Doppler Effect with acceleration Problem

[zorro]

Homework Statement

A source S and a detector D are initially at a distance of x=1km. Both start moving towards one another with same acceleration a = 10m/s². Frequency of source is f = 2000Hz. Find the frequency observed by the detector at time t=4s. Speed of sound in air is v=300m/s

The Attempt at a Solution

I found out the velocity of detector at the end of 4s as 40m/s.
f1 = f(v+v_d/v-v_s)
If we substitute v_s as 40m/s, the answer comes wrong.
Any help appreciated.

 

[hikaru1221]

10m/s^2? That's just like free falling! Maybe big airplane? Never mind then
The formulas f = f_o(v+v_d)/v and f = f_ov/(v-v_s) only apply to cases where the velocities of source and detector remain constant. I suggest that you have a look at the proofs of those formulas, find out where those proofs IMPLICITLY assume that those velocities are constant and then come back to this problem You will have to make some changes to the formulas before applying them to this particular problem Another hint: The question doesn't give you the initial distance for no use.

 

[zorro]

Well I know the concept behind the derivation of the formula.
The apparent frequency of the sound heard is due to the difference in time interval between two successive wavefronts coming from the source.

In this problem,
the separation between the source and the detector after t=4s is 840m.
We have to find the freq. of sound heard at this time.
The time lapse between two successive wavefronts is 1/2000 s
So basically we have to find out the time interval between the wavefront reaching the detector at t=4s and the one just before it which is offcourse less than 1/2000.

 

[zorro]

Provide me with some more hints please

 

[hikaru1221]

Hehe, you have ASSUMED that the velocities are constant Now an MCQ See the picture. The black lines denote wave fronts. Suppose that instance is at t=4s. Which wave front does D (detector) "capture" and read?
A. (1)
B. (2)
C. (3)
D. (4) ?!
At what time is that wave front emitted from S (source)? Is it at t=4s? Or earlier? Or later (?!)?

 

[zorro]

The detector will capture 1 ,2 ,3 wavefronts one after the other.
It captures 1 first. It was emitted from the source at time less than 4s.
Is it right?

 

[hikaru1221]

Sorry for late reply; I didn't see your reply.
Yes, you're right

So basically what D receives is the wavefront that S emits earlier. To make it easier to understand, assume that S can emit 2 consecutive wavefronts while its speed is still nearly the same (then another couple of wavefronts when S jumps to another speed and so forth). So the distance between the 2 consecutive wavefronts is determined by the speed of S at the time of emission t1. This distance is simply the wavelength of the signal emitted by S at the time of emission t1. Then when D detects this signal, what D receives is the signal of wavelength (or frequency, by the relation lambda = c/f) corresponding to the time t1, while D is at the time t2 > t1!

In the case of constant speeds, since the speed of S doesn't change, there is no difference in the signal at time t1 or t2. Therefore even if you plug in the speed of S at time t2, it still gives you the right answer.

On the other hand, in this problem, the speeds of S at t1 and t2 are different, and so are the signals. Therefore you cannot plug in the speed of S at time t2; it should be the speed of S at time t1.

 

[zorro]

Sorry for late reply; I didn't see your reply.
So the distance between the 2 consecutive wavefronts is determined by the speed of S at the time of emission t1. This distance is simply the wavelength of the signal emitted by S at the time of emission t1. Then when D detects this signal, what D receives is the signal of wavelength (or frequency, by the relation lambda = c/f) corresponding to the time t1, while D is at the time t2 > t1!

In this problem, we don't know what is the speed of source at time t1. If the question had been to find the frequency of sound observed by the detector corresponding to the first wavefront emitted, things would have been quite easier.
We don't know which two consecutive wavefronts reach the detector in time t= 4 + Δt s
When is the wavefront that reaches the detector at t=4s emitted by the source? Certainly its not at t=0 but at some other time between 0 and 4s. This problem is spinning my head!

 

[hikaru1221]

In this problem, we don't know what is the speed of source at time t1.

