The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y_1(x, t)=4 (kx-ω t) and y_2(x, t)=2 (kx-ω t+(2 π)/(3)) , are : (Take the angular frequency of initial waves same

Correct answer: D. [2 3 , (π)/(6)] aligned & array l A= 2^2+4^2+2 × 2 × 4 × 120^ = 12 =2 3 φ= 2 120^ 4+2 120^ = 3 3 = 1 3 φ=(π)/(6) array aligned

JEE Main 20258 AprilEvening Shift

JEE Main 2025 · 8 April · Evening Shift · Physics · Oscillations

The amplitude and phase of a wave that is formed by the...

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,

\mathrm{y}_1(\mathrm{x}, \mathrm{t})=4 \sin (\mathrm{kx}-\omega \mathrm{t})

and

\mathrm{y}_2(\mathrm{x}, \mathrm{t})=2 \sin \left(\mathrm{kx}-\omega \mathrm{t}+\frac{2 \pi}{3}\right)

, are

:

(Take the angular frequency of initial waves same as

\omega

)

D. $\left[2 \sqrt{3}, \frac{\pi}{6}\right]$

\begin{aligned} & \begin{array}{l}\mathrm{A}=\sqrt{2^2+4^2+2 \times 2 \times 4 \times \cos 120^{\circ}} \\ \quad=\sqrt{12}=2 \sqrt{3} \\ \tan \phi=\frac{2 \sin 120^{\circ}}{4+2 \cos 120^{\circ}}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \\ \phi=\frac{\pi}{6}\end{array}\end{aligned}