Doppler effect and acceleration of source

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.
  • #1
orangephysik
11
0
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.
 
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  • #2
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,
:welcome: ##\qquad## !​

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 ?

##\ ##
 
  • #3
BvU said:
Hello @orangephysik,
:welcome: ##\qquad## !​

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?
 
  • #4
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 :smile: !

##\ ##
 
  • #5
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 :smile: !

##\ ##
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 :biggrin:
 
  • #6
Hoping you're right is one thing. What is needed to convince you that you are right :wink: ?

Well done!

##\ ##
 
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Likes DeBangis21 and orangephysik

1. What is the Doppler effect?

The Doppler effect is a phenomenon in which the perceived frequency of a wave, such as sound or light, changes when the source of the wave is moving relative to the observer. This results in a shift in the wavelength and can be observed in everyday situations, such as the change in pitch of a siren as an ambulance passes by.

2. How does the Doppler effect relate to the acceleration of the source?

The acceleration of the source can affect the perceived frequency of a wave due to the Doppler effect. When the source is accelerating towards the observer, the frequency of the wave will increase, resulting in a higher pitch. Conversely, when the source is accelerating away from the observer, the frequency will decrease, resulting in a lower pitch.

3. What is the difference between the Doppler effect and the Doppler shift?

The Doppler effect refers to the change in frequency of a wave due to relative motion between the source and observer. The Doppler shift, on the other hand, refers specifically to the change in frequency of electromagnetic waves, such as light, due to the Doppler effect.

4. Can the Doppler effect be observed in all types of waves?

Yes, the Doppler effect can be observed in all types of waves, including sound, light, and even water waves. However, the amount of frequency shift may vary depending on the type of wave and the relative velocity between the source and observer.

5. How is the Doppler effect used in practical applications?

The Doppler effect has many practical applications, including in radar and sonar systems, where it is used to determine the velocity of objects by measuring the frequency shift of reflected waves. It is also used in medical imaging, such as ultrasound, to measure blood flow and detect abnormalities. Additionally, the Doppler effect is used in astronomy to study the motion of celestial objects and in speed detection devices, such as police radar guns.

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