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
1
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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FAQ: Doppler effect and acceleration of source

What is the Doppler effect?

The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. It is commonly observed with sound waves, but it also applies to electromagnetic waves such as light.

How does the acceleration of the source affect the Doppler effect?

When the source of the waves is accelerating, the frequency shift observed is not constant but changes over time. The rate of change of frequency depends on the rate of acceleration and the direction of the source's motion relative to the observer.

Can the Doppler effect be observed with light waves?

Yes, the Doppler effect can be observed with light waves, and it is known as the relativistic Doppler effect when considering high velocities close to the speed of light. This effect is used in astronomy to determine the movement of stars and galaxies.

What is the formula for the Doppler effect when the source is accelerating?

The formula for the Doppler effect in the case of an accelerating source is more complex than for a constant velocity. It often requires integration to account for the continuously changing velocity, but a simplified version for small accelerations is given by \( f' = f \left( \frac{c \pm v(t)}{c \mp v(t)} \right) \), where \( f \) is the emitted frequency, \( f' \) is the observed frequency, \( c \) is the speed of the wave, and \( v(t) \) is the velocity of the source at time \( t \).

What are some practical applications of the Doppler effect with accelerating sources?

Practical applications include radar and sonar systems, where the motion of objects can be tracked even if they are accelerating. In medical imaging, the Doppler effect is used in ultrasound to measure blood flow velocity. In astrophysics, it helps in understanding the dynamics of celestial bodies and their acceleration.

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