A closed orgain pipe of length `l_(0)` is resonating in `5^(th)` harmonic mode with rod clamped at two points `l` and `3l` from one end. If the length of the rod is `4l` and it is vibrating in first overtone. Find the length of the rod. [Velocity of sound in air `=v_(s)` Young's modulus for the rod `Y` and density `rho`]

Answer

`(12)/(5) (l_(0))/( v_(s)) sqrt((y)/(rho))`

Explanation

For section `AB`
`l = ( 2 n - 1) (lambda)/(4) , where n = 1, 2, 3,…`
`f_(1) = (v)/( lambda) = (( 2 n -1 )v)/( 4 l)`
For section `BC`
` 2l = m (lambda)/(2) , where m = 1 , 2, 3 , …`
`f_(2) = (vm)/( 4l)`
from ` (3) + (4) f_(1) = f_(2)`
`(( 2n -1 ))/( 4 l) = ( mv)/( 4 l)`
` n = ( m + 1)/( 2)`
`m = 1 , n = 1` for fundamental , `m = 3 , n = 2`
first overtone ` f_(2) = ( mv)/( 4l) = ( 3v)/( 4l)`
Given that ` f = f_(2)`
` ( 5 v_(s))/( 4 l_(0)) = ( 3v)/( 4 l)`
`l = ( 3v l_(0))/( 5 v_(s))`
Therefore , length of the rod ` = 4 l = ( 12)/(5) (l_(0))/(v_(s)) sqrt((Y)/( rho))`