Question

Two waves are represented by $y_1=a\sin \left(\omega t+\frac{\pi }{6}\right)$ and $y_2=a\cos \omega t$ What will be their resultant amplitude?

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

Answer 1

Answer: (C)

Given waves are $y_1=a\sin \left(\omega t+\frac{\pi }{6}\right)$ $y_2=a\cos \omega t$ $=a\sin \left(\omega t+\frac{\pi }{2}\right)$ Therefore, phase difference between two waves is $\text{o }|\text{ =}\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$ The resultant amplitude is given by $\therefore$ $A^2=a^2+a^2+2a.a\cos \frac{\pi }{3}$ $=a^2+a^2+2a^2.\frac{1}{2}$ $=3a^2$ Hence, $A=\sqrt{3}a$