A) \[2a\,\,\cos \,\theta \]
B) \[\sqrt{2}a\,\,\cos \,\theta \]
C) \[4a\,\,\cos \,\frac{\theta }{2}\]
D) \[\sqrt{2}a\,\,\cos \,\frac{\theta }{2}\]
Correct Answer: C
Solution :
Key Idea: Resultant displacement is sum of individual displacements. Given \[{{y}_{1}}\] and \[{{y}_{2}}\] the displacement at a point x at any instant\[t\] Then by principle of superposition, the resultant displacement at that point is \[y={{y}_{1}}+{{y}_{2}}\] Given, \[{{y}_{1}}=2a\,\sin \left( \omega t-kx \right),\] \[{{y}_{2}}=2a\,\sin \left( \omega t-k\,x-\theta \right)\] \[\therefore \]\[y=2a\left[ \sin \left( \omega t-kx \right)+\sin \left( \omega t-kx-\theta \right) \right]\] Using the formula \[\sin A+\sin \,B=2\sin \,\frac{A+B}{2}\cos \frac{A-B}{2}\] \[y=2a\left[ 2\,\sin \frac{\left( \omega t-kx+\omega t-kx-\theta \right)}{2}\times \cos \frac{\omega t-kx-\left( \omega t-kx-\theta \right)}{2} \right]\] \[\therefore \] \[y=4a\,\,\cos \frac{\theta }{2}\sin \left( \omega t-kx-\frac{\theta }{2} \right)\] ?(1) Also, standard equation of wave is \[y=A\,\sin \left( \omega t-kx \right)\] ...(2) Comparing Eqs. (1) and (2), we get \[A=4\,a\,\,\cos \frac{\theta }{2}\]
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