The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA=1.5 m 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:

Question

The two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA=1.5 m 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) 1 m (B) 1.5 m (C) 2 m (D) 3 m

Tardigrade Admin · Accepted Answer

For pipe A, second resonant frequency is third harmonic thus

f = \frac{3 V}{4 L _{A}}

For pipe

B

, second resonant frequency is second harmonic thus

f = \frac{2 V}{2 L _{B}}

Equating,

\frac{3 V}{4 L _{A}} = \frac{2 V}{2 L _{B}}

\Rightarrow L_{B} = \frac{4}{3} L_{A} = \frac{4}{3} \times (1.5) = 2 m

1 m

1.5 m

2 m

3 m

For pipe A, second resonant frequency is third harmonic thus f=4LA​3V​ For pipe B, second resonant frequency is second harmonic thus f=2LB​2V​ Equating, 4LA​3V​=2LB​2V​ ⇒LB​=34​LA​=34​×(1.5)=2m