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.
The air column here is closed at one end because the surface of water acts as a wall. Which of the following is correct ?

Options

A. At the closed end of air column , there is a displacement node and also a pressure node
B. At the closed end of the air column , there is a displacement node and also a pressure antinode
C. At the closed end of the air column , there is a displacement antinode and a pressure node
D. At the closed end of the air column , there is a displacement antinode and also a pressure antinode

Answer

B

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`