We don't know immediately, but we may know. Do some math
The initial distance between D and S is d. At time t₁, S moves by a distance A = at₁²/2 and emits the wavefront that later meets D at t=4s. This wavefront moves with constant speed v = 300m/s. So by t=t₂=4s, this wavefront moves by a distance B = v(t₂-t₁). Also at t=4s, D moves by a distance C = at₂²/2. We must have: d = A + B + C, as D meets the wavefront at t=4s. Now solve for t₁

 

[zorro]

I got the answer. But I have a doubt.
Consider this question-
A source emitting a sound of frequency f is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v.

here I proceed as follows-
Let the initial distance between them be d.
at t=0, the first wavefront is emitted by the source.
Time taken by the w.f. to travel the distance d, t1 = d/v
Now the second w.f. will be emitted after a time 1/f. During this time, the source moves through a distance given by s = 0.5a/f²
So the time taken by second w.f. to reach the observer is t2 = 1/f + (d-s)/v
The time interval between two succesive wavefronts is t2-t1
Hence the frequency heard by the observer is 1/(t2-t1)

Why can't we proceed in the previous problem by this way? Can we do this (above) problem by your method (finding the velocity of the source at some time)?

 

[hikaru1221]

I got the answer. But I have a doubt.
Consider this question-
A source emitting a sound of frequency f is placed at a large distance from an observer. The source starts moving towards the observer with a uniform acceleration a. Find the frequency heard by the observer corresponding to the wave emitted just after the source starts. The speed of sound in the medium is v.

here I proceed as follows-
Let the initial distance between them be d.
at t=0, the first wavefront is emitted by the source.
Time taken by the w.f. to travel the distance d, t1 = d/v
Now the second w.f. will be emitted after a time 1/f. During this time, the source moves through a distance given by s = 0.5a/f²
So the time taken by second w.f. to reach the observer is t2 = 1/f + (d-s)/v
The time interval between two succesive wavefronts is t2-t1
Hence the frequency heard by the observer is 1/(t2-t1)

Why can't we proceed in the previous problem by this way? Can we do this (above) problem by your method (finding the velocity of the source at some time)?

This method is simple wrong for both problems. The real phenomenon is not like that. Have a look at what I wrote when making the assumption about the 2 consecutive wavefronts:

To make it easier to understand, assume that S can emit 2 consecutive wavefronts while its speed is still nearly the same (then another couple of wavefronts when S jumps to another speed and so forth).

That is, this is just to make it easier to understand, but not what really happens. When S moves with acceleration, the frequency observed changes continuously: at time t, D receives a signal with frequency f; at another time t', D receives another signal with another frequency f', even if t' = t + 1/f or t' = t + 1/(1000f). Therefore in fact, the two wavefronts (or signals) "carry" different frequencies. But for simplicity, I assumed that two consecutive wavefronts are so close to each other that the difference in their frequencies is negligible.

Things become really abstract when some quantity changes continuously, especially when it comes to Doppler effect where people usually think of the shape of the emitted wave. However that thinking does not work here, as the shape of the wave changes continuously. But let's put it this way: S emits a wavefront A and then if S suddenly keeps its speed unchanged right after that, all of the arguments that lead to the formula of the constant-speed case must be applicable. That means, D "sees" a wavefront A with frequency f = f_o/(v-v_s) where v_s is the speed of S when S emits wavefront A. So you get the idea?

 

[zorro]

This method is simple wrong for both problems. The real phenomenon is not like that.

I don't agree with you because the answer I get for the second problem matches with my book's answer.

When S moves with acceleration, the frequency observed changes continuously: at time t, D receives a signal with frequency f; at another time t', D receives another signal with another frequency f', even if t' = t + 1/f or t' = t + 1/(1000f). Therefore in fact, the two wavefronts (or signals) "carry" different frequencies. But for simplicity, I assumed that two consecutive wavefronts are so close to each other that the difference in their frequencies is negligible.

These are my views (based on whatever knowledge I have)-
I agree that the two successive wavefronts carry different frequencies - which is the basic cause of the Doppler Effect.
The frequency of sound heard by the detector is nothing but the rate of interception of wavefronts. In the second problem, we just need to find the rate corresponding to the first two wavefronts emitted by the source.
Similarly, we can extend this argument to the first problem, where we need to find the rate of interception of the two wavefronts corresponding to the one at time t=4s and the one just before it.

 

[rl.bhat]

Consider this line of thinking.
The source is emitting 2000 waves per second. In four seconds it sends 8000 waves. When the source and the director start moving, after 4 seconds the first wavefront reaches the detector at a distance 840 m from the source. All 8000 waves are cramped in 840 m. So the average apparent wavelength is 840/8000 = 0.105 m.
At that instant the velocity of the sound with respect to the detector is (v + vd). Hence the apparent frequency f' = 340/0.105 Hz. Is it the correct answer?

