A) \[\frac{\lambda }{4\pi }\]
B) \[\frac{2\lambda }{\pi }\]
C) \[\frac{\lambda }{2\pi }\]
D) \[\lambda \]
Correct Answer: A
Solution :
Work done\[y=a\sin \frac{2\pi }{\lambda }(vt-x)\] ?(i) Differentiating Eq (i) w,r,t.t, we get \[\frac{dy}{dt}=\frac{2\pi va}{\lambda }\cos \frac{2\pi }{\lambda }(vt-x)\] Now, maximum velocity is obtained when \[\cos \frac{2\pi }{\lambda }(vt-x)=1\] \[\therefore \] \[{{v}_{\max }}={{\left( \frac{dy}{dt} \right)}_{\max }}=\frac{2\pi va}{\lambda }\] but \[{{v}_{\max }}=\frac{v}{2}\] (given) \[\therefore \] \[\frac{v}{2}=\frac{2\pi va}{\lambda }\] \[\Rightarrow \] \[a=\frac{\lambda }{4\pi }\]
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