The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of open organ pipe is 80cm then the length of the closed organ pipe will be
A
30cm
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B
35cm
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C
40cm
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D
20cm
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Solution
The correct option is D20cm Length of open organ pipe, l=80cm=0.8m So, first overtone (n=2), f1=nv2l=2v2×0.8=5v4 Length of closed organ pipe, l=? So, fundamental frequency (n=1), f2=nv4l=v4l Given, f1=f2 Hence, 5v4=v4l ⇒l=0.2m=20cm