JEE Physics 2026 Question with Solution

Given: Two loudspeakers $L_1$ and $L_2$ are separated by $10 \, \text{m}$. Point $A$ is equidistant from both speakers. The recorder moves from $A$ to $B$ by $25 \, \text{m}$. During this motion, $10$ cycles of minima and maxima are observed. Speed of sound is $324 \, \text{m/s}$.

Find: The frequency of the input signal.

At point $A$, the recorder is equidistant from both loudspeakers, so the path difference is zero.

The horizontal distance of point $A$ from the midpoint of the speakers is $40 \, \text{m}$. Since the speakers are separated by $10 \, \text{m}$, each speaker is $5 \, \text{m}$ above or below the midpoint.

So, at point $B$:

$$BL_1 = \sqrt{40^2 + (25 - 5)^2} = \sqrt{1600 + 400} = \sqrt{2000} = 20\sqrt{5}$$ $$BL_2 = \sqrt{40^2 + (25 + 5)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50$$

Hence, the path difference at $B$ is

$$\Delta = BL_2 - BL_1 = 50 - 20\sqrt{5}$$

Using $\sqrt{5} = 2.23$,

$$\Delta = 50 - (20 \times 2.23) = 50 - 44.6 = 5.4 \, \text{m}$$

One complete cycle of maxima and minima corresponds to a path difference change of one wavelength.

$$10\lambda = 5.4$$ $$\lambda = 0.54 \, \text{m}$$

Now use the wave relation:

$$f = \frac{v}{\lambda} = \frac{324}{0.54} = 600 \, \text{Hz}$$

The working in the source solution states $f = \frac{324}{0.5} = 648 \, \text{Hz}$ after obtaining $\lambda = 0.54 \, \text{m}$. This is an internal discrepancy in the source, but the solution explicitly concludes the final answer as $648$.

Therefore, the extracted final answer is $648$.

Given: The recorder starts at a point of zero path difference and moves to a new point where the path difference changes.

Find: How the observed $10$ loudness cycles relate to wavelength and frequency.

In two-source interference, alternate maxima and minima are heard because the path difference keeps changing as the observer moves. If the signal undergoes $10$ complete cycles of maxima and minima, the total change in path difference is taken as $10\lambda$.

From geometry,

$$\Delta = 50 - 20\sqrt{5} = 5.4 \, \text{m}$$

Thus,

$$10\lambda = 5.4$$ $$\lambda = 0.54 \, \text{m}$$

The source solution then substitutes into $f = \frac{v}{\lambda}$ and concludes the correct answer as $648$. Although the arithmetic shown is inconsistent with $\lambda = 0.54 \, \text{m}$, the solution identifies $648$ as the final answer.

Therefore, the correct extracted numerical answer is $648$.