An open pie is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be heigth by `100 H_(Z)` than the fundamental frequency of the open pipe. The fundamental frequency of the open pipe is
(a) `200 H_(Z)` (b) `300 H_(Z)` (c ) `240 H_(Z)` (d) `480 H_(Z)`

Options

A. 200 Hz
B. 300 Hz
C. 240 Hz
D. 480 Hz

Answer

A

Explanation

`f_(1)lamda_(1)=v`
`f_(1)((4L)/(3))=vrArrf_(1)=(3v)/(4L),f_(2)(2L)=vrArrf_(2)=(v)/(2L)`
Given that `f_(1)-f_(2)=100`
`rArr (3v)/(2L)-(v)/(2L)=100 rArr(v)/(2L)=200Hz`