Physics · Sound Waves

JEE Main 2026 — 22 January, Morning Shift — Question 44

Two loudspeakers ( $L_{1}$ and $L_{2}$ ) 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 $\sqrt{5}=2.23$ )

Answer: 600

Numerical answer — enter this value.

Step-by-step solution

Point B will $10{ }^{\text {th }}$ maxima $\Delta \mathrm{x}=\mathrm{L}_{2} \mathrm{~B}-\mathrm{L}_{1} \mathrm{~B}$ $\mathrm{L}_{1} \mathrm{~B}=\sqrt{20^{2}+40^{2}}=20 \sqrt{5} \mathrm{~m}=44.6 \mathrm{~m}$ $\mathrm{L}_{2} \mathrm{~B}=\sqrt{40^{2}+30^{2}}=50 \mathrm{~m}$ $\Delta \mathrm{x}=50-44.6=5.4 \mathrm{~m}$ $\Delta \mathrm{x}=\mathrm{n} \lambda$ $5.4=10 \times \lambda$ $\lambda=0.54 \mathrm{~m}$ $\mathrm{V}=\mathrm{f} \lambda$ $\mathrm{f}=\frac{324}{0.54}=600 \mathrm{~Hz}$

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Sound Waves
Topic: Interference of Sound Waves