Asked by GINSENG

Calculate the length of a closed organ pipe with a 680 Hz fundamental frequency. What is the wavelength of the 4th overtone?
Answered by Damon
Half a wave fits in the pipe at fundamental, amplitude is zero at both ends.
I do not know what you are using for the speed of sound. Call it s meters/second
period of wave T= 1/680 seconds
wave goes distance 2 L in 1/680 seconds
so
2 L = s * T = s/680
L = s /1360

first harmonic, one whole wave fits in pipe
second harmonic, 1.5 waves fit in pipe
third harmonic 2 waves fit in pipe
fourth harmonic 2.5 waves fit in pipe
so wavelength = that 2 L we had /2.5


Answered by Damon
sorry I have it closed at both ends
Usually open at one, closed at the other
fundamental 1/4 wave fits in pipe
4 L = s * T = s/680 = wavelength of fundamental
L = s /2720

first harmonic, 3/4 wave fits in pipe
second harmonic, 1 1/4 waves fit in pipe
third harmonic 1 3/4 waves fit in pipe
fourth harmonic 2 1/4 waves fit in pipe
so wavelength = original wavelength we had /2.25
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