Question

Two loudspeakers $L_1$ and $L_2$ driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at $D$ records a series of maxima and minima. If the speed of sound is $330m{s}^{-1}$ then the frequency at which the first maximum is observed is :

• A. $165Hz$
• B. $330Hz$
• C. $495Hz$
• D. $660Hz$

Answer 1

Answer: (B)

$330Hz$

Path difference $\Delta x=\sqrt{9^2+{40}^2}-40=1$ m
Hence phase change = $\delta =\frac{2\pi }{\lambda }\Delta x=\frac{2\pi }{\lambda }$
For the first maximum $\delta =\frac{2\pi }{\lambda }=2\pi$
Hence $\lambda =1\text{m}$
Frequency$=\nu =\frac{v}{\lambda }=330\text{Hz}$
Ans: B