A) \[{{y}_{O}}\pi /2\]
B) \[2y0\pi \]
C) \[{{y}_{0}}\pi \]
D) \[{{y}_{0}}\pi /4\]
Correct Answer: A
Solution :
\[y={{y}_{0}}\sin 2\pi ft\] Maximum particle velocity\[=A\omega ={{y}_{0}}\omega \] Wave velocity\[=\frac{\omega }{k}\] where, \[k=\frac{2\pi }{\lambda }\] According to question \[{{y}_{0}}\omega =4\frac{\omega }{k}\] \[{{y}_{0}}=\frac{4}{k}\] \[\Rightarrow \] \[{{y}_{0}}=\frac{\lambda }{2\pi }\times 4\] \[\therefore \] \[\lambda =\frac{\pi {{y}_{0}}}{2}\]
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