A) \[y=A\,\sin \,k(x-vt)\]
B) \[y=2A\,\sin \,kx\cos \omega t\]
C) \[y=A\,\cos \,2\pi \left( \frac{kx}{\lambda }-\frac{t}{T} \right)\]
D) \[y=A\,\cos \frac{2\pi vt}{\lambda }\]
Correct Answer: B
Solution :
Stationary wave is represented as \[y=a\sin \frac{2\pi }{\lambda }(vt-x)\] (from left to right) or \[y=a\sin \frac{2\pi }{\lambda }(vt+x)\] (from right to left, reflected wave) Hence, option is wrong. Stationary wave also can be represented as \[y=2a\sin \frac{2\pi x}{\lambda }\cos \frac{2\pi t}{T}\] Also, it can be shown as \[y=2a\sin kx\cos \omega t\] \[\left( \because \,k=\frac{2\pi }{\lambda }and\omega =\frac{2\pi }{T} \right)\] Therefore, option is correct. It is quite clear that in option . \[y=A\cos 2\pi \left( \frac{kx}{\lambda }-\frac{t}{T} \right)\] is wrong because correct equation is \[y=A\sin 2\pi \left( \frac{kx}{\lambda }-\frac{t}{T} \right)\] and option is obviously wrong.
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