# Doppler shift for accelerated motion

1. Jun 17, 2011

### maa105

I was talking to one of my friends the other day about doppler effect and we went into the discussion of accelerating object and doppler effect. we both agreed that the frequency shift would vary in time. then we argued about the frequency shift for different observers and here is were it got interesting, after arguing for some time we took an example:

say there is a sound source emitting frequency f0=1kHz that starts accelerating from the origin from rest with acceleration a to the right at time t=0s. say also we have a observer at x=-1m, what I argued that at time 1/300s (time taken sound to travel from the sound source at origin to x=-1) the observer will start recording a frequency of f0 and then this frequency will start declining with time. similarly an observer at x=-2m will start recording frequency f0 at time 2/300s and that frequency will start declining with time. (see fig1 for illustration)
Fig1

a consequence from the above image is that if say you take a specific point in time t=T where T>2/300s, at this specific time instance T each observer would register different frequency (see fig2). this too made sense since the signal received by observer one at t=T was emitted by the source at an earlier point in time than the one received by observer 2 at the same time t=T (due to propagation delay [the time it takes the signal to reach the observer] and due to the fact that the speed of sound is finite), which the speed of the source when it emitted the wave that was received at t=T by observer one is different than the speed of the source when it emitted the wave that was received at same time t=T by observer two. and due to this speed difference I reasoned that there should be a frequency shift between the two observers at time t=T (or any time instance for that matters)
Fig2

This reasoning made sense to me, but then my friend told me I was wrong and pulled of a pencil and paper and manipulated some equations to prove that both observers at x=-1 and x=-2 receive the same frequency at all points in time. his formulas made sense but then what is wrong in my previous reasoning?

Fig3 illustration of case study

Last edited by a moderator: Apr 26, 2017
2. Jun 17, 2011

### Drakkith

Staff Emeritus
I really don't know, but I don't see anything wrong with your reasoning. Without seeing whatever math he did I can't see why he thinks that. And what's up with Fig 3? What is that supposed to represent?

3. Jun 17, 2011

### maa105

first are you guys seeing the images cause I hosted them on my google site account

ok.
The third image is the image of the whole setup.

I'll include an image of the formulas in a very clear and easy to read format and another clear text copy (in case the image link fails)

the formulas:
in image format:

Transmitted signal:
s(t)= A_t sin(2πf_0 t)u(t) - Where u(t) is the unit step function
s_r (t)= A_r sin(2πf_0 (t-∆t))u(t-∆t)

∆t=d/c Where d is the distance between transmitter and receiver
And c is the speed of sound

After substitution we get:
s_r (t)= A_r sin(2πf_0 (t-d/c))u(t-∆t)

Now d is not fixed. Using kinematic formulas we get:
d=1/2 at^2+v_0 t+x_0

=> s_r (t)= A_r sin(2πf_0 (t-(1/2 at^2+v_0 t+x_0)/c))u(t-∆t)
=> s_r (t)= A_r sin(2πf_0 (t-1/2c at^2-v_0/c t-x_0/c))u(t-∆t)
=> s_r (t)= A_r sin(2πf_0 (-1/2c at^2+(1-v_0/c)t-x_0/c))u(t-∆t)
=> s_r (t)= A_r sin(2πf_0 (-1/2c at^2+(1-v_0/c)t)-(2πf_0 x_0)/c)u(t-∆t)
=> s_r (t)= A_r sin(2πf_0 t(-1/2c at+1-v_0/c)-(2πf_0 x_0)/c)u(t-∆t)

Now from the above we see that the frequency at each point in time is:
f=f_0 (-1/2c at+1-v_0/c)

This means that the frequency is the same at each point in time no matter what x0 is!
The only place x0 enters in the picture is in the phase of the signal: φ=-(2πf_0 x_0)/c
This means that observe one and observer two will always register the same frequency.
The received signals only differ in phase φ_1=-(2πf_0 x_1)/c,φ_2=-(2πf_0 x_2)/c

SO WHAT GIVES WHAT IS WRONG IN MY INITIAL REASONING?

regards

Last edited by a moderator: Apr 26, 2017