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
B
60
C
120
D
0

Solution

## The correct option is A $$30 ^ { \circ }$$Let the resultant wave be $$E=E0' \sin (wt+\phi)$$./ 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 $$\sin \cos +\sin (B)=2\sin \left (\dfrac {A+B}{2}\right) \cos \left (\dfrac {B-A}{2}\right)$$$$\sin wt +\sin (wt +60^o)=2 \sin (wt +30^o).\cos (wt)$$So, $$E=2E_0 \cos (wt) \sin (wt+30^o)$$So, $$E_0=2E_0 \cos wt$$ and $$\phi =30^o$$Option $$A$$ is correctPhysics

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