Two loudspeakers ($L_1$ and $L_2$) are placed with a separation of $10 \, \text{m}$, as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. A voice recorder, initially at point $A$, at equidistance to both loudspeakers, is moved by $25 \, \text{m}$ along the line $AB$ 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 $324 \, \text{m/s}$ and $ \sqrt{5} = 2.23 $)

Question

Two loudspeakers ($L_1$ and $L_2$) are placed with a separation of $10 \, \text{m}$, as shown in the figure. Both speakers are fed with an audio input signal of the same frequency with constant volume. A voice recorder, initially at point $A$, at equidistance to both loudspeakers, is moved by $25 \, \text{m}$ along the line $AB$ 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 $324 \, \text{m/s}$ and $ \sqrt{5} = 2.23 $)

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At point $A$, the recorder is equidistant from both loudspeakers, so the path difference is zero. As the recorder moves from $A$ to $B$, the path difference between the two sound waves changes, producing alternating maxima and minima due to interference. Step 1: Determine the change in path difference. Horizontal distance of point $A$ from the midpoint of the speakers is $40 \, \text{m}$. Vertical displacement from $A$ to $B$ is $25 \, \text{m}$. Distance of $B$ from the upper speaker $L_1$: $$ BL_1 = \sqrt{40^2 + (25 - 5)^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5} $$ Distance of $B$ from the lower speaker $L_2$: $$ BL_2 = \sqrt{40^2 + (25 + 5)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 $$ Hence, path difference at point $B$ is: $$ \Delta = BL_2 - BL_1 = 50 - 20\sqrt{5} $$ Using $ \sqrt{5} = 2.23 $: $$ \Delta = 50 - (20 \times 2.23) = 50 - 44.6 = 5.4 \, \text{m} $$ Step 2: Relate path difference to number of cycles. One complete cycle of maxima and minima corresponds to a path difference change of one wavelength $ \lambda $. Given number of cycles $ = 10 $: $$ 10\lambda = 5.4 $$ $$ \lambda = 0.54 \, \text{m} $$ Step 3: Calculate frequency. Using $ v = f\lambda $: $$ f = \frac{v}{\lambda} = \frac{324}{0.5} $$ $$ f = 648 \, \text{Hz} $$ Final Answer: $$ \boxed{648} $$

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Correct Answer: 648

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