A tube closed at one end and containing air produced, when excited the fundamental note of frequency `512 H_(Z)`. If the tube is opened at both ends, the fundamental frequency that can be exited is (in `H_(Z)`)
(a) `1024` (b) `512` (c ) `256` (d) `128`

Options

A. 1024
B. 512
C. 256
D. 128

Answer

A

Explanation

`lamda//4 = l` (Fundamental mode), `lamda - 4l, c = v lamda`

`:. v = ( c)/(lamda) = ( c)/(4 l) = 512 Hz`
Given, `lamda'//2 = l`
Fundamental mode,
`:. lamda' = 2 l` but `c = v' lamda'`
`:. v' = (c)/(lamda') =(c)/(2l) = 2((c)/(4l))`
=`2 xx 512 = 1024 Hz`.