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How to find f0 when given highest and lowest frequency

  • Thread starter HHH
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  • #26
HHH
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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
 
  • #27
HHH
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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?
 
  • #28
davenn
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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
 
  • #29
Orodruin
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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.
 
  • #30
davenn
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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
 
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  • #31
Orodruin
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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.
 
  • #32
HHH
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I dont understand all this... just in grade 11 physics. Thanks for the help though
 
  • #33
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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)
 
  • #34
Khashishi
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velocity has a direction. not just speed.
 
  • #35
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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.
 
  • #36
HHH
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Yeah i think the 50 was the sound bouncing off the wall.
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.
Do you mean solving by substitution?
 
  • #37
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Do you mean solving by substitution?
There are many ways to solve a system of two equations, and I don't like very specific names for simple algebraic operations on them, but the answer is probably "yes".

Yeah i think the 50 was the sound bouncing off the wall.
That should not give 50 Hz.
 
  • #38
HHH
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That should not give 50 Hz.
any idea on what the 50Hz and 200Hz was? It came up multiple times when the teacher was swinging the device.
 
  • #39
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Something oscillating with the frequency of the power grid (depending on where you live)?
Or some random other source of sound. Where does the number 200 come from? If you have even more other frequencies, it just shows the values are not reliable.
 
  • #40
HHH
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Something oscillating with the frequency of the power grid (depending on where you live)?
Or some random other source of sound. Where does the number 200 come from? If you have even more other frequencies, it just shows the values are not reliable.
We did it in a classroom, how can an app pick up power grid thing. Also, during the lab we kept getting different frequencies ranging between 50 to 640, teacher said use the highest and lowest.
 
  • #41
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158 m/s is very obviously wrong. It is about 600 km/h; things moving at such speeds require a lot of energy pumped into them just to keep moving through the atmosphere, and that energy is then dissipated into heat and sound - a lot of sound, making the entire exercise rather pointless.

A reasonable speed would not exceed 20 m/s, then the min/max observed frequencies should differ by about 10%.
 
  • #42
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how can an app pick up power grid thing
Various objects can transfer electric currents to mechanical vibrations. All sorts of coils love to do that.
teacher said use the highest and lowest.
That is a bad idea.
 
  • #43
HHH
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That is a bad idea.
Yeah i agree, i think he just wants to see of our math is correct. Thanks for the help
 

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