Question

Two waves are represented by $x_1=A\sin (\omega t+\frac{\pi }{6})$ and $x_2=A\cos \omega t$ then the phase difference between them is :

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\pi$

Answer 1

The correct option is C $\frac{\pi }{3}$

The general equation of the waves given is

$x_1=A\sin \left(\omega t+\frac{\pi }{6}\right)$

$x_2=A\cos \left(\omega t\right)$

Then $x_1$ can be written as

$x_1=A\cos \left(\frac{\pi }{2}-\omega t-\frac{\pi }{6}\right)$

On comparing and solving we get the phase difference is$\frac{\pi }{3}$