The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y 1 (x, t)=4 sin (k x-ω t) and y 2 (x, t)=2 sin (k x-ω t+2 π/3 ) , are : (Take the angular frequency of initial waves same as ω )

Physics · Transverse waves

JEE Main 2025 — 8 April, Evening Shift — Question 53

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_{1}$ $(x, t)=4 \sin (k x-\omega t)$ and $y_{2}(x, t)=2$ $\sin \left(k x-\omega t+\frac{2 \pi}{3}\right)$ , are : (Take the angular frequency of initial waves same as $\omega$ )

Option A:
$[2 \sqrt{3}, \pi / 6]$
Correct
Option B:
$[6,2 \pi / 3]$
Option C:
$[6, \pi / 3]$
Option D:
$[\sqrt{3}, \pi / 6]$

Answer: A

Step-by-step solution

$A=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi}$

$\phi=\frac{2 \pi}{3}$

$A=\sqrt{4^{2}+2^{2}+2 \times 4 \times 2 \cos \frac{2 \pi}{3}}=2 \sqrt{3}$

$\tan \alpha=\frac{2 \sin \phi}{4+2 \cos \phi}=\frac{1}{\sqrt{3}}$

$\alpha=\pi / 6$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Transverse waves
Topic: super position, Reflection and Transmission of waves

Two strings with circular cross section and made of same material are stretched to have same amount of tension. A transverse wave is then…

Answer: A

Step-by-step solution

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