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. $1.5m$
• B. $3m$
• C. $2m$
• D. $1m$

Answer 1

The correct option is C $2m$

Step 1: Draw a model diagram of the given problem.


Step 2: Find the length of the pipe B.

Let consider $A$ and $B$ are two pipes,
The length of the pipe $A\text{i}s{L}_A=1.5m$ and length of the pipe $B\text{i}s{L}_B=?$

For pipe $A$, second resonant frequency is third harmonic,
$\Rightarrow f_1=\frac{3v}{4L_A}$

For pipe $B$, second resonant frequency is second harmonic,
$\Rightarrow f_2=\frac{2v}{2L_B}$

Equating,
$\Rightarrow f_1=f_2$
$\Rightarrow \frac{3v}{4L_A}=\frac{2v}{2L_B}$
$\Rightarrow L_B=\frac{4}{3}{L}_A$
$\Rightarrow L_B=\frac{4}{3}\left(1.5m\right)$
$L_B=2m$

Final Answer: (𝒄)