A) a
B) \[\sqrt{2}a\]
C) \[\sqrt{3}a\]
D) 2a
Correct Answer: C
Solution :
Given waves are \[{{y}_{1}}=a\,\sin \left( \omega t+\frac{\pi }{6} \right)\] \[{{y}_{2}}=a\,\cos \,\omega t\] \[=a\,\sin \left( \omega t+\frac{\pi }{2} \right)\] Therefore, phase difference between two waves is \[\text{o }\!\!|\!\!\text{ =}\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}\] The resultant amplitude is given by \[\therefore \] \[{{A}^{2}}={{a}^{2}}+{{a}^{2}}+2a.a\,\cos \,\frac{\pi }{3}\] \[={{a}^{2}}+{{a}^{2}}+2{{a}^{2}}.\frac{1}{2}\] \[=3{{a}^{2}}\] Hence, \[A=\sqrt{3}a\]
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