A) \[\frac{A}{\sqrt{2}},\frac{\omega }{2}\]
B) \[\frac{A}{\sqrt{2}},\omega \]
C) \[\sqrt{2A,}\frac{\omega }{2}\]
D) \[\sqrt{2}A,\omega \]
Correct Answer: D
Solution :
Let \[{{y}_{1}}=A\sin (\omega t)\] and \[{{y}_{2}}=A\sin \left( \omega t+\frac{\pi }{2} \right)\] Resultant amplitude \[{{R}^{2}}={{A}^{2}}+{{A}^{2}}+2{{A}^{2}}\cos \left( \frac{\pi }{2} \right)\] \[{{R}^{2}}=2{{A}^{2}}+2{{A}^{2}}\times 0\] \[{{R}^{2}}=2{{A}^{2}}\] \[\Rightarrow \] \[R=\sqrt{2}A\] However, both will have the same frequency on superimposing.
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