Two harmonic waves moving in the same direction superimpose to form a wave x= a cos (1.5 t ) cos (50.5t) where t is in seconds. Find the period with which they beat. (close to nearest integer)

Physics · Sound Waves

JEE Main 2025 — 7 April, Morning Shift — Question 62

Two harmonic waves moving in the same direction superimpose to form a wave $x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos$ (50.5t) where $t$ is in seconds. Find the period with which they beat. (close to nearest integer)

Option A:
1 s
Option B:
6 s
Option C:
4 s
Option D:
2 s
Correct

Answer: D

Step-by-step solution

$x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})$

Clearly, $\begin{aligned} x= & \frac{a}{2} \cos (50.5 t+1.5 t)+\frac{a}{2} \cos (50.5 t-1.5 t) \Rightarrow \\ & x=\frac{a}{2} \cos (52 t)+\frac{a}{2} \cos (49 t) \\ & f_{1}=\frac{52}{2 \pi} \text { and } f_{2}=\frac{49}{2 \pi} \Rightarrow \\ & \Delta f=\frac{3}{2 \pi} \\ & \Delta T=\frac{1}{\Delta f}=\frac{2 \pi}{3}=2.09 \Rightarrow \\ & \Delta T \approx 2 \mathrm{~s} \end{aligned}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Sound Waves
Topic: Interference in Time - Beats and Beat Frequency