The Doppler Effect - Finding Velocity of Source

In summary: Hz = 392 HzGood. Now set those two expressions equal to each other and solve for the source velocity.In summary, the fire truck is moving at a velocity of approximately 17.5 m/s based on the difference in frequency measured by the detector and the speed of sound.
  • #1
harujina
77
1

Homework Statement



A fire truck emitting a 450 Hz signal passes by a stationary detector.
The difference in frequency measured by the detector is 58 Hz.
If the speed of sound is 345 m/s, how fast is the fire truck moving?

Homework Equations



frequency of observer = (velocity of sound + velocity of detector / velocity of sound + velocity of source) f0

The Attempt at a Solution



f obs (frequency of observer) = 450 Hz + 58 Hz = 508 Hz
v sound (velocity of sound) = 345 m/s
v detector = 0
f0 (frequency of source) = 450 Hz
v source = ?

How would I solve for velocity of source? I tried plugging all of the above variables into the equation and isolating the variable but I got a totally wrong answer and I'm not sure why.
 
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  • #2
Question 1: is the truck moving towards or away from the detector?
Question 2: what equation did you use and what was the answer you got?
 
  • #3
Careful with the sign of the source velocity.
 
  • #4
wreckemtech said:
Question 1: is the truck moving towards or away from the detector?
Question 2: what equation did you use and what was the answer you got?

That's the thing... it doesn't say. My teacher told me to just use it as it is going away from the detector.

Doc Al said:
Careful with the sign of the source velocity.
I used + for the source going away from the observer, yet I still got it wrong?

This is what I did:

508 Hz = (345 m/s + 0 / 345 + v source) 450 Hz
508 Hz / 450 Hz = 345 m/s / 345 m/s + v source
1.13 Hz (345 m/s) = 345 m/s / v source

and... everything went downhill from there...
 
  • #5
It says that the truck is passing by.
It may be that the 58 Hz is the difference in the frequency between the frequency measured when the truck approaches and the one measured when it is going away.

Anyway, your algebra is not right. The last line in the above does not look good.
Check again.
 
  • #6
harujina said:
I used + for the source going away from the observer, yet I still got it wrong?

This is what I did:

508 Hz = (345 m/s + 0 / 345 + v source) 450 Hz
508 Hz / 450 Hz = 345 m/s / 345 m/s + v source
1.13 Hz (345 m/s) = 345 m/s / v source

and... everything went downhill from there...
That won't work, since you are multiplying the original frequency by a fraction less than one and setting it equal to a larger frequency.

I think nasu picked up on the key point: that the truck passes by the observer. So at first it is coming towards the observer (higher frequency observed) then it is going away from the observer (lower frequency observed). Set up equations for both coming and going, taking care of the signs of the source velocity in each segment.
 
  • #7
I still don't get it...
Or I do, but I'm still getting something wrong with my algebra.
What do I do in any case, to find the velocity of the source?
 
  • #8
harujina said:
I still don't get it...
Or I do, but I'm still getting something wrong with my algebra.
What do I do in any case, to find the velocity of the source?
Do this:
(1) Write an expression for the higher frequency observed when the truck is approaching.
(2) Write an expression for the lower frequency observed when the truck is moving away.
(3) Compare those two and then you can solve for the source velocity.
 
  • #9
nasu said:
It says that the truck is passing by.
It may be that the 58 Hz is the difference in the frequency between the frequency measured when the truck approaches and the one measured when it is going away.

Anyway, your algebra is not right. The last line in the above does not look good.
Check again.
Wait, so... difference meaning it would be half for each?
i.e. f obs = 450 Hz + (58 Hz / 2); = 450 Hz + 29 Hz = 479 Hz (truck going away)
f obs = 450 - 29 Hz = 421 Hz (truck approaching)

Doc Al said:
Do this:
(1) Write an expression for the higher frequency observed when the truck is approaching.
(2) Write an expression for the lower frequency observed when the truck is moving away.
(3) Compare those two and then you can solve for the source velocity.

... if the above is correct:
(1) 421 Hz = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
(2) 479 Hz = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away
(3) ?
 
