Sound Wave Doppler effect question

  • B
  • Thread starter jha192001
  • Start date
  • #1
22
2

Main Question or Discussion Point

I have seen few examples on Doppler effect and i am confused about one such.
We are standing on ground.
If the source of sound S moves and Object O is stationary. We would presume the frequency as well as wavelength of sound be changed to the obeject O.
But if O moves towards or away from S. Only wavelength of sound is changed.
Is this effect also seen in light wave Doppler effect?
what is this phenomenon? I am not able to find on it clearly, what changes and what not ?
Thanking you
 

Answers and Replies

  • #2
davenn
Science Advisor
Gold Member
2019 Award
9,057
6,778
If the source of sound S moves and Object O is stationary. We would presume the frequency as well as wavelength of sound be changed to the obeject O.
yes

But if O moves towards or away from S. Only wavelength of sound is changed.
no, why would you think that "

if the wavelength changes, the frequency must also change, for the observer

If the sound source is moving and the observer is stationary or the opposite … moving observer and stationary sound source
the effect is the same

ohhh and that is the same for light or any other EM radiation
think of a police radar, laser or microwave, source is stationary and the car is moving
The wavelength shortens ( for car travelling towards the radar) and the frequency increases
 
  • #3
Merlin3189
Homework Helper
Gold Member
1,417
570
...
If the sound source is moving and the observer is stationary or the opposite … moving observer and stationary sound source
the effect is the same

ohhh and that is the same for light or any other EM radiation ...
Are you not thinking solely of EM waves?
I thought it was generally recognised that there is a clear difference for waves dependent on a medium such as air.

I wonder if what the op is thinking about is that, when she is stationary she observes a consistent set of frequency, velocity and wavelength. The observed frequency is higher than the moving source claims to emit, but fits with the observed wavelength and known speed of propagation in that medium.
When the observer is moving she sees again a higher frequency on her oscilloscope or frequency meter than that emitted by the stationary source (though not changed by the same amount), but may be in a quandry about measuring the wavelength and speed of propagation. She knows she is moving through the medium and could take that into account in her measurements. If she does, she sees that the speed and wavelength measured on the ground are inconsistent with the uncorrected frequency she is observing.

This is the opposite of what op claims of course. I need to know how op is measuring f and λ.
 
  • #4
davenn
Science Advisor
Gold Member
2019 Award
9,057
6,778
Are you not thinking solely of EM waves?
I thought it was generally recognised that there is a clear difference for waves dependent on a medium such as air.
Sound or EM
As far as I'm aware, the Doppler "effect" is the same, tho the process is slightly different ... medium/no medium

upload_2018-10-2_10-6-4.png



edit ... ohhh and as an afterthought .....

And of course, you can do the same thing with sound as what is being done in the lower section of the image with the laser/microwave radar :smile:
 

Attachments

Last edited:
  • #5
Merlin3189
Homework Helper
Gold Member
1,417
570
There are obvious differences if one of you travels faster than the speed of sound (which is why you don't have to worry for EM waves).
Source approaches > SoS, you don't hear anything until passes, then you hear lower f as it moves away.
You approach stationary source at > SoS, you hear raised f until it passes, then nothing more.

Otherwise there are quantitative diffs.
eg. , speed of sound 330 m/s, source 1 kHz
Stationary source,you move towards source at 165 m/s, you hear 1.5 kHz.
Stationary observer, source move towards you at 165 m/s, you hear 2 kHz.
(Please check by your own methods. I don't know any simple method, but I found online ## f_o = f_s \frac {(V + V_s)} {(V + V_o)} ## and they seem to agree with those as far as I understand it.)

I can't check how this works for EM, as without a reference frame, especially for time, I'm stuck. All I know there is that it can't make any difference who moves, so it must be different from sound.
 
  • #6
olivermsun
Science Advisor
1,244
117
The difference for waves in a medium is that ##c## is fixed to the medium. Either way, you figure out the change in wave crests encountered per time by the (possibly moving) observer.
 
  • #7
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,463
1,163
There are obvious differences if one of you travels faster than the speed of sound (which is why you don't have to worry for EM waves).
Source approaches > SoS, you don't hear anything until passes, then you hear lower f as it moves away.
You approach stationary source at > SoS, you hear raised f until it passes, then nothing more.

