A vertical pipe open at both ends is partially submerged in water . A tuning fork is unknown frequency is placed near the top of the pipe and made to vibrate . The pipe can be moved up and down and thus length of air column the pipe can be adjusted. For definite lengths of air column in the pipe, standing waves will be set up as a result of superposition of sound waves travelling in opposite directions. Smallest value of length of air column , for which sound intensity is maximum is `10 cm`[ take speed of sound , `v = 344 m//s`].
Answer the following questions.
Length of air column for third resonance will be

Options

A. `30 cm`
B. `45 cm`
C. `20 cm`
D. `50 cm`

Answer

D

Explanation

Displacement node corresponds to pressure antinode .
`L = (lambda_(0))/(4)`
`lambda_(0) = 4 L = 40 cm` ( First harmonic)
`v = lambda f`
`f = (v)/(lambda) = ( 344)/( 40 xx 10^(-2)) = 860 Hz`
For the second resonance
`( 3lambda_(0))/(4) = L`
`L = 30 cm`
For the third resonance `5 lambda_(0)//4 = L`
`L = 50 cm`
Also , ` v = v//lambda`
`3 rd` harmonic is `2 nd` overtone .
Hence , frequency for `2 nd` overtone
` = ( 5 v)/( 4 L) = 4300 Hz`