An open and a closed pipe have same length . The ratio of frequency of their `n th` overtone is

Options

A. `( n + 1)/( 2n + 1)`
B. ` 2 ( n + 1)/( 2n + 1)`
C. `( n)/( 2n + 1)`
D. `( n + 1)/( 2n)`

Answer

B

Explanation

Let `l` be the length of the pipe and `v` the speed of sound . Then frequency of open organ pipe of `n th` overtone is
`f_(1) = ( n + 1) (v)/( 2 l)`
and frequency of closed organ pipe of `n th` overtone
`f_(2) = ( 2 n + 1) (v)/( 4 l)`
Therefore , the desired ratio is
`(f_(1))/(f_(2)) = ( 2 (n + 1))/(( 2 n + 1))`