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Doppler Effect inconsistent in terms of relative velocities

by memoguy
Tags: doppler, effect, inconsistent, relative, terms
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memoguy
#1
Mar16-14, 08:14 PM
P: 28
Hi all!

Let me use an example to make this clear. There is a car travelling directly toward a man at 10m/s. The car is pressing down its horn, producing a frequency of 100hz. The speed of sound is 330 m/s. What frequency does the man hear?

Ok, so we can use the equation:
f_observer=( (v+v_o)/(v-v_s ) ) * f_source
Where v=330, v_o=0 & v_s = 10
Thus f_observer = 103.125

But, if the car was still and the observer moved toward the car at 10m/s we could say:
f_observer=( (v+v_o)/(v-v_s ) ) * f_source
Where v=330, v_o=10, v_s =0
thus f_observer = 103.0303....

103.0303 does not equal 103.125.

Why is there a difference, surely these two things could be the same.
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UltrafastPED
#2
Mar16-14, 08:29 PM
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Your formula does not agree with Hyperphysics:
http://hyperphysics.phy-astr.gsu.edu...d/dopp.html#c2
AlephZero
#3
Mar16-14, 08:55 PM
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The two situations are not the same. In one, the sound source is moving relative to the air. In the other, the observer is moving relative to the air.

To see the difference, think about extreme situations, like
(1) the source is moving towards the observer at twice the speed of sound (660 m/s)
(2) the observer is moving towards the source at 660 m/s

dauto
#4
Mar17-14, 10:43 AM
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Doppler Effect inconsistent in terms of relative velocities

Quote Quote by memoguy View Post
surely these two things could be the same.
No, they are not the same. Sound moves through air so all speeds are measured relative to air which gives us an absolute reference frame. Motion is not relative here. Compare that with the relativistic Doppler effect for electromagnetic waves (light).
[tex]f_o=f_s \sqrt \frac{c+v_{rel}}{c-v_{rel}},[/tex]
where [itex]v_{rel}[/itex] is the relative speed between source and observer. Here there is no medium and the speed of the source is measured relative to the observer. Motion is relative here.
Malverin
#5
Mar17-14, 12:17 PM
P: 123
Quote Quote by memoguy View Post
Hi all!

Let me use an example to make this clear. There is a car travelling directly toward a man at 10m/s. The car is pressing down its horn, producing a frequency of 100hz. The speed of sound is 330 m/s. What frequency does the man hear?

Ok, so we can use the equation:
f_observer=( (v+v_o)/(v-v_s ) ) * f_source
Where v=330, v_o=0 & v_s = 10
Thus f_observer = 103.125

But, if the car was still and the observer moved toward the car at 10m/s we could say:
f_observer=( (v+v_o)/(v-v_s ) ) * f_source
Where v=330, v_o=10, v_s =0
thus f_observer = 103.0303....

103.0303 does not equal 103.125.

Why is there a difference, surely these two things could be the same.
Sound waves, travel with constant speed in the medium (when pressure, temperature, humidity.. are constant). So when the source is moving it emits a wave, travel some distance and emits again (time between emissions is time period of the wave.)
So the source gets closer to the observer and the second wave has shorter distance to go and it arrives sooner than if the source was stationary.
The speed of the wave relative to observer is the same as this with stationary source(wave can not move in the medium slower or faster ) => speed of sound in the medium.

When the source is stationary and observer is moving, the speed of wave relative to him is the sum of his speed and waves speed (they are traveling against each other and wave travels with the speed of sound in the medium)



http://hyperphysics.phy-astr.gsu.edu...ound/dopp.html


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