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

Answer

A

Explanation

For both ends open, fundamental frequency
`(2 lambda_1)/4=1implies lambda_1= 21`
`therefore v_1=(c)/(lamda_1)=(c)/(2 1)`…….(i)
Fond one end closed the third harmonic
`(3lambda_2)/4=1implies lambda_2=(4 1)/3`
`v_2=(c)/(lambda_2)=(3c)/(4 1)`
Given `v_2-v_1=100`
From Eqs. (i) and (ii) `(v_2)/(v_1)=(3//4)/(1//2)=3/2`
On solving , ew get `v_1=200 Hz`