A pipe is closed at one end by a membrane which may be considered a seat of displacement node and is set to sonic oscillations of frequency `2000 Hz`. Find the velocity of sound if on moving the piston , resonance occurs at the interval of `8.5 cm`.

Explanation

The two ends of the pipe are the seats of nodes just as in a stretched string . Hence , the air inside emits the full series of harmonics . Resonance occurs when the frequency of the membrane matches with one of the harmonics of the air column in the pipe . Let the `n th` thrmonic of the column of length `I` ( here `I` is also the distance of the piston from the membrane ) resonate with the tuning fork . Then
`n xx ( c)/( 2I) = v`
where ` v =` frequency of the tuning fork
`nc = 2//v`
Let the air column resonate again when the piston is moved by `x` away from the membrane , that is , the distance of the piston is `(I + x)` . Now the membrane resonates with `( n + 1) th` harmonic of the new column of air .
Therefore , `( n + 1) ( c )/( 2( I + x)) = v`
or ` ( n + 1) c = 2 v (I + x)`
Subtracting ` c = 2 v x = 2 xx 2000 xx 8.5 xx 10^(-2)`
`= 340 m//s`