  • #10
No, the difference is not necessarily split even
Why don't you follow the steps suggested by Doc Al?
 
  • #11
harujina said:
Wait, so... difference meaning it would be half for each?
i.e. f obs = 450 Hz + (58 Hz / 2); = 450 Hz + 29 Hz = 479 Hz (truck going away)
f obs = 450 - 29 Hz = 421 Hz (truck approaching)
No. Don't try to guess the frequencies by assuming that half the difference goes to each.
Also: Which gives the higher observed frequency? Truck approaching the observer or truck going away?

... if the above is correct:
(1) 421 Hz = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
(2) 479 Hz = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away
(3) ?
No. Instead:
(1) F1 = ...
(2) F2 = ...

Don't guess what those frequencies are, but use what you know about how they are related.
 
  • #12
Doc Al said:
No. Don't try to guess the frequencies by assuming that half the difference goes to each.
Also: Which gives the higher observed frequency? Truck approaching the observer or truck going away?


No. Instead:
(1) F1 = ...
(2) F2 = ...

Don't guess what those frequencies are, but use what you know about how they are related.
Oh, oops. The truck approaching the observer should be the higher frequency...

(1) F1 = 450 Hz + 58 Hz = 508 Hz
(2) F2 = 450 Hz - 58 Hz = 392 Hz
 
  • #13
harujina said:
Oh, oops. The truck approaching the observer should be the higher frequency...

(1) F1 = 450 Hz + 58 Hz = 508 Hz
(2) F2 = 450 Hz - 58 Hz = 392 Hz
Stop assuming you know the observed frequencies!

Treat F1 and F2 as unknowns.
 
  • #14
(1) F1 = 450 Hz + x = ?
(2) F2 = 450 Hz - y = ?

...Like this?
 
  • #15
harujina said:
(1) F1 = 450 Hz + x = ?
(2) F2 = 450 Hz - y = ?

...Like this?
No.

The right hand sides of your equations for F1 and F2 were correct here:
harujina said:
(1) 421 Hz = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
(2) 479 Hz = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away
 
  • #16
Doc Al said:
No.

The right hand sides of your equations for F1 and F2 were correct here:
Oh, I see.

(1) F1 = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
(2) F2 = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away

So there are three unknown variables here? The v source, F1, and F2...
 
  • #17
harujina said:
Oh, I see.

(1) F1 = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
(2) F2 = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away
Right.

So there are three unknown variables here? The v source, F1, and F2...
But there's another fact you can use, from the problem statement. How do F1 and F2 relate?
 
  • #18
Doc Al said:
Right.


But there's another fact you can use, from the problem statement. How do F1 and F2 relate?

F1 - F2 = 58 Hz ?
 
  • #19
harujina said:
F1 - F2 = 58 Hz ?
Exactly!
 

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon that occurs when there is a relative motion between a source of waves and an observer. It causes a change in the frequency of the waves perceived by the observer, which results in a perceived change in the pitch of the sound or the color of the light.

2. How does the Doppler Effect relate to finding the velocity of a source?

The Doppler Effect can be used to determine the velocity of a source by measuring the change in frequency of the waves. The greater the relative motion between the source and the observer, the greater the change in frequency, and thus, the higher the velocity of the source.

3. What type of waves does the Doppler Effect apply to?

The Doppler Effect can be observed in all types of waves, including sound, light, and even water waves. It is not limited to a specific type of wave, but rather to the relative motion between the source and the observer.

4. How is the Doppler Effect used in real-life applications?

The Doppler Effect has many practical applications, such as in radar technology, where it is used to calculate the velocity of moving objects. It is also used in medical imaging, such as ultrasound, to measure blood flow velocity. Additionally, astronomers use the Doppler Effect to determine the velocity and direction of stars and galaxies.

5. Are there any limitations to using the Doppler Effect to find the velocity of a source?

Yes, there are some limitations to using the Doppler Effect. It only works if there is a relative motion between the source and the observer, and the velocity must be constant. In some cases, the velocity of the source may not be directly towards or away from the observer, which can affect the accuracy of the measurements. Additionally, if there are multiple sources or multiple observers, the calculations can become more complex.

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