A) \[\frac{\pi }{2}\]
B) \[\pi \]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
Let \[{{X}_{1}}=A\sin \omega t\] \[{{X}_{2}}={{X}_{0}}+A\sin (\omega t+\phi )\] \[\left| {{X}_{2}}+{{X}_{1}} \right|={{X}_{0}}+A\sin (\omega t=\phi )+A\sin \omega t\] \[{{\left| {{X}_{2}}+{{X}_{1}} \right|}_{\max }}={{X}_{0}}+\sqrt{{{A}^{2}}+{{A}^{2}}+2A\,A\cos \varphi }\] \[={{X}_{0}}+2A\] \[\Rightarrow {{X}_{0}}+2A\sin \frac{\phi }{2}={{X}_{0}}+2A\] \[\sin \frac{\phi }{2}=1=\sin \frac{\pi }{2},\] \[\phi =\pi \]
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