- #31
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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:
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.
$$
1/f_0 = 0.5(1/f_h + 1/f_\ell).
$$
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.