The displacement equations of… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

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The displacement equations of two interfering waves are given by ${y}_{1}=10\mathrm{sin}(\omega t+\frac{\pi }{3})\mathrm{cm}$ , ${y}_{2}=5[\mathrm{sin}(\omega t)+\sqrt{3}\mathrm{cos}\omega t]\mathrm{cm}$ respectively. The amplitude of the resultant wave is _____ $\mathrm{cm}$ .

Official previous-year question

Held on 31 Jan 2023 · Verified 6 Jul 2026.

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Solution

Displacement equation of second wave can be written as, ${y}_{2}=5(\mathrm{sin}\omega t+\sqrt{3}\mathrm{cos}\omega t) =10(\mathrm{sin}\omega t\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \mathrm{cos}\omega t) =10(\mathrm{sin}\omega t\times \mathrm{cos}\frac{\pi }{3}+\mathrm{sin}\frac{\pi }{3}\times \mathrm{cos}\omega t)$

$=10\mathrm{sin}(\omega t+\frac{\pi }{3})$

As we can see, the phase difference between the waves is $0$ . Therefore, resultant amplitude will be sum of the amplitudes of the waves.

So $A={A}_{1}+{A}_{2}=10\mathrm{cm}+10\mathrm{cm}=20\mathrm{cm}$

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