A) \[\lambda =2\pi {{y}_{0}}\]
B) \[\lambda =\pi {{y}_{0}}/3\]
C) \[\lambda =\pi {{y}_{0}}/2\]
D) \[\lambda =\pi {{y}_{0}}\]
Correct Answer: D
Solution :
On comparing the given equation with standard equation\[y=a\sin \frac{2\pi }{\lambda }(vt-x)\]. It is clear that wave speed \[{{(v)}_{wave}}=v\] and maximum particle velocity \[{{({{v}_{\max }})}_{particle}}=a\omega \]\[={{y}_{0}}\times \]co-efficient of t \[={{y}_{0}}\times \frac{2\pi v}{\lambda }\] \[\because \,\,\,{{({{v}_{\max }})}_{particle}}=2{{(\omega )}_{wave}}\] Þ \[\frac{a\times 2\pi v}{\lambda }=2v\] Þ \[\lambda =\pi {{y}_{0}}\]
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