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A railway engine passes by the platform at a speed of 36 km/h blowing its whistle having a frequency of 660 Hz. Calculate the difference in the frequencies of the whistle heard by a person standing on the platform as the engine goes past the person.
Correct answer is '20'. Can you explain this answer?
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A railway engine passes by the platform at a speed of 36 km/h blowing ...
According to Doppler effect when source moves toward observer. The frequency increases, hence apparent frequency is given as
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A railway engine passes by the platform at a speed of 36 km/h blowing ...
Calculation of the Doppler effect:
The Doppler effect is a phenomenon that occurs when there is relative motion between a source of sound and an observer. It causes a shift in the frequency of the sound heard by the observer. The formula for calculating the Doppler effect is:
f' = (v + vo) / (v + vs) * f
where:
f' is the frequency heard by the observer,
v is the speed of sound in air,
vo is the velocity of the observer,
vs is the velocity of the source of sound, and
f is the frequency of the source of sound.
Given data:
Speed of the engine (source of sound): 36 km/h
Frequency of the whistle: 660 Hz
Calculating the velocity of the source:
The velocity of the source is given as 36 km/h. We need to convert it to m/s.
36 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 10 m/s
Calculating the difference in frequencies:
To calculate the difference in frequencies, we need to consider the velocity of the observer as well. Since the person is standing on the platform, the velocity of the observer is assumed to be zero.
Using the formula for the Doppler effect:
f' = (v + vo) / (v + vs) * f
We substitute the given values:
f' = (v + 0) / (v + 10) * 660 Hz
Since the speed of sound in air is approximately 343 m/s, we substitute that value for v:
f' = (343 m/s + 0) / (343 m/s + 10 m/s) * 660 Hz
Simplifying the expression:
f' = 343 / 353 * 660 Hz
Calculating the value:
f' ≈ 639.83 Hz
The difference in frequencies is the absolute difference between the original frequency (660 Hz) and the frequency heard by the observer (639.83 Hz):
Difference = |660 Hz - 639.83 Hz|
Difference ≈ 20.17 Hz
Rounded to the nearest whole number, the difference in frequencies is 20 Hz.
The Doppler effect is a phenomenon that occurs when there is relative motion between a source of sound and an observer. It causes a shift in the frequency of the sound heard by the observer. The formula for calculating the Doppler effect is:
f' = (v + vo) / (v + vs) * f
where:
f' is the frequency heard by the observer,
v is the speed of sound in air,
vo is the velocity of the observer,
vs is the velocity of the source of sound, and
f is the frequency of the source of sound.
Given data:
Speed of the engine (source of sound): 36 km/h
Frequency of the whistle: 660 Hz
Calculating the velocity of the source:
The velocity of the source is given as 36 km/h. We need to convert it to m/s.
36 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 10 m/s
Calculating the difference in frequencies:
To calculate the difference in frequencies, we need to consider the velocity of the observer as well. Since the person is standing on the platform, the velocity of the observer is assumed to be zero.
Using the formula for the Doppler effect:
f' = (v + vo) / (v + vs) * f
We substitute the given values:
f' = (v + 0) / (v + 10) * 660 Hz
Since the speed of sound in air is approximately 343 m/s, we substitute that value for v:
f' = (343 m/s + 0) / (343 m/s + 10 m/s) * 660 Hz
Simplifying the expression:
f' = 343 / 353 * 660 Hz
Calculating the value:
f' ≈ 639.83 Hz
The difference in frequencies is the absolute difference between the original frequency (660 Hz) and the frequency heard by the observer (639.83 Hz):
Difference = |660 Hz - 639.83 Hz|
Difference ≈ 20.17 Hz
Rounded to the nearest whole number, the difference in frequencies is 20 Hz.
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A railway engine passes by the platform at a speed of 36 km/h blowing ...
According to Doppler effect when source moves toward observer. The frequency increases, hence apparent frequency is given as
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A railway engine passes by the platform at a speed of 36 km/h blowing its whistle having a frequency of 660 Hz. Calculate the difference in the frequencies of the whistle heard by a person standing on the platform as the engine goes past the person.Correct answer is '20'. Can you explain this answer?
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