Question

The resultant amplitude of a vibrating particle by the superposition of the two waves $y_1=a\sin (\omega t+\frac{\pi }{3})$ and $y_2=a\sin \omega t$ is :

• A. $a$
• B. $\sqrt{2}a$
• C. $2a$
• D. $\sqrt{3}a$

Answer 1

The correct option is D $\sqrt{3}a$

$y_1=a\sin (\omega t+\frac{\pi }{3})y_2=a\sin \omega t\theta =\omega t+\frac{\pi }{3}-\omega t\theta =\frac{\pi }{3}$

Resultant amplitude $=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \theta }=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \frac{\pi }{3}}=\sqrt{3}a$