A) \[\lambda =2\,\pi {{y}_{0}}\]
B) \[\lambda =\frac{\pi {{v}_{0}}}{3}\]
C) \[\lambda =\frac{\pi {{y}_{0}}}{2}\]
D) \[\lambda =\pi {{y}_{0}}\]
Correct Answer: D
Solution :
The given equation is \[y={{y}_{0}}\sin \,\frac{2\pi }{\lambda }(vt-x)\] .....(i) In the wave equation v is the particle velocity. Differentiating Eq. (i) with respect to time, \[u=\frac{dy}{dx}={{y}_{0}}\frac{2\pi v}{\lambda }\cos \frac{2\pi }{\lambda }(vt-x)\] Maximum particle velocity, \[{{u}_{\max }}={{y}_{0}}\frac{2\pi v}{\lambda }\] Now it is given that, maximum particle velocity \[=2\times \] wave velocity \[or{{y}_{0}}\frac{2\pi v}{\lambda }=2v\] \[or\lambda =\pi {{y}_{0}}\]
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