# Doppler effect and acceleration of source

• orangephysik
In summary: Hoping you're right is one thing. What is needed to convince you that you are right :wink: ?Well done!In summary, the conversation involved finding the acceleration and time for a source moving towards us with a constant acceleration. The wavelength of the source was calculated using the formula f_0 = v_phasefront / λ, and the final speed of the source was found to be 171.5 m/s. Using this, along with the equations v(t) = a*t and s(t) = 1/2 * at^2, the acceleration was determined to be 8.65 m/s and the time it takes for the source to reach us was found to be 19.83 seconds.
orangephysik
Homework Statement
A source at rest is at a distance of s_0 = 1.7 km from you and you measure a frequency of f_0 = 520 Hz. At time t = 0 the source moves directly towards you with a constant acceleration. When the source reaches you, you measure a frequency of f_1 = 1040 Hz.

a) What is the acceleration of the source and at which time t does the source reach you?
Relevant Equations
Measured frequency, f_Source * = f_0 (1/ (1 - [v_source / v_phasefront] )
Hi. I need help with part a).
I calculated the wavelength of the source by using the formula f_0 = v_phasefront / λ and got λ = (343 m/s) / (520 Hz) = 0.6596 m.
And then I set up an equation for the velocity of the source v(t) = a*t (with v(t = 0 )= 0 m/s) and s(t) = 1/2 * at^2 + s_0. But I just have no idea how I can find the acceleration with these information.

orangephysik said:
Homework Statement:: A source at rest is at a distance of s_0 = 1.7 km from you and you measure a frequency of f_0 = 520 Hz. At time t = 0 the source moves directly towards you with a constant acceleration. When the source reaches you, you measure a frequency of f_1 = 1040 Hz.

a) What is the acceleration of the source and at which time t does the source reach you?
Relevant Equations:: Measured frequency, f_Source * = f_0 (1/ (1 - [v_source / v_phasefront] )

Hi. I need help with part a).
I calculated the wavelength of the source by using the formula f_0 = v_phasefront / λ and got λ = (343 m/s) / (520 Hz) = 0.6596 m.
And then I set up an equation for the velocity of the source v(t) = a*t (with v(t = 0 )= 0 m/s) and s(t) = 1/2 * at^2 + s_0.

Hello @orangephysik,

1. I don't think there's anything relativistic about this exercise, but don't mind to be proven wrong.

orangephysik said:
But I just have no idea how I can find the acceleration with these information.

2. Well, what information haven't you used yet ? Something with Doppler ?

##\ ##

BvU said:
Hello @orangephysik,

1. I don't think there's anything relativistic about this exercise, but don't mind to be proven wrong.
2. Well, what information haven't you used yet ? Something with Doppler ?

##\ ##
I had f_0 = v_phasefront / λ and got λ = (343 m/s) / (520 Hz) = 0.6596 m, and this is the wavelength of the source when it was at rest.
With the same formula, f_1 = v_phasefront / λ' , so λ' = (343 m/s) / (1040 Hz) = 0.3298 m, this is the wavelength when the source reaches me.

Now using
Measured frequency, f_Source * = f_0 (1/ (1 - [v_source / v_phasefront] )

1040 Hz = 520 Hz * (1/ [1 - (v_source / 343 m/s) ]), rearranging I got v_source = 171.5 m/s.

I also know λ' = λ - v_source * T (whereby T is the period). Plugging in the values to find T, I got T = 1.923 ms.

So the acceleration must be a = v_source / T = 8.92*10^4 m/(s^2). Is this the correct way of solving the question?

orangephysik said:
Is this the correct way of solving the question?
Actually there are two questions. One is asking for the acceleration of the source, the other for a time.
I agree the first step is to find the final speed of the source. Looks OK.
But then you derive a T in a cumbersome way (it is 1/520) and you seem to think this period is equal to the time the source needs to reach you. Why ? 2 ms to cover what distance again ?

The equations you set up at the end of post #1 look a lot more promising to me !

##\ ##

BvU said:
Actually there are two questions. One is asking for the acceleration of the source, the other for a time.
I agree the first step is to find the final speed of the source. Looks OK.
But then you derive a T in a cumbersome way (it is 1/520) and you seem to think this period is equal to the time the source needs to reach you. Why ? 2 ms to cover what distance again ?

The equations you set up at the end of post #1 look a lot more promising to me !

##\ ##
Oh right. The period T is the time it takes for the sound waves to travel a distance of λ - λ' = 0.3298 m.

So since acceleration is a = Δv/Δt, I already have Δv since v_0 = 0 m/s and now I just need Δt. But the question implies that I can find the acceleration without knowing Δt.

I have v(t) = a*t ⇔ 171.5 m/s = a*t ⇔ t = (171.5 m/s) / a (whereby t is the time it takes to reach me, since 171.5 m/s is the speed when it reaches me)

I also have s(t) = 1/2 * at2
I know that it has travelled 1.7 km when it reaches me, so 1.7 km = 1/2 * at2
⇔ t = √[(3.4 km)/a]

Setting these two equations for t equal, I get a = 8.65 m/s, which means t = 19.83 s.

I hope this time I'm right

Hoping you're right is one thing. What is needed to convince you that you are right ?

Well done!

##\ ##

DeBangis21 and orangephysik

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