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

For both ends open, fundamental frequency
`(2 lamda_(1))/(4) =l rArr lamda_(1)=2l`
`:. v_(1) =(c)/(lamda_(1))=(c)/(2l)`

For one end closed the third harmonic
`(3 lamda_(2))/(4)=l rArr lamda_(2) =(4l)/(3)`
`v_(2)=(c)/(lamda_(2)) = (3c)/( 4l)`
Given `v_(2) - v_(1) = 100`
From Eqs. (i) and (ii)
`(v_(2))/(v_(1)) = (3//4)/(1//2) = (3)/(2)`
On solving, we get `v_(1) = 200 Hz`.