Assignment 12 Query

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course Phy 122

2/24 3 pm

012. `Query 10

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Question: `q**** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?

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Your Solution:

Well we know that when the wavelength decreases, more half waves can fit on the string. So if the wavelength is halfed than there would be 2 half-waves. If the wavelength was halfed again after that, then there would be 4 half-waves.

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Given Solution: `q** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc..

So you get

1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be

1 * 1/2 `lambda = L so `lambda = 2 L.

For 2 wavelengths fit into the string you get

2 * 1/2 `lambda = L so `lambda = L.

For 3 wavelengths you get

3 * 1/2 `lambda = L so `lambda = 2/3 L; etc.

Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc..

FOR A STRING FREE AT ONE END:

The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **

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Self-critique (if necessary):

I think I did this right, not completely sure though.

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The wavelength doesn't get halved. What happens is that for each higher harmonic you fit two more half-waves to the length of the string.

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Self-critique Rating:3

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Question: `q**** Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?

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Your solution:

You take the velocity and divide it by the wavelength.

confidence rating #$&*:

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Given Solution:

`a** The frequency is the number of crests passing per unit of time.

We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second.

So frequency is equal to the wave velocity divided by the wavelength. **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `q **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?

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Your solution:

You take th tension and divide it by the mass per unit length.

confidence rating #$&*:

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Given Solution:

`a** We divide tension by mass per unit length:

v = sqrt ( tension / (mass/length) ). **

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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&#This looks good. See my notes. Let me know if you have any questions. &#