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 \[y=0.08\sin \frac{2\pi }{\lambda }(200-x)\] ?(i) Comparing Eq. (i) with \[y=a\sin (\omega t-kx)\] ?(ii) we have \[\omega =\frac{2\pi }{\lambda }\times 200=\frac{400\pi }{\lambda }\] \[k=\frac{2\pi }{\lambda }\] Hence, wave velocity \[v=\frac{\omega }{k}=\frac{400\pi /\lambda }{2\pi /\lambda }=200\,m/s\]You need to login to perform this action.
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