A) \[2a\sin \text{ }\!\!\theta\!\!\text{ }\]
B) \[\sqrt{2}a\sin \text{ }\!\!\theta\!\!\text{ }\]
C) \[4a\sin \frac{\text{ }\!\!\theta\!\!\text{ }}{2}\]
D) \[\sqrt{2}a\sin \frac{\text{ }\!\!\theta\!\!\text{ }}{2}\]
Correct Answer: C
Solution :
The given equations of motion are \[{{y}_{1}}=2a\sin (\omega t-kx)\] \[{{y}_{2}}=2a\sin (\omega t-kx-\theta )\] Now, the resultant equation of wave is given by \[y={{y}_{1}}+{{y}_{2}}\] \[=2a\sin (\omega t-kx)+2a\sin (\omega t-kx-\theta )\] \[y=2a\left[ 2\sin \frac{(\omega t-kx-\omega t-kx-\theta )}{2} \right.\] \[\left. \times \cos \frac{\omega t-kx-(\omega t-kx-\theta )}{2} \right]\] \[y=4a\cos \frac{\theta }{2}\sin \left( \omega t-kx-\frac{\theta }{2} \right)\] ?.(1) Now, comparing Eq. (1) with\[y=A\sin (\omega t-kx),\]we have Resultant amplitude \[A=4a\cos \frac{\theta }{2}\]
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