Otherwise there are quantitative diffs.
eg. , speed of sound 330 m/s, source 1 kHz
Stationary source,you move towards source at 165 m/s, you hear 1.5 kHz.
Stationary observer, source move towards you at 165 m/s, you hear 2 kHz.
(Please check by your own methods. I don't know any simple method, but I found online ## f_o = f_s \frac {(V + V_s)} {(V + V_o)} ## and they seem to agree with those as far as I understand it.)

I can't check how this works for EM, as without a reference frame, especially for time, I'm stuck. All I know there is that it can't make any difference who moves, so it must be different from sound.
For EMR in a vacuum you use ## f_o = f_s \sqrt{ \frac {(1+\frac{v}{c})} {(1-\frac{v}{c})}} ##

where v is the relative velocity between source and receiver, and is positive if they are approaching each other.

One way of looking at it is to treat the pure Doppler effect factor like you are considering a stationary observer and moving source, which gives
## f_o = f_s \frac {(c + v_s)} {c} ## or ## f_o= f_s \left (1+\frac{v}{c} \right )##
(v = vs)

Then you apply the time dilation factor to the source frequency to get
## f_o = f_s \frac{1+\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}} ##

## f_o = f_s \frac{1+\frac{v}{c}}{\sqrt{(1-\frac{v}{c}) (1+\frac{v}{c})}} ##

## f_o = f_s \sqrt{ \frac {(1+\frac{v}{c})} {(1-\frac{v}{c})}} ##
 
  • #8
Merlin3189
Homework Helper
Gold Member
1,417
570
Thanks. I wasn't asking, just pointing out my limited knowledge, but it's very kind of you to help me along. I've never been able to get any sort of understanding of relativity and I don't like relying formulae that I can't work out for myself.

I need to check up on 'time dilation factor' and also to convince myself that it's valid to start with the 'pure' Doppler formula. Otherwise it seems plausible. If my maths doesn't let me down, I think I get the same result if I start with the moving observer formula, which is a good sign! (Assuming I'm right in taking the equivalent case to v=vs, is v= -vo ) I suppose I should also try v= vs - vo and see what I get.
 
  • #9
22
2
Sorry sir i am still not fully satisfied.
For now lets only say about sound as a wave.(Not EM)
The same question again.
If only observer moves or only source. Whats the changes in both cases in the frequency,wavelength,speed of sounds as. heard by the object/observer. I know frequency changes by diffrent amount as per motion of people wrt medium. But please elaborate on this more.
Thanking You.
 
  • #10
194
117
A simpler way to grasp the concept(maybe):

Think simply in terms of 'time of flight' between emitter and receiver and use the positive peaks of an emitted sinusoid as your 'ruler.'

If the emitter/receiver are stationary WRT each other, then emitted = received. The distance is constant. All peaks travel the same distance.

If the emitter/receiver are moving toward each other, then the distance is getting shorter over time. The first positive peak of a sinusoidal waveform travels 'farther' than the 2nd, and so on...
The effect is that the peaks arrive (at the receiver) 'closer together' because the later peaks don't have to travel as far. If the rate of closure is constant, then the 'shift' is constant - that's doppler.

If the emitter/receiver are moving apart, then the distance is increasing; the peaks arrive 'farther apart.'
 
  • #11
22
2
A simpler way to grasp the concept(maybe):

Think simply in terms of 'time of flight' between emitter and receiver and use the positive peaks of an emitted sinusoid as your 'ruler.'

If the emitter/receiver are stationary WRT each other, then emitted = received. The distance is constant. All peaks travel the same distance.

If the emitter/receiver are moving toward each other, then the distance is getting shorter over time. The first positive peak of a sinusoidal waveform travels 'farther' than the 2nd, and so on...
The effect is that the peaks arrive (at the receiver) 'closer together' because the later peaks don't have to travel as far. If the rate of closure is constant, then the 'shift' is constant - that's doppler.

If the emitter/receiver are moving apart, then the distance is increasing; the peaks arrive 'farther apart.'
Yes sir, But isnt that the explanation of the Doppler Effect? The question i put forward was the change the wave would go through in the two diffrent case mentioned above. Why doppler effect predicts diffrently on if the *Observer* is moving and on Source is only moving.

