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Phy 122
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problem set 6.18 question
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I'm working problem set 6 number 18. In the solution section I don't understand where the 10 meters comes from when we are calculating the wavelenghts in a certain distance. That certain distance being 10 meters in this case. I pasted the problem in question below. I surrounded the 10 meters in question with ???? in front and end.
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Problem
Two identical strings under identical tensions are positioned with their far ends attached to the right earring of a volunteer. The string. which has mass density 3.5 grams/meter and is under a tension of 7.4 Newtons (not quite enough to cause discomfort in the volunteer, whose head is held stationary).
The other ends of the strings undergo simple harmonic oscillation of amplitude .74 cm. The simple harmonic motion is induced in both strings by the same harmonic oscillator, so these ends oscillate in phase. However, the oscillator is attached to the far end of one string at a distance 10 cm further from the volunteer's earring than that of the other.
Which of the following frequencies will be most likely to cause the volunteer more discomfort than the others, and which the least: 211.4 Hz, 369.9 Hz, 105.7 Hz or 52.85 Hz?
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 7.4 N and `mu the mass density 3.5 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 2114 m/s.
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 2114 m/s / ( 211.4 cycles/sec) = 10 meters,
`lambda2 = 2114 m/s / ( 369.9 cycles/sec) = 5.715 meters,
`lambda3 = 2114 m/s / ( 105.7 cycles/sec) = 20 meters,
`lambda4 = 2114 m/s / ( 52.85 cycles/sec) = 40 meters.
We next calculate the number of wavelengths in the distance ????10 meters????:
For 211.4 Hz we see that the wave on the longer string lags the other by 10 meters / ( 10 meters/cycle) = 1 cycle(s).
For 369.9 Hz we see that the wave on the longer string lags the other by 10 meters / ( 5.715 meters/cycle) = 1.749 cycle(s).
For 105.7 Hz we see that the wave on the longer string lags the other by 10 meters / ( 20 meters/cycle) = .5 cycle(s).
For 52.85 Hz we see that the wave on the longer string lags the other by 10 meters / ( 40 meters/cycle) = .25 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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How was 10 meters choosen for the calculations of number of wavelengths in the distance 10 meters? Thanks
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The problem states a difference of 10 cm in the lengths of the strings. The unit of that difference should have been meters, not cm.
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