#$&*
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.
** Question Form_labelMessages **
Question 6
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For introductory problem set, set 6 waves, number 18
when the velocity is calculated, the equation is sqrt(T/mu) but the squareroot is not taken in any of the examples. I'm sure i'm missing something but I do not understand why it's not taken. Thanks.
Example below
Set 56 Problem number 18
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Problem
Two strings are under identical 6 Newtons tensions and are positioned with their far ends attached to the tongue ring of a volunteer. The strings have mass density 13 grams/meter and are each under a tension of 6 Newtons (not quite enough hurt). One string is 7 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 32.98 Hz, 115.4 Hz, 82.45 Hz or 131.9 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 6 N and `mu the mass density 13 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 461.5 m/s.
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 461.5 m/s / ( 32.98 cycles/sec) = 13.99 meters,
`lambda2 = 461.5 m/s / ( 115.4 cycles/sec) = 3.999 meters,
`lambda3 = 461.5 m/s / ( 82.45 cycles/sec) = 5.597 meters,
`lambda4 = 461.5 m/s / ( 131.9 cycles/sec) = 3.498 meters.
We next calculate the number of wavelengths in the distance 7 meters:
For 32.98 Hz we see that the wave on the longer string lags the other by 7 meters / ( 13.99 meters/cycle) = .5003 cycle(s).
For 115.4 Hz we see that the wave on the longer string lags the other by 7 meters / ( 3.999 meters/cycle) = 1.75 cycle(s).
For 82.45 Hz we see that the wave on the longer string lags the other by 7 meters / ( 5.597 meters/cycle) = 1.25 cycle(s).
For 131.9 Hz we see that the wave on the longer string lags the other by 7 meters / ( 3.498 meters/cycle) = 2.001 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.
@&
The problems are solved by an algorithm, and it appears that the algorithm didn't take the square root.
That's my fault, of course, since I wrote the algorithm.
Note that there are arithmetic errors on a few of the problems in the introductory set. Some I have chosen to leave in, to see who is paying attention.
However I've not been aware of the error on this problem. Nobody has mentioned it in the 15 years it's been up there.
So thanks for finding this glitch.
The velocity should be about 21.5 m/s, the wavelengths around .7, .2, .4 and .16 meters, and the lags in distance will also differ accordingly.
*@
#$&*
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.
** Question Form_labelMessages **
Question 6
** **
** **
** **
For introductory problem set, set 6 waves, number 18
when the velocity is calculated, the equation is sqrt(T/mu) but the squareroot is not taken in any of the examples. I'm sure i'm missing something but I do not understand why it's not taken. Thanks.
Example below
Set 56 Problem number 18
--------------------------------------------------------------------------------
Problem
Two strings are under identical 6 Newtons tensions and are positioned with their far ends attached to the tongue ring of a volunteer. The strings have mass density 13 grams/meter and are each under a tension of 6 Newtons (not quite enough hurt). One string is 7 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 32.98 Hz, 115.4 Hz, 82.45 Hz or 131.9 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 6 N and `mu the mass density 13 grams/ meter * (1 kg / 1000 grams). A simple calculation shows that v = 461.5 m/s.
For each frequency f the wavelength is `lambda = v / f, so we obtain wavelengths
`lambda1 = 461.5 m/s / ( 32.98 cycles/sec) = 13.99 meters,
`lambda2 = 461.5 m/s / ( 115.4 cycles/sec) = 3.999 meters,
`lambda3 = 461.5 m/s / ( 82.45 cycles/sec) = 5.597 meters,
`lambda4 = 461.5 m/s / ( 131.9 cycles/sec) = 3.498 meters.
We next calculate the number of wavelengths in the distance 7 meters:
For 32.98 Hz we see that the wave on the longer string lags the other by 7 meters / ( 13.99 meters/cycle) = .5003 cycle(s).
For 115.4 Hz we see that the wave on the longer string lags the other by 7 meters / ( 3.999 meters/cycle) = 1.75 cycle(s).
For 82.45 Hz we see that the wave on the longer string lags the other by 7 meters / ( 5.597 meters/cycle) = 1.25 cycle(s).
For 131.9 Hz we see that the wave on the longer string lags the other by 7 meters / ( 3.498 meters/cycle) = 2.001 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.
@&
The problems are solved by an algorithm, and it appears that the algorithm didn't take the square root.
That's my fault, of course, since I wrote the algorithm.
Note that there are arithmetic errors on a few of the problems in the introductory set. Some I have chosen to leave in, to see who is paying attention.
However I've not been aware of the error on this problem. Nobody has mentioned it in the 15 years it's been up there.
So thanks for finding this glitch.
The velocity should be about 21.5 m/s, the wavelengths around .7, .2, .4 and .16 meters, and the lags in distance will also differ accordingly.
*@