<< Post edited by a Mentor to fix the all-caps sections >>
 
Last edited by a moderator:
  • #12
194
117
OK.
You apparently want to examine the wave 'independent' of the observer?

Think about it in the terms that I suggested. Sketch a snapshot of an emitted wave moving through the air. For a stationary emitter The emitted wavelength (from the POV of the emitter) will be identical to the 'standing' wavelength. If the emitter is moving in the same direction as the emitted wave, the peaks 'bunch' as described above. The peaks are closer together in space; The 'standing' frequency is higher. If the emitter is moving away, it's farther/lower.

For a stationary emitter and moving receiver, the 'standing' and emitted wavelengths are the same (I said that already). The moving receiver causes the peaks to be 'intercepted' by the receiver at a different rate than they were emitted.

In the case of the moving emitter, the 'objective' frequency of the wave in space is shifted; For the moving receiver, the shift is 'apparent at the receiver.' These cases are indistinguishable at the receiver.

OR: maybe I still don't understand your question.
 
  • #13
Janus
Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,463
1,163
Yes sir, But isnt that the explanation of the Doppler Effect? The question i put forward was the change the wave would go through in the two diffrent case mentioned above. Why doppler effect predicts diffrently on if the *Observer* is moving and on Source is only moving.

<< Post edited by a Mentor to fix the all-caps sections >>
Assuming a non-relativistic Doppler effect of sound through the air:
Given:
speed of sound through air 343 m/s
Source frequency :10 hz
Relative speed between Source and receiver: 171.5 m/sec
Starting distance between Source and receiver: 343 m.

Source moving relative to air, receiver not:
The front of a sound wave leaves the source when it is 343 meters from the receiver at t = 0
This will arrive at the receiver at t = 1 sec.
At t= 0.10 sec, the back of the wave (start of the second wave) leaves the source. By this time it has moved 17.15 meters closer to the receiver, and will take an additional (343-17.15)/343 = .95 sec to reach the receiver, arriving at t = 1.05 sec. This is just 0.05 sec later than the front of the wave arrived. Thus the source measures a period of 0.05 sec for the wave and a frequency of 20 hz for the wave.
As far as wavelength goes, When the back end of the wave leaves the source at t =0.10 sec, the front end has moved 34.3 m relative to where it was emitted. This means it is just 34.3-17.15 = 171.5 m ahead of the back of the wave when it is emitted, the wavelength has been halved.

Receiver moving relative to air and source not:
Front of wave is still emitted when source and receiver are 343 m apart and at t=0. wave travels at 343 m/s towards receiver. Receiver is moving at 34.3 m/sec towards source ( and relative to the air). Front of wave and receiver meet 343(343+171.5) 0.666667... sec later at t = 2/3 sec...
Back of wave leaves source at t = .01, and when the receiver is 17.15 meters closer to the source. It will take 0.633333.. sec to meet up with the receiver, and they meet at t= 0.733333.. sec, or 0.066666... sec after the front of the wave, a wave period of 0.666666... sec equates to a frequency of ~ 15 hz, 3/4 of what he measured when it was the source moving relative to the air.
Wavelength: Since the source never moves relative to the air, the wavelength is just 34.3 m (the distance the wave travels relative to the air in one second). However, since the receiver is moving at 171.5 m/s relative to the air in one direction and the wave is moving at 343 m/sec relative to it in the other, the speed of the wave is 514.5 m/s relative to the receiver. 514.5/34.3 = 15 hz, the same answer we got above.
 
  • #14
467
119
I think the OP's question is quite a deep one. In essence she ('she' is presumed - forgive if incorrectly!) is struggling to deal with the fact that for light there is no "ether" in which the wave propagates. And thus no distinction between observer moving/source stationary or observer stationary/source moving. In the case of sound waves there is an "ether" namely the air in which the wave propagates.
 

Related Threads for: Sound Wave Doppler effect question

Replies
16
Views
5K
  • Last Post
Replies
8
Views
716
  • Last Post
Replies
1
Views
1K
Replies
9
Views
1K
Replies
8
Views
3K
  • Last Post
Replies
6
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
Top