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 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

Answer 1

Answer: (C)

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