A) 2
B) \[\sqrt{3}\]
C) \[\sqrt{2}\]
D) 1
Correct Answer: B
Solution :
For a wave travelling with amplitude a and angular velocity \[\omega \] at time t, the displacement y is given by \[y=a\,\sin \,\omega t\] For phase difference \[\text{o }\!\!|\!\!\text{ }\] the general equation is \[y=a\sin (\omega t+o|)\] Given equations are \[{{y}_{1}}=\sin (\omega t+\frac{\pi }{3})\]and \[{{y}_{2}}=\sin \omega t\] Comparing with standard equation, we have \[{{a}_{1}}=1,\,{{a}_{2}}=1,\,\text{o }\!\!|\!\!\text{ =}\frac{\pi }{3}={{60}^{o}}\] Therefore, resultant amplitude is \[a=\sqrt{a_{1}^{2}+a_{2}^{2}+2{{a}_{1}}{{a}_{2}}\cos \text{o }\!\!|\!\!\text{ }}\] \[a=\sqrt{1+1+2\times 0.5}=\sqrt{3}\]
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