Standing waves

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Second overtone of a closed pipe is same as the second overtone of a open pipe. Then the ratio of first overtone of the closed pipe to the third overtone of the open pipe is

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Solution

$\begin{array}{l} 5 . \frac{v}{4 l_{C}} = 6 . \frac{v}{4 l_{O}} \Rightarrow \frac{l_{C}}{l_{O}} = \frac{5}{6} \\ ∴ \frac{F i r s t o v e r t o n e o f c l o s e d p i p e}{Third o v e r t o n e o f t h e open p i p e} = \frac{3 . \frac{v}{4 l_{c}}}{8 . \frac{v}{4 l_{0}}} = \frac{3}{8} . \frac{l_{0}}{l_{c}} = \frac{3}{8} . \frac{6}{5} = \frac{9}{20} \end{array}$

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Standing waves

Remember concepts with our Masterclasses.

Second overtone of a closed pipe is same as the second overtone of a open pipe. Then the ratio of first overtone of the closed pipe to the third overtone of the open pipe is

5.v4lC=6.v4lO⇒lClO=56∴First overtone of closed pipeThird overtone of the open pipe=3.v4lc8.v4l0=38.l0lc=38.65=920

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