Two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA = 1.5m and one open end..

Q: Two pipes are submerged in sea water, arranged as shown in figure. Pipe A with length LA = 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:

Numerical

(a) 1 m

(b) 1.5 m

(c) 2 m

(d) 3 m

Ans: (c)

Sol: $\large 3 \frac{v}{4 L_A} = 2 \frac{v}{2 L_B}$

$\large 3 \frac{1}{4 L_A} = \frac{1}{ L_B}$

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

$\large L_B = \frac{4 \times 1.5}{3} = 2 m$