Question

An open and closed organ pipe have the same length. The ratio of $p^{th}$ mode of frequency of vibration of air in two pipes is

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

Answer 1

The correct option is B $\frac{2p}{2p-1}$

Let the velocity of sound wave and the length of the organ pipes be $v$ and $L$ respectively.

Frequency of $pth$ mode of vibration of air in an open organ pipe is given as $\nu =\frac{pv}{2L}$

where, $p=1,2,3,.......$

Frequency of $p^{th}$ mode of vibration of air in a closed organ pipe is given as ${\nu }^{\prime }=\frac{(2p-1)v}{4L}$

where, $p=1,2,3,.......$

$\therefore$ Ratio of the frequencies, $\frac{\nu }{{\nu }^{\prime }}=\frac{\frac{pv}{2L}}{\frac{(2p-1)v}{4L}}$

$⟹$$\frac{\nu }{{\nu }^{\prime }}=\frac{2p}{2p-1}$