Question

Two speakers S₁ and S₂, driven by the same amplifier, are placed at y = 1.0 m and y = −1.0 m. The speakers vibrate in phase at 600 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 in air is 330 m s⁻¹. (a) At what angle θ will the intensity of sound drop to a minimum for the first time? (b) At what angle will he hear a maximum of sound intensity for the first time? (c) If he continues to walk along the line, how many more can he hear?

Answer 1

Given:
Frequency of source f = 600 Hz
Speed of sound in air v = 330 m/s
$v=f\lambda$
$\therefore \lambda =\frac{v}{f}$=$\frac{330}{600}=0.5\mathrm{mm}$
Let the man travel a distance of $\left(y\right)$ parallel to the y-axis and let ^{$\left(d\right)$} be the distance between the two speakers. The man is standing at a distance of ^{$\left(D\right)$} from the origin.
The path difference (x) between the two sound waves reaching the man is given by:
$x={\mathrm{S}}_2\mathrm{Q}-{\mathrm{S}}_1\mathrm{Q}=\frac{yd}{D}$
Angle made by man with the origin:
$\theta =\frac{y}{D}$
Given:
d = 2 m

(a) For minimum intensity:
The destructive interference of sound (minimum intensity) takes place if the path difference is an odd integral
multiple of half of the wavelength.

$$\begin{gathered} \therefore x=(2n+1)\left(\frac{\lambda }{2}\right) \\ \mathrm{For}\left(\mathrm{n}=0\right) \\ \therefore \frac{yd}{D}=\frac{\lambda }{2} \\ \left(\because \theta =\frac{y}{D}\right) \\ \therefore \theta d=\frac{\lambda }{2} \\ \therefore \theta =\frac{\lambda }{2d}=\frac{0.55}{4}=0.1375\mathrm{rad} \\ \Rightarrow \theta =0.1375\times (57.1)°=7.9° \end{gathered}$$


(b) For maximum intensity:
The constructive interference of sound (maximum intensity) takes place if the path difference is an integral multiple of the wavelength.
$$\begin{gathered} x=n\lambda \\ \mathrm{For}\left(n=1\right): \\ \Rightarrow \frac{yd}{D}=\lambda \\ \Rightarrow \theta =\frac{\lambda }{d} \\ \Rightarrow \theta =\frac{0.55}{2}=0.275\mathrm{rad} \\ \therefore \mathrm{\theta }=16° \end{gathered}$$
(c) The more number of maxima is given by the path difference:
$$\begin{gathered} \frac{yd}{D}=2\lambda ,3\lambda ,4\lambda ,..... \\ \Rightarrow \frac{y}{D}=\mathrm{\theta }=32°,64°,128° \end{gathered}$$

He will hear two more maxima at 32° and 64° because the maximum value of θ may be 90°.