12ν/5
Hence the fractional change in apparent frequency with respect to initial frequency = (Change in apparent frequency)÷(Initial apparent frequency)
=(12ν/5)÷(3ν)
=4/5
=0.8
68. Two electric trains run at the same speed of 72 km/h along the same track and in the same direction with a separation of 2.4 km between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of 500 m from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at 500 Hz and the speed of sound in air is 340m/s, find the frequencies heard by the person.
ANSWER: The speed of train u = 72 km/h
→u = 72000/3600 m/s =20 m/s.
 |
| Diagram for Q-68 |
The distance of the person from each of the trains
=√(500²+1200²) =√(1690000) =√(1300²) =1300 m. {From the right-angled triangle}
If α be the angle between the track and the line joining the person with the train, the instantaneous speed of train towards the person, u'=u cosα.
=20*1200/1300 m/s
=240/13 m/s
=18.46 m/s
The frequency of the whistle = ν = 500 Hz
Speed of sound in air = V = 340 m/s
Hence the frequency heard by the person from the approaching train = Vν/(V-u')
=340*500/(340-18.46)
=529 Hz
And the frequency heard by the person from the leaving train = Vν/(V+u')
=340*500/(340+18.46)
=474 Hz
69. A violin player riding on a slow train plays a 440 Hz note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears 4 beats per second. The speed of sound in air = 340 m/s.
(a) Calculate the speed of the train.
(b) What beat frequency is heard by the player in the train?
ANSWER: (a) The apparent frequency heard by the standing person will be more than 440 Hz due to Doppler's effect. Since he hears 4 beats per second, the apparent frequency of the violin in the train, ν' = 440+4 Hz =444 Hz.
Here ν =440 Hz
V = 340 m/s
u =?
ν' = Vν/(V-u)
→444 =340*440/(340-u)
→340-u = 340*440/444 =336.94
→u = 340 - 336.94 m/s = 3.06 m/s
=3.06*3600/1000 km/h ≈11 km/h
(b) For the person in the train, the source on the ground is stationary and observer approaches.
The apparent frequency heard by the person in the train ν'' =ν(V+u)/V
= 440*(340+3.06)/340 Hz
=440*1.009 Hz
=443.96 Hz
The beat frequency heard when the train approaches =443.96 - 440 = 3.96 beats/s.
i.e. a little less than 4 beats/s
70. Two identical tuning forks vibrating at the same frequency 256 Hz are kept fixed at some distance apart. A listener runs between the forks at a speed of 3.0 m/s so that he approaches one tuning fork and recedes from the other (figure 16-E10). Find the beat frequency observed by the listener. The speed of sound in air = 332 m/s.
|
| Figure for Q-70 |
ANSWER: Speed of sound in air V = 332 m/s,
The speed of the observer, u = 3.0 m/s.
The frequency of the sources ν = 256 Hz.
The sources are stationary and observer moving, hence the apparent frequency heard by the observer from the front tuning fork (approaching)
ν' = (V+u)ν/V
and the apparent frequency heard by the observer from the receding source
ν" = (V-u)ν/V
Hence the beat frequency observed by the listener
=ν' - ν"
=(V+u)ν/V - (V-u)ν/V
=2uν/V
=2*3.0*256/332 Hz
=4.6 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
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
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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-------------------------------------------------------------------------------
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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--------------------------------------------------------------------------------------------------------------
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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