Question

The displacement equation for two waves undergoing superposition are $y_1=4\sin \omega t,y_2=3\sin (\omega t+\pi /2),$ then resultant amplitude will be:

• A. $5cm$
• B. $7cm$
• C. $1cm$
• D. $0cm$

Answer 1

Answer: (A)

Equations of the waves is $y_1=4\sin \omega t$ , $y_2=3\sin \left(\omega t+\frac{\pi }{2}\right)$ Comparing these equations with $y=a\sin (\omega t+\varphi )$ $a_1=4,a_2=3,\varphi =\frac{\pi }{2}$ Resultant amplitude $R=\sqrt{{a_1}^2+{a_2}^2+2a_1{a}_2\cos \varphi }$ $=\sqrt{4^2+3^2+2\times 4\times 3\cos \frac{\pi }{2}}$ $=\sqrt{16+9+24\times 0}$ $=5cm$