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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Ring Discomfort
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Set 6 Problem 18 states:
A simple harmonic oscillator drives the ends of two strings, identical except for a length difference of 12 cm, at frequency 51.09 Hz, then at 43.79 Hz, then at 29.19 Hz and finally at 36.49 Hz at an amplitude of .85 cm.
The strings are under identical tensions and are positioned with their far ends attached to the nose ring of a volunteer. The strings have mass density 13.7 grams/meter and are each under a tension of 4.8 Newtons (not quite enough hurt).
Which of the frequencies will be most likely to cause the volunteer more discomfort than the others, and which the least?
Solution
If the peaks of the two waves arrive at the ear of the volunteer simultaneously, whatever discomfort they cause will be maximized, since they will reinforce one another. If a 'peaks' of one wave arrive simultaneously with the 'valleys' of the other, the two will 'cancel out' and cause minimum discomfort.
If the difference in the distances traveled by the two waves is 1, 2, 3, ... complete wavelengths then, since the two start in phase, they will arrive in phase. If the difference in distances is 1/2, 3/2, 5/2, ... complete wavelengths then the peaks of one will arrive along with the valleys of the other.
We calculate the wavelengths corresponding to the four given frequencies:
We first calculate the wave velocity, which is v = `sqrt( T / `mu ), where T is the tension 4.8 N and `mu the mass density 13.7 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 350.3 m/s.
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 350.3 m/s / ( 51.09 cycles/sec) = 6.856 meters,
`lambda2 = 350.3 m/s / ( 43.79 cycles/sec) = 7.999 meters,
`lambda3 = 350.3 m/s / ( 29.19 cycles/sec) = 12 meters,
`lambda4 = 350.3 m/s / ( 36.49 cycles/sec) = 9.599 meters.
We next calculate the number of wavelengths in the distance 12 meters:
For 51.09 Hz we see that the wave on the longer string lags the other by 12 meters / ( 6.856 meters/cycle) = 1.75 cycle(s).
For 43.79 Hz we see that the wave on the longer string lags the other by 12 meters / ( 7.999 meters/cycle) = 1.5 cycle(s).
For 29.19 Hz we see that the wave on the longer string lags the other by 12 meters / ( 12 meters/cycle) = 1 cycle(s).
For 36.49 Hz we see that the wave on the longer string lags the other by 12 meters / ( 9.599 meters/cycle) = 1.25 cycle(s).
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Some of the variable values seem to come out wrong:
And the length of the string doesn't convert from cm to meters.
Using the solution as a template and redoing the calculations, I get:
We first calculate the wave velocity, which is v = `sqrt( T / `mu ), where T is the tension 4.8 N and `mu the mass density 13.7 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = `sqrt(4.8N/.0137kg/m) = 18.718 m/s
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I agree with this velocity; my mental calculation came out around 18 m/s.
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For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 18.718 m/ss / ( 51.09 cycles/sec) = .366 meters,
`lambda2 = 18.718 m/s / ( 43.79 cycles/sec) = .427 meters,
`lambda3 = 18.718 m/s / ( 29.19 cycles/sec) = .641 meters,
`lambda4 = 18.718 m/s / ( 36.49 cycles/sec) = .513 meters.
We next calculate the number of wavelengths in the distance 12 cm:
For 51.09 Hz we see that the wave on the longer string lags the other by. 12 meters / ( .366 meters/cycle) = .327 cycle(s).
For 43.79 Hz we see that the wave on the longer string lags the other by .12 meters / ( .427 meters/cycle) = .281 cycle(s).
For 29.19 Hz we see that the wave on the longer string lags the other by .12 meters / ( .641 meters/cycle) = .187 cycle(s).
For 36.49 Hz we see that the wave on the longer string lags the other by .12 meters / ( .513 meters/cycle) = .233 cycle(s).
These are not at all like the neat integers and integers +.5 I was hoping to get, but I think my calculations are correct.
In this case since .327 cycles is closest to .5, I would say 51.09 Hz was likely to minimize discomfort.
Since .187 was closest to 0, I would say 29.19 Hz was likely to maximize discomfort.
???Is my thinking correct???
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Yes, Very good.
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See my brief note.
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