Q72 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
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 $\sqrt{5}=2.23$)
Correct answer: 600
Solution
Point B will $10^{\text {th }}$ maxima
$\Delta x=L_2 B-L_1 B$
$L_1 B=\sqrt{20^2+40^2}=20 \sqrt{5} \mathrm{~m}=44.6 \mathrm{~m}$
$L_2 B=\sqrt{40^2+30^2}=50 \mathrm{~m}$
$\Delta x=50-44.6=5.4 \mathrm{~m}$
$\Delta x=n \lambda$
$5.4=10 \times \lambda$
$\lambda=0.54 \mathrm{~m}$
$V=f \lambda$
$f=\frac{324}{0.54}=600 \mathrm{~Hz}$
$\Delta x=L_2 B-L_1 B$
$L_1 B=\sqrt{20^2+40^2}=20 \sqrt{5} \mathrm{~m}=44.6 \mathrm{~m}$
$L_2 B=\sqrt{40^2+30^2}=50 \mathrm{~m}$
$\Delta x=50-44.6=5.4 \mathrm{~m}$
$\Delta x=n \lambda$
$5.4=10 \times \lambda$
$\lambda=0.54 \mathrm{~m}$
$V=f \lambda$
$f=\frac{324}{0.54}=600 \mathrm{~Hz}$
Hence, the correct answer is 600.