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How to find f0 when given highest and lowest frequency

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  • #1
HHH
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< Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template was used >

How would you determine f0 (frequency of the source) and Vsource when given highest and lowest frequency

Highest Frequency = 640 Hz
Lowest Frequency = 50 Hz

I cant use the Doppler effect equation because there are two unknown variables
 
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Answers and Replies

  • #2
Khashishi
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Your question is very unclear.
 
  • #3
HHH
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Your question is very unclear.
Basically we did a lab, in which our teacher had a device emitting a sound. He attached the device to a string and started swinging it above his head. He told us to use an app, to determine the highest and lowest frequency, then based on that he told us that we should be able to find the speed at which he was spinning the device and the original frequency of the device (f0).

We got 640 Hz for the highest frequency (moving towards)
and we got 50 Hz for the lowest frequency (moving away)
 
  • #4
Khashishi
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Why do you think the Doppler formula won't work?
 
  • #5
davenn
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I cant use the Doppler effect equation because there are two unknown variables
which variables did you think were unknown ?
 
  • #6
HHH
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which variables did you think were unknown ?
I dont have f0 and Vsource, which are two variables

Fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0

i have 2 values for Fobs (640 and 50 - high and low frequency)
i have local speed of sound = 344.12 m/s
I have Vdetector = 0 m/s
I dont have f0 and Vsource so i dont know how to solve this with two unknown variables
 
  • #7
Khashishi
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to solve two unknowns, you can use two equations, or you can use one equation at two data points.
 
  • #8
HHH
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to solve two unknowns, you can use two equations, or you can use one equation at two data points.
Do you mean by using substitution? Also doesn't the frequency difference seem to large (between high and low)?
 
  • #9
davenn
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my maths is shockingly bad..... it helps to draw pic's

doppler.GIF


the Vsource at points 1 and 3 are constant just have + or - values

For an instant of time at point 2 that forward/reverse velocity relative to the observer is zero
and at that point Fo is heard

with the velocity being constant, I would expect the Fo at point 2 to be half way between 50 and 640 ?

some one please correct me if I am wrong.

if so you now have Fo and can work out the Vs at points 1 and 3 which = the overall Vs of the spinning source


Dave
 
  • #10
HHH
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my maths is shockingly bad..... it helps to draw pic's

View attachment 75826

the Vsource at points 1 and 3 are constant just have + or - values

For an instant of time at point 2 that forward/reverse velocity relative to the observer is zero
and at that point Fo is heard

with the velocity being constant, I would expect the Fo at point 2 to be half way between 50 and 640 ?
So basically i can find f0 by finding the average between the two frequencies?
f0 = fhigh + flow / 2
f0 = 640 +50 /2
f0 = 690/2
f0 = 345Hz

Then what would i do next with that value?
 
  • #11
Khashishi
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Don't ask questions before you try to solve it with the help people have given you.
 
  • #12
davenn
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Then what would i do next with that value?
so now you have everything except Vs

what are you going to do with your Doppler formula to solve for Vs instead of the usual Fobs ?
 
  • #13
HHH
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I used the highest frequency and plug it in as fobs and solved for Vsource, however when i do this i get 158 m/s. When i try the same thing with the lowest frequency (plugging it in as fobs) i get 2000 m/s. Both these values seem to high and both these values should also be the same, it doesnt make any sense
 
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  • #14
davenn
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show us the formula and working you used to solve for Vsource
and we can see if it is correct

there is a reason why you got a difference.

Hint one of your answers is correct, but for the other answer you forgot to take something into account
do you know what that was ?


Dave
 
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  • #15
HHH
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1. Using highest frequency
fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0
640= ((344.12 + 0) / (344.12- Vsource)) 345
640/345 = 344.12/344.12-Vsource
1.885 = 344.12/344.12-Vsource
344.12 = 638.367 - 1.885Vsource
344.12 - 638.367 = -1.885Vsource
-294.247= -1.885Vsource
Vsource = 158.6 m/s

1. Using lowest frequency
fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0
50 = ((344.12 + 0) / (344.12 + Vsource)) 345
50/345 = 344.12/344.12+Vsource
0.1449 = 344.12/344.12+Vsource
344.12 = 49.87 + 0.1449Vsource
344.12 - 49.87 = 0.1449Vsource
294.24 = 0.1449Vsource
Vsource = 2030.64 m/s
 
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  • #16
HHH
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show us the formula and working you used to solve for Vsource
and we can see if it is correct

there is a reason why you got a difference.

Hint one of your answers is correct, but for the other answer you forgot to take something into account
do you know what that was ?


Dave
I am not sure what i did wrong.... it seems right to me
 
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  • #17
davenn
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OK to lessen the complication you only need to work out the Vs for the increase in freq ( as the source is coming towards you

you got the right answer 158 m/s

taking into account that the velocity is constant, the velocity of the source as it is moving away from you is still 158 m/s its just going to be negative

( was trying myself to justify the same result and haven't been able to do that haha)
someone may show us both how to use the 50Hz and get the -158 m/s ... as I said my maths suck LOL

here's an example graphic I took out of a physics paper that showed the rearranged formula to solve for Vs


doppler2.gif
 
  • #18
HHH
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Isnt the speed of 158 m/s way to fast... for some swinging something above their head?
Also, can anyone explain why it is different when moving towards and moving away... i still don't understand it.
 
  • #19
davenn
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Isnt the speed of 158 m/s way to fast... for some swinging something above their head?
we can only go on the observed sound freq's given in the problem ( they may or may not be realistic) its irrelevant, we both were able to come up with a correct solvable answer :)

Also, can anyone explain why it is different when moving towards and moving away... i still don't understand it.
your moving away speed of ~ 2000m/s is incorrect. As I said, because the source velocity is constant the moving away speed is the same, just negative

Dave
 
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  • #20
HHH
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Thanks for the help
 
  • #21
davenn
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no probs ....
but before you go .....
 
  • #22
HHH
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yes?
 
  • #23
davenn
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sorry

I thought I had worked out how to come up with 50Hz

need to ponder that a bit more

if I do I will post it in here :)

Dave
 
  • #24
HHH
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I wanted to ask you something else...

what if i did

fidfference = f1 - f2
640 - 50 = ((344.12/344.12-Vsource)345) - ((344.12/344.12+Vsource)345)
then factor everything out... would that be right, or the method used before?

i get 197.4 m/s and -599 m/s
 
  • #25
davenn
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i get 197.4 m/s and -599 m/s
which is obviously incorrect compared to the result using the formula in that example

the 158 m/s is correct :)
 

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