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