Waves

JEE Main level, single correct, medium

115 of 142

Question 115 of 142

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

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Answer: (D) 2 s

The given equation can be written as
x=a2cos⁡[1.5+50.5]t+a2cos⁡[50.5−1.5]\mathrm{x}=\frac{\mathrm{a}}{2} \cos [1.5+50.5] \mathrm{t}+\frac{\mathrm{a}}{2} \cos [50.5-1.5]
x=a2cos⁡[52t]+a2cos⁡[49t]x=\frac{a}{2} \cos [52 t]+\frac{a}{2} \cos [49 t]
Here, 2πf1&2πf2=492 \pi \mathrm{f}_1 \& 2 \pi \mathrm{f}_2=49
f1=522π,f2=492π\mathrm{f}_1=\frac{52}{2 \pi}, \mathrm{f}_2=\frac{49}{2 \pi}
∴fBat =f1−f2=32π Hz\therefore \mathrm{f}_{\text {Bat }}=\mathrm{f}_1-\mathrm{f}_2=\frac{3}{2 \pi} \mathrm{~Hz}
∴TBeat =1fBeat =2π3sec\therefore \mathrm{T}_{\text {Beat }}=\frac{1}{\mathrm{f}_{\text {Beat }}}=\frac{2 \pi}{3} \mathrm{sec}
=2.09sec≈2sec=2.09 \mathrm{sec} \approx 2 \mathrm{sec}

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