A closed organ pipe has length `L`. The air in it is vibrating in thirf overtone with maximum ampulitude a . The amplitude at distance`(L)/(7)` from closed of the pipe is

Options

A. `a`
B. `a//2`
C. `( a sqrt(3))/(2)`
D. zero

Answer

A

Explanation

Figure S7 . 25 shows variation of displacement of particle in a closed organ pipe for ` 3 rd` overtone.
For third overtone
`l = ( 7 lambda)/(4) or lambda = ( 4 l)/( 7) or (lambda)/(4) = (l)/(7)`
Hence the amplitude at `P` at a distance `l//7` from closed end is `'a'` because there is an antinode at that point.
Alternate : Because there is node at `x = 0` the displacement amplitude as function of `x` can be written as
`A = a sin kx = a sin ( 2 pi)/(lambda) x`
For third overtone
`l = ( 7 lambda)/(4) or lambda = ( 4 l)/( 7)`
`A = a sin ( 2pi 7 x)/( 4 l)`
At `x = (l)/(7) rArr A = a`