Question

The two pipes are submerged in sea water, arranged as shown in figure. Pipe A, with length L_A = 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 the pipe A sets up resonance in pipe B, which has two open ends. The resonance is at the second lowest resonant frequency of the pipe B. The length of the pipe B is:

• A. 1m
• B. 1.5m
• C. 2 m
• D. 3 m

Answer 1

Answer: (C)

2 m

Answer 2

For pipe A, second resonant frequency is third harmonic thus f = $\frac{3V}{4L_{A}}$

For pipe B, second resonant frequency is second harmonic thus f = $\frac{2V}{2L_{B}}$

Equating, $\frac{3V}{4L_{A}}$= $\frac{2V}{2L_{B}}$

L_B = $\frac{4}{3}$L_A

= $\frac{4}{3}$.(1.5) = 2m.