Question 78: Waves

Question

Two loudspeakers $\left(L_1\right.$ and $\left.L_2\right)$ are placed with a separation of 10 m , as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point $A$, at equidistance to both loud speakers, is moved by 25 m along the line $A B$ while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is $\_\_\_\_$ Hz (Speed of sound in air is $324 \mathrm{~m} / \mathrm{s}$ and $\sqrt{5}=2.23$ )

Accepted Answer

When sound waves from two coherent sources (like the two loudspeakers) meet, they interfere with each other.
Constructive Interference (Maxima) occurs when the waves are in phase.
The path difference, $\Delta \mathrm{x}$ is an integer multiple of the wavelength ( $\lambda$ ).
$$ \Delta \mathrm{x}=\mathrm{n} \lambda \ldots(	ext { where } \mathrm{n}=0,1,2, \ldots) $$
Destructive Interference (Minima) occurs when the waves are out of phase. This happens when the path difference is a half-integer multiple of the wavelength.
$$ \Delta \mathrm{x}=\left(\mathrm{n}+\frac{1}{2}ight) \lambda $$
It is given that A is equidistant from both speakers $\left(\mathrm{L}_{1 \mathrm{~A}}=\mathrm{L}_{2 \mathrm{~A}}ight)$. Therefore, the path difference at A is zero $\left(\Delta \mathrm{x}_{\mathrm{A}}=0ight)$. This corresponds to the central maximum ( $\mathrm{n}=0$ ).
As the recorder moves towards $B$, the distance to $L_2$ increases more than the distance to $L_1$, so the path difference ( $\Delta \mathrm{x}$ ) steadily increases.
The signal undergoes 10 cycles of minima and maxima. One complete cycle means moving from one maximum to the next consecutive maximum, which corresponds to an increase in path difference by exactly one wavelength ( $\lambda$ ).
Since it goes through 10 full cycles starting from the central maximum, the path difference at point B must be exactly 10 wavelengths.
$$ \Delta \mathrm{x}_{\mathrm{B}}=10 \lambda $$
The path length from $L_1$ to $B$ is,
$$ \mathrm{x}_{\mathrm{L} 1}=\sqrt{40^2+(25-5)^2} $$
$$\Rightarrow $$ $\mathrm{x}_{\mathrm{L} 1}=\sqrt{40^2+20^2}=\sqrt{2000}$
$$\Rightarrow $$ $$ \mathrm{x}_{\mathrm{L} 1}=20 \sqrt{5} \mathrm{~m} $$
The path length from $L_2$ to $B$ is,
$$ \mathrm{x}_{\mathrm{L} 2}=\sqrt{(40)^2+(25+5)^2}=\sqrt{1600+900} $$
$$\Rightarrow $$ $$ \mathrm{x}_{\mathrm{L} 2}=\sqrt{2500}=50 \mathrm{~m} $$
So, the path difference is,
$$ \Delta \mathrm{x}_{\mathrm{B}}=\mathrm{x}_{\mathrm{L} 2}-\mathrm{x}_{\mathrm{L} 1} $$
$$\Rightarrow $$ $$ \Delta \mathrm{x}_{\mathrm{B}}=50-20 \sqrt{5} $$
We are given $\sqrt{5}=2.23$.
$$ \Delta \mathrm{x}_{\mathrm{B}}=50-20(2.23) $$
$$\Rightarrow $$ $\Delta \mathrm{x}_{\mathrm{B}}=50-44.6$
$$\Rightarrow $$ $\Delta \mathrm{x}_{\mathrm{B}}=5.4 \mathrm{~m}$
$\Rightarrow 10 \lambda=\Delta \mathrm{x}_{\mathrm{B}}$
$$ \Rightarrow \lambda=0.54 \mathrm{~m} $$
Using the relation between wave speed (v), frequency (f), and wavelength ($\lambda$) :
$$ \mathrm{v}=\mathrm{f} \cdot \lambda $$
We are given the speed of sound $\mathrm{v}=324 \mathrm{~m} / \mathrm{s}$.
$$ \mathrm{f}=\frac{\mathrm{v}}{\lambda}=\frac{324}{0.54}=600 \mathrm{~Hz} $$
Therefore, the frequency of the input signal is 600.

Waves

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