Question

An oscillator of frequency $680\text{Hz}$ drives two speakers. The speakers are fixed on a vertical pole at a distance $3\text{m}$ from each other as shown in figure. A person whose height is almost the same as that of the lower speaker walks towards the lower speaker in a direction perpendicular to the pole. Assuming that there is no reflection of sound from the ground and sound speed is $v=340\text{m/s}$, choose the correct option(s):

• A. The person hears the first minimum sound intensity at $18\text{m}$ away from the pole.
• B. The person hears the second minimum intensity at a distance of $5.6\text{m}$ from the speaker.
• C. The person hears a minimum sound intensity for a total of $6$ times.
• D. The person hears a maximum sound intensity for a total of $8$ times.

Answer 1

Answer: (A), (B), (C)

The person hears a minimum sound intensity for a total of $6$ times.

Frequency of source = $680\text{Hz}$
and $v=340\text{m/s}$
From formula
wavelength $\left(\lambda \right)=\frac{v}{f}=\frac{340}{680}=0.5\text{m}$
Let the person be at distance $d$ when the observed first minimum intensity.
Hence, path difference $\left(\Delta x\right)=\frac{\lambda }{2}$
$\Rightarrow S_{2}A-S_{1}A=\frac{\lambda }{2}$
$\Rightarrow \sqrt{d^2+9}-d=\frac{\lambda }{2}=\frac{1}{4}$
$\Rightarrow \sqrt{d^2+9}=d+\frac{1}{4}$
$\Rightarrow d=17.87\approx 17.9\text{m}$


For second time,
Destructive interference occurs at $\frac{3}{2}\lambda$
So
$\Delta x=\frac{3}{2}\lambda$
$\Rightarrow \sqrt{d^2+9}-d=\frac{3}{2}\times \frac{1}{2}$
$\Rightarrow \sqrt{d^2+9}=\frac{3}{4}+d$
$\Rightarrow d=5.6\text{m}$

For maximum (i.e constructive interference)
path difference $=n\lambda$
If final maximum is obtained at foot of the pole,
Distance between the sources $=3=n\lambda$
$\Rightarrow 3=n\left(\frac{1}{2}\right)$
$\Rightarrow n=6$
Therefore, total $6$ maximum and $6$ minimum intensity will be formed.
Option (a), (b), (c) are correct.