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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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problem set 6.18 v question
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Working prblem 6.18 I don't understand where the wave velocity of 1972m/s comes from. I have pasted the actual problem below with the 1972m/s in question surrounded by ???? front and end. I think the formula for wave velocity is v=sqrt(T/mu) where mu is in kg/m and T is in Newtons. That being the case I have T=7.1N. mu is given as 3.6g/m so I convert to .0036kg/m for equation. Now I have v=sqrt(T/mu)=sqrt(7.1N/(.0036kg/m))=44.41m/s not 1972m/s.
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It looks like the program didn't take the square root--or, rather, I neglected to tell it to do so.
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Set 56 Problem number 18
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Problem
Two strings are under identical 7.1 Newtons tensions and are positioned with their far ends attached to the tongue ring of a volunteer. The strings have mass density 3.6 grams/meter and are each under a tension of 7.1 Newtons (not quite enough hurt). One string is 6 cm longer than the other.
The 'near' ends of the strings are attached to the same simple harmonic oscillator. The oscillator can be driven at 575.2 Hz, 410.9 Hz, 164.3 Hz or 657.4 Hz. At a given amplitude of oscillation, which frequency would be most likely to cause the volunteer discomfort, 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 7.1 N and `mu the mass density 3.6 grams/ meter * (1 kg / 1000 grams). ????A simple calculation shows that v = 1972 m/s.????
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 1972 m/s / ( 575.2 cycles/sec) = 3.428 meters,
`lambda2 = 1972 m/s / ( 410.9 cycles/sec) = 4.799 meters,
`lambda3 = 1972 m/s / ( 164.3 cycles/sec) = 12 meters,
`lambda4 = 1972 m/s / ( 657.4 cycles/sec) = 2.999 meters.
We next calculate the number of wavelengths in the distance 6 meters:
For 575.2 Hz we see that the wave on the longer string lags the other by 6 meters / ( 3.428 meters/cycle) = 1.75 cycle(s).
For 410.9 Hz we see that the wave on the longer string lags the other by 6 meters / ( 4.799 meters/cycle) = 1.25 cycle(s).
For 164.3 Hz we see that the wave on the longer string lags the other by 6 meters / ( 12 meters/cycle) = .5 cycle(s).
For 657.4 Hz we see that the wave on the longer string lags the other by 6 meters / ( 2.999 meters/cycle) = 2 cycle(s).
From these results it is clear which frequency gives us a whole number (e.g., 1, 2, 3, ... ) of wavelengths of path difference, which gives an integer plus a half-integer number (e.g, .5, 1.5, 2.5, ...), and which give integer plus or minute quarter-integer numbers (e.g., .25, .75, 1.25, 1.75, ...). The first will reinforce and cause maximum effect on the ear; the second will cancel and cause minimum effect; and the last will cause an intermediate effect.
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Can you step me through the calc to find the 1972m/s? Thanks.
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I must have left out the instruction to take the square root when I coded this version of the problem.
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