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 :
The given equations of motion are \[{{y}_{1}}=2a\,\,\sin (\omega t-ku)\] ??(i) \[{{y}_{2}}=2a\,\,\sin (\omega t-k\theta )\] ??(ii) Hence the equation of resultant wave is given by \[y={{y}_{1}}+{{y}_{2}}\] It is known as super position. \[\Rightarrow \]\[y=2a\,\sin \,(\omega t-ku)+2a\,\sin \,(\omega t-ku-\theta )\] \[\Rightarrow \] \[y=2a\left[ \frac{2\sin \,(\omega t-ku+\omega t-ku-\theta )}{2} \right.\times \] \[\left. \cos .\frac{\omega t-ku-(\omega t-ku-\theta )}{2} \right]\] \[\Rightarrow \] \[y=4a\,\cos \,\frac{\theta }{2}\,\sin \,(\omega t-ku-\frac{\theta }{2})\] ...(iii) Again comparing equation (iii) with \[y=A\,\sin \,(\omega t-ku)\] we have, Resultant amplitude \[(A)=4a\,\cos \,\frac{\theta }{2}.\]
You need to login to perform this action.
You will be redirected in
3 sec
