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

Correct answer: D. 2 s The given equation can be written as x= a 2 [1.5+50.5] t+ a 2 [50.5-1.5] x=(a)/(2) [52 t]+(a)/(2) [49 t] Here, 2 π f_1 \& 2 π f_2=49 f_1=(52)/(2 π), f_2=(49)/(2 π) f_ Bat =f_1-f_2=(3)/(2 π) ~Hz T_ Beat = 1 f_ Beat =(2 π)/(3) sec =2.09 sec 2 sec

JEE Main 20257 AprilMorning Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Morning Shift
Subject: Physics
Chapter: Waves and Sound

JEE Main 2025 · 7 April · Morning Shift · Physics · Waves and Sound

Two harmonic waves moving in the same direction superimpose to form...

Two harmonic waves moving in the same direction superimpose to form a wave

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

where

t

is in seconds. Find the period with which they beat (close to nearest integer)

Options

A6 s
B4 s
C1 s
D2 s

Correct answer

D. 2 s

Step-by-step solution

The given equation can be written as

\mathrm{x}=\frac{\mathrm{a}}{2} \cos [1.5+50.5] \mathrm{t}+\frac{\mathrm{a}}{2} \cos [50.5-1.5]

x=\frac{a}{2} \cos [52 t]+\frac{a}{2} \cos [49 t]

Here,

2 \pi \mathrm{f}_1 \& 2 \pi \mathrm{f}_2=49

\mathrm{f}_1=\frac{52}{2 \pi}, \mathrm{f}_2=\frac{49}{2 \pi}

\therefore \mathrm{f}_{\text{Bat }}=\mathrm{f}_1-\mathrm{f}_2=\frac{3}{2 \pi} \mathrm{~Hz}

\therefore \mathrm{T}_{\text{Beat }}=\frac{1}{\mathrm{f}_{\text{Beat }}}=\frac{2 \pi}{3} \mathrm{sec}

=2.09 \mathrm{sec} \approx 2 \mathrm{sec}

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