A) \[\frac{2\pi a}{\sqrt{5}}cm\]
B) \[\frac{2\pi a}{5}cm\]
C) \[\frac{\pi a}{5}cm\]
D) \[2\sqrt{5}\pi a\,cm\]
Correct Answer: B
Solution :
: \[y=a\sin 2\pi \left( pt-\frac{x}{\lambda } \right)\] where\[p=\]frequency, \[2\pi p=\omega ,p=\frac{v}{\lambda }\] wave velocity \[=v\] particle velocity\[=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}\] Maximum particle velocity\[{{V}_{m}}=\omega a\] ...(i) Given : \[{{V}_{m}}=5v\] \[\therefore \]\[\omega a=5v\] \[\Rightarrow \] \[(2\pi p)a=5(p\lambda )\] or \[2\pi a=5\lambda \] or \[\lambda =\frac{2\pi a}{5}cm\]
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