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  4. If the length of an open organ pipe is 33.3 cm, then the frequency of fifth overtone (v5 = 333m/s)

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Q. If the length of an open organ pipe is $33.3\,$ cm, then the frequency of fifth overtone $(v_5 = 333m/s)$

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Solution:

$f_{0}=\frac{V}{2l}$
$=\frac{333}{2 \times 33.3 \times 10^{-2}}=\frac{1000}{2}=500 Hz$
frequency for 5 th overtone
$\phi =6 f_{0}$
$=6 \times 500$
$=3000\, Hz$