A) 4A
B) 2A
C) \[\sqrt{2}A\]
D) A
Correct Answer: C
Solution :
Given, \[{{y}_{1}}=A\sin (kx-\omega t)\] \[{{y}_{2}}=A\cos (kx-\omega t)\] or \[{{y}_{2}}=A\sin \left( kx-\omega t+\frac{\pi }{2} \right)\] Phase difference of two waves \[=\frac{\pi }{2}\] \[\therefore \]Resultant amplitude \[R=\sqrt{{{A}^{2}}+{{A}^{2}}+2AA\cos \phi }\] \[=\sqrt{{{A}^{2}}+{{A}^{2}}+2{{A}^{2}}\cos \frac{\pi }{2}}\] \[=\sqrt{2{{A}^{2}}}\] \[\left( \because \cos \frac{\pi }{2}=0 \right)\] \[R=\sqrt{2}A\]
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