JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(\omega\))
- A \(\left[6, \frac{2 \pi}{3}\right]\)
- B \(\left[6, \frac{\pi}{3}\right]\)
- C \(\left[\sqrt{3}, \frac{\pi}{6}\right]\)
- D \(\left[2 \sqrt{3}, \frac{\pi}{6}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[2 \sqrt{3}, \frac{\pi}{6}\right]\)
Step-by-step Solution
Detailed explanation
\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}
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