Question

Two speakers $S_1\text{and}{S}_2$ are derived from the same amplifier and placed at the same distance from $x$-axis as shown in figure. The speakers vibrate in phase at $680\text{Hz}$. A man stands at a point on the $x$-axis at a very large distance from the origin and starts moving parallel to the $y$-axis. The speed of sound is $340\text{m/s}$. Then, at what angle $\theta$ with the $x-$ axis will the intensity of sound drop to a minimum for the first time?

• A. ${5.4}^{\circ }$
• B. ${9.3}^{\circ }$
• C. ${7.2}^{\circ }$
• D. ${3.6}^{\circ }$

Answer 1

The correct option is D ${3.6}^{\circ }$

The man is at a large distance from the speaker. Let the intensity of sound fall to a minimum when the man reaches $Q$. The path difference between waves starting from $S_1$ and $S_2$, and reaching $Q$ is.
Path difference $\left(\Delta x\right)=S_{2}Q-S_{1}Q$


From figure,
Path difference $\left(\Delta x\right)=S_{2}T$
In $\Delta S_1{S}_{2}T$,
$\sin \theta =\frac{S_{2}T}{S_1{S}_2}$
$\Rightarrow S_{2}T=S_1{S}_2\sin \theta =4\sin \theta \approx 4\theta$
(Because $\theta$ is very small as man is far away)
So,
$\Delta x=4\theta \left(1\right)$
and we know.
$\lambda =\frac{v}{f}=\frac{340}{680}=\frac{1}{2}\text{m}$
For first minimum intensity of sound where destructive interference occurs,
$\Delta x=\frac{\lambda }{2}=\frac{1}{4}$
$\Rightarrow 4\theta =\frac{1}{4}$
$\Rightarrow \theta =\frac{1}{16}=0.0625\text{rad}={3.58}^{\circ }\approx {3.6}^{\circ }$