Question

Two waves $E_1=E_0\sin \omega t$ and $E_2=E_0\sin (\omega t+60)$ superimpose each other. Find out initial phase of resultant wave?

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. $0^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

Let the resultant wave be

$E=E{0}^{\prime }\sin (wt+\varphi )$./ Then

$E=E_1+E_2$

$E=E_0\sin wt+E_0\sin (wt+{60}^o)$

$E=E_0(\sin wt+\sin (wt+{60}^o)$

As $\sin \cos +\sin \left(B\right)=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{B-A}{2}\right)$

$\sin wt+\sin (wt+{60}^o)=2\sin (wt+{30}^o).\cos \left(wt\right)$

So, $E=2E_0\cos \left(wt\right)\sin (wt+{30}^o)$

So, $E_0=2E_0\cos wt$ and $\varphi ={30}^o$

Option $A$ is correct