How to find f0 when given highest and lowest frequency

[HHH]

< 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 can't use the Doppler effect equation because there are two unknown variables

 

[Khashishi]

Your question is very unclear.

 

[HHH]

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)

 

[Khashishi]

Why do you think the Doppler formula won't work?

 

[davenn]

I can't use the Doppler effect equation because there are two unknown variables

which variables did you think were unknown ?

 

[HHH]

which variables did you think were unknown ?

I don't 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 don't have f0 and Vsource so i don't know how to solve this with two unknown variables

 

[Khashishi]

to solve two unknowns, you can use two equations, or you can use one equation at two data points.

 

[HHH]

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)?

 

[davenn]

my maths is shockingly bad... it helps to draw pic's

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 sourceDave

 

[HHH]

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?

 

[Khashishi]

Don't ask questions before you try to solve it with the help people have given you.

 

[davenn]

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 ?

 

[HHH]

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 doesn't make any sense

 

[davenn]

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

 

[HHH]

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

 

[HHH]

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

 

[davenn]

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

 

[HHH]

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.

 

[davenn]

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

 

[HHH]

Thanks for the help

 

[davenn]

no probs ...
but before you go ...

 

[HHH]

yes?

 

[davenn]

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

 

[HHH]

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

 

[davenn]

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 :)

 

[HHH]

I also think the reason the two values are off is because of sources of error. During the lab he was swinging it around and the readings on the app kept changing (probably because the speed at which he was swinging it wasn't constant). He just told us to use highest and lowest values, could that be the reason for the difference? He told us to assume it was constant though after

 

[HHH]

btw i also changed the low value to 260Hz... other people in my class got a low frequency closer to that ( i think the 50 might have been an outlier) . What are your thoughts on what i said about sources of error?

 

[davenn]

its doesn't really matter what the low and high frequencies are
you gave 50 Hz and 640 Hz and we found the Fo and the Vs ( 345Hz and 158 m/s)

if you change the low freq to 260 Hz, then the Fo will be different as will the Vs
you can use the same method and find the Fo and the Vs

cheers
Dave

 

[Orodruin]

If the source is moving, it is the mean of the wavelengths that will be the natural wavelength of the source. This means that the mean inverse of the frequencies is the inverse of the natural frequency. In general:
$$
f = f_0/(1-v/c)
$$
Where the velocity is the velocity of the source towards the observer.

In general, you have two equations (one for the frequency of the source moving away and one for moving towards the observer) and two unknowns and should be able to solve for both natural frequency and velocity of the source.

 

[davenn]

If the source is moving, it is the mean of the wavelengths that will be the natural wavelength of the source. This means that the mean inverse of the frequencies is the inverse of the natural frequency. In general:
$$
f = f_0/(1-v/c)
$$
Where the velocity is the velocity of the source towards the observer.

In general, you have two equations (one for the frequency of the source moving away and one for moving towards the observer) and two unknowns and should be able to solve for both natural frequency and velocity of the source.

HI Orodruin

pleased to have you and matterwave looking over this for us :)
let me do those calcs and see the result

lets step back a bit ... .we were only given the 2 observed freq's 50Hz moving away and 640 Hz moving towards the observer. From that we need to fins Fo and Vsource

ok I assume
f = the freq heard by observer and Fo is the original ( natural) freq of the source ?

so given that we have F = 640 Hz, how do we rearrange the formula to solve for Fo ?

Dave

 

[Orodruin]

You can do the same thing you did for the mean frequency, but for the mean inverse frequency instead, which is proportional to the wavelength. You would have:
$$
1/f_0 = 0.5(1/f_h + 1/f_\ell).
$$
which gives
$$
f_0 = 2\frac{f_h f_\ell}{f_h + f_\ell}.
$$
I am not going to go into more detail as the OP should try to solve the problem from here. Instead let me just observe some nice things about this expression:

• When $f_h \to \infty$, $f_0 \to 2f_\ell$. This has a very physical interpretation, $f_h$ goes to infinity when the source is moving at the speed of sound, when this happens the wavelength of the sound from the receding source will be twice that of the stationary source, resulting in the frequency of the stationary source being twice that of the receding one.
• When $f_h = f_\ell$, $f_0 = f_h = f_\ell$, the source is not moving.

We can also examine how reasonable the 158 m/s is. I am of course guessing a bit on your teacher's setup but it should be relatively reasonable. I am guessing the rope had a length of order 1 m, anything larger would be difficult to handle and might be dangerous for closeby students. If the source has a mass of say 0.1 kg, the resulting centripetal force in the rope would have to be around 2.5 kN, which corresponds to the same force necessary to lift 250 kg. Is the teacher able to hold the rope against such a force? I am going to be doubtful.

Now, the corrected frequency shift gives me something of the same order of magnitude even for the corrected lower frequency ... The reason is of course that in order for the frequency to be significantly affected, you need to move at a velocity which is a significant fraction of the speed of sound, in this case 344 m/s. I cannot speak for the quality of the data, but as reasoned above without knowing more of the setup, I am slightly sceptical.

 

[HHH]

I don't understand all this... just in grade 11 physics. Thanks for the help though

 

[nasu]

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?

No, fo is not the average between the two frequencies.
You need to write the equation for the two cases and solve the two equations simultaneously for the two unknowns (fo and v)

 

[Khashishi]

velocity has a direction. not just speed.

 

[mfb]

The 50 Hz value is wrong. You might have picked up the frequency of the power grid if you are not in the Americas, or some other source of sound. There is no way to swing the sound source fast enough to get such a large frequency range and 640 Hz is much more realistic for the experiment than 50 Hz.

Anyway, if you want to find a solution for those values, set up both equations for doppler shift and solve for both variables. Orodruin's suggestion is possible as well, but probably more complicated to understand.