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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
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B
60
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C
120
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D
0
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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 correct


Physics

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