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course Phy 242
questions repeated from query 11
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):
Good. The same question was on Query 11.
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Self-critique Rating:
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Good.
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