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The two pipes are submerged in sea water...


The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length `L_A=1.5m` and one open end, contains a small sound source that sets up the standing wave with the second lowest resonant frequency of that pipe. Sound from pipe A sets up resonance in pipe B, which has both ends open. The resonance is at the second lowest resonant frequency of pipe B. The length of the pipe B is

A

1m

B

1.5 m

C

2 m

D

3 m

Text Solution

Verified by Experts

The correct Answer is:
C
For pipe A, second resonant frequency is third harmonic thus `f=(3V)/(4L_A)`
For pipe B, second resonant frequency is second harmonic thus `f=(2V)/(2L_B)`
Equating `(3V)/(4L_A)=(2V)/(2L_B)`
`impliesL_B=(4)/(3)L_A=(4)/(3)xx(1.5)=2m`
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