Question

An open and closed organ pipe have same length. The ratio of pth mode of frequency of vibration of two pipes is

• A. $1$
• B. $p$
• C. $p(2p+1)$
• D. $\frac{2p}{(2p-1)}$

Answer 1

Answer: (D)

$\frac{2p}{(2p-1)}$

For closed pipe, $p^{th}$ mode frequency is:
${\nu }_{cp}=\frac{v}{4l}(2p-1)$

For open pipe, $p^{th}$ mode frequency is:
${\nu }_{op}=\frac{v}{2l}p$

Thus, $\frac{{\nu }_{op}}{{\nu }_{cp}}=\frac{2p}{2p-1}$