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JEE Mains · Physics · STD 11 - 14. waves and sound

दो प्रगामी सरल आवर्त तरंगों, \(\mathrm{y}_1(\mathrm{x}, \mathrm{t})=4 \sin (\mathrm{kx}-\omega \mathrm{t})\) तथा \(\mathrm{y}_2(\mathrm{x}, \mathrm{t})=2 \sin \left(\mathrm{kx}-\omega \mathrm{t}+\frac{2 \pi}{3}\right)\) के अध्यारोपण द्वारा निर्मित तरंग का आयाम और कला (फेज़) \(:\)
(प्रारंभिक तरंगों की कोणीय आवृत्ति \(\omega\) के समान मानें)

  1. A \(\left[6, \frac{2 \pi}{3}\right]\)
  2. B \(\left[6, \frac{\pi}{3}\right]\)
  3. C \(\left[\sqrt{3}, \frac{\pi}{6}\right]\)
  4. D \(\left[2 \sqrt{3}, \frac{\pi}{6}\right]\)
Verified Solution

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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