A) \[{{y}_{1}}=2a\,\sin \,(\omega t-kx)\] \[{{y}_{2}}=2a\,\sin \,(\omega t-kx-\theta )\] The amplitude of the medium particle will be: \[2a\,\cos \,\theta \]
B) \[\sqrt{2\,}a\,\cos \,\theta \]
C) \[4a\,\cos \,\frac{\theta }{2}\]
D) \[\sqrt{2\,}a\,\cos \,\frac{\theta }{2}\]
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 \[y={{y}_{1}}+{{y}_{2}}\] \[=2a\,\sin \,(\omega t-kx)+2a\,\sin \,(\omega t-kx-\theta )\] \[y=2a\left[ 2\sin \frac{(\omega t-kx+\omega t-kx-\theta )}{2} \right.\left. \times \cos \frac{\omega t-kx-(\omega t-kx-\theta )}{2} \right]\] Now comparing equation (1) with \[y=A\,\sin \,(\omega t-kx),\,\]we have Resultant amplitude \[A=4a\,\cos \frac{\theta }{2}\]
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