Doppler effect for accelerated source

[Patrikp]

Homework Statement

At the initial moment source and detector are located in same point and are both in rest. Source is released into free fall and detector remains in rest. Source has frequency f_s=512 Hz and speed of sound is v=340 m/s . When detector detects frquency f =485 Hz , what distance has the source travelled?

Homework Equations

$$f=f_s \left( \frac{v_s}{v_s+v} \right)$$

The Attempt at a Solution

My book uses the formula I gave above, for which I know it´s true when there is no acceleration. I tried to get to formula myself but i aways get that time between registered waves at the detector is increasing so frequency should be decreasing? If it should then again I do not know how to go about calculating that kind of frequency.
Thanks and sorry for bad english

 

[alw34]

I'm unsure exactly your issue, but it may be the sign convention of the formula.
Does the sign convention explanation here help you?

https://en.wikipedia.org/wiki/Doppler_effect#General

 

[Patrikp]

I'm unsure exactly your issue, but it may be the sign convention of the formula.
Does the sign convention explanation here help you?

https://en.wikipedia.org/wiki/Doppler_effect#General

No, that is not the issue.
I am having trouble understanding why this formula works even when the source is accelerating ( does not have constant velocity ).

 

[alw34]

The frequency shift is relative velocity dependent...at an instant... between emitter and receiver. It does not depend on acceleration at that instant except for overall elapsed time...v = at, d = vt.

Suppose the emitter and receiver moved apart, then briefly back towards each other, and then apart again...during that time frequency will be plus and minus, but ,say, five minutes later, the only thing that matters is the final/instantaneous relative velocity between them. You can think of a cop on the highway with his Doppler radar...the only thing that counts is your velocity at the moment of measurement, not the pattern of your acceleration nor even the route you took.

 

[Patrikp]

The frequency shift is relative velocity dependent...at an instant... between emitter and receiver. It does not depend on acceleration at that instant except for overall elapsed time...v = at, d = vt.

Suppose the emitter and receiver moved apart, then briefly back towards each other, and then apart again...during that time frequency will be plus and minus, but ,say, five minutes later, the only thing that matters is the final/instantaneous relative velocity between them. You can think of a cop on the highway with his Doppler radar...the only thing that counts is your velocity at the moment of measurement, not the pattern of your acceleration nor even the route you took.

I don´t understand. Can you describe that mathematically?

 

[alw34]

Plug known quantities you posted into the formula...

You may have to adjust a sign as source and emitter are moving apart.

which variable do you then solve for?

 

[Patrikp]

Plug known quantities you posted into the formula...

You may have to adjust a sign as source and emitter are moving apart.

which variable do you then solve for?

I don`t understand why this formula works in this case. Can you describe mathematically why does it work even though the source accelerates? That`s my problem.

 

[haruspex]

No, that is not the issue.
I am having trouble understanding why this formula works even when the source is accelerating ( does not have constant velocity ).

In the time a given wave (one cycle, say) is emitted, the speed of the source will change very little. Once emitted, it is unaffected by subsequent accelerations of the source.
You might need to take into account that by the time the detector detects the given frequency the source will have fallen a bit further.

 

[Patrikp]

In the time a given wave (one cycle, say) is emitted, the speed of the source will change very little. Once emitted, it is unaffected by subsequent accelerations of the source.
You might need to take into account that by the time the detector detects the given frequency the source will have fallen a bit further.

Ok, but is the above formula correct in this case or is just approximation? Can you help me derive it in this specific case?

 

[haruspex]

Ok, but is the above formula correct in this case or is just approximation? Can you help me derive it in this specific case?

The formula is correct, and the derivation is no different. You just have to ignore the tiny increase in velocity during one cycle.

 

[alw34]

You can think of a cop on the highway with his Doppler radar...the only thing that counts is your velocity at the moment of measurement, not the pattern of your acceleration nor even the route you took.

I am unable to think of a clearer example than this.

Posts #8, #9 are consistent with this concept.

Perhaps another way to say the same thing: What makes you think the velocity and the frequency in your formula are valid only for an unchanging single set of parameters? If the formula works at one moment, for some set of 'v's and 'f's', it also works at all other instants for those parameters at those other times; it is an instantaneous relationship.

Another viewpoint,same idea: If the relative motion of the emitter and source vary over time, those prior waves will have already passed the receiver; all the receiver detects are the waves based on the moment of the calculation.

 

[physics user1]

I think this could help:

the frequency heard changes in time as velocity does, you can see them as functions of time, because the frequency is related to the velocity and the velocity to the time, when the source is not in accelerated motion the velocity is constant in time and so does the frequency heard, when is in accelerated motion the velocity isn't constant and so the frequency, that equation you wrote express the frequency in function of the velocity at any time, f (t)= fs (vs / (vs + v (t) )

You can see it as a composed function g (f (x))

When v (t) Is constant f (t) Is too when v (t) Is accelerated both of them vary, you will notice a change in frequency while the source falls

So you can calculate the distance travelled, don't think about the lag because of the time needed by the sound wave to arrive to the observer