A) \[400\sqrt{2}\,m/s\]
B) \[200\sqrt{2}\,m/s\]
C) 400 m/s
D) 200 m/s
Correct Answer: D
Solution :
Key Idea: Compare the given wave equation with standard equation of travelling wave moving along negative x-direction. The given wave equation is given by \[y=0.08\sin \frac{2\pi }{\lambda }\,(200\,t-x)\] ? (i) Comparing Eq. (i) with \[y=a\sin \,(\omega \,t-k\,x)\] ? (ii) we have \[\omega =\frac{2\pi }{\lambda }\times 200=\frac{400\pi }{\lambda }\] \[k=\frac{2\pi }{\lambda }\] Hence, wave velocity \[v=\frac{\omega }{k}=\frac{400\,\pi /\lambda }{2\pi /\lambda }=200\,\,m/s\]
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