 

[hikaru1221]

I don't agree with you because the answer I get for the second problem matches with my book's answer.

Sorry, then I have to disagree with unscientific statement based on no scientific reasoning.

These are my views (based on whatever knowledge I have)-
I agree that the two successive wavefronts carry different frequencies - which is the basic cause of the Doppler Effect.

No, this is NOT the cause of Doppler effect. The cause of Doppler effect is that there is RELATIVE MOTION between S and D. In the case that S and D moves with constant speeds, the wavefronts "carry" the same frequency as viewed in the reference frame of S (f_o)/ ground (f_ov/(v-vs)) / D ((f_o(v+vd)/(v-vs)).

The frequency of sound heard by the detector is nothing but the rate of interception of wavefronts. In the second problem, we just need to find the rate corresponding to the first two wavefronts emitted by the source.

The concept of wavefront is just for the ease of understanding / imagining the phenomenon. It has nothing to do with what is observed. For example, you switch on your detector in a time less than a period, so your detector "sees" the first wavefront but not the second one, then your detector doesn't even detect anything?

 

[zorro]

Consider this line of thinking.
The source is emitting 2000 waves per second. In four seconds it sends 8000 waves. When the source and the director start moving, after 4 seconds the first wavefront reaches the detector at a distance 840 m from the source. All 8000 waves are cramped in 840 m. So the average apparent wavelength is 840/8000 = 0.105 m.
At that instant the velocity of the sound with respect to the detector is (v + vd). Hence the apparent frequency f' = 340/0.105 Hz. Is it the correct answer?

No the correct answer is 2340.8 Hz

No, this is NOT the cause of Doppler effect. The cause of Doppler effect is that there is RELATIVE MOTION between S and D. In the case that S and D moves with constant speeds, the wavefronts "carry" the same frequency as viewed in the reference frame of S (f_o)/ ground (f_ov/(v-vs)) / D ((f_o(v+vd)/(v-vs)).

The concept of wavefront is just for the ease of understanding / imagining the phenomenon. It has nothing to do with what is observed. For example, you switch on your detector in a time less than a period, so your detector "sees" the first wavefront but not the second one, then your detector doesn't even detect anything?

hmm...thanks for clearing my misconception. I had wrong notion of the cause of Doppler Effect. Can you name the book from where you studied high school physics

 

[hikaru1221]

hmm...thanks for clearing my misconception. I had wrong notion of the cause of Doppler Effect. Can you name the book from where you studied high school physics

Textbook in my country
You can refer to some popular introductory university textbooks such as University Physics (Freedman, Young), Fundamentals of Physics (Halliday), etc. But they definitely won't explain everything clearly to you. Most of the times the rest is left for you to think Actually I find Wikipedia sometimes really ample, so if you don't major in physics, you should not waste money buying those thick books. Instead of finding here and there for books, you will find sitting alone thinking really worth it. By the way, those introductory textbooks won't provide you the answer for most of your problems; those problems of yours are quite advanced at the introductory level.
MIT OpenCourseWare is also a good idea. The lecture notes there are very concise and contain accurate knowledge as well as logical arguments, and thus, worth reading, though brief
Now I really regret that I never thought of exploiting the Internet while in high school

 

[zorro]

Textbook in my country
You can refer to some popular introductory university textbooks such as University Physics (Freedman, Young), Fundamentals of Physics (Halliday), etc. But they definitely won't explain everything clearly to you. Most of the times the rest is left for you to think Actually I find Wikipedia sometimes really ample, so if you don't major in physics, you should not waste money buying those thick books. Instead of finding here and there for books, you will find sitting alone thinking really worth it. By the way, those introductory textbooks won't provide you the answer for most of your problems; those problems of yours are quite advanced at the introductory level.
MIT OpenCourseWare is also a good idea. The lecture notes there are very concise and contain accurate knowledge as well as logical arguments, and thus, worth reading, though brief

Yeah I have Halliday-Resnick-Walker Fundamentals of Physics.
You are absolutely right, Physics is a subject which requires more thinking that reading.
MIT OpenCourseWare is a good idea, I will read those lecture notes. Thanks alot!