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A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio
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Verified Answer
The correct answer is:
3:4
If is given that
First over tone of closed pipe $=$ First over tone of open pipe $\quad 3\left(\frac{v}{4 l_1}\right)=2\left(\frac{v}{2 l_2}\right) ;$ where $l_1$ and $l_2$ are the lengths of closed and open organ pipes hence $\frac{I_1}{I_2}=\frac{3}{4}$
First over tone of closed pipe $=$ First over tone of open pipe $\quad 3\left(\frac{v}{4 l_1}\right)=2\left(\frac{v}{2 l_2}\right) ;$ where $l_1$ and $l_2$ are the lengths of closed and open organ pipes hence $\frac{I_1}{I_2}=\frac{3}{4}$
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