question_answer

A body moves along a straight line \[\mathbf{OAB}\]simple harmonically. It has at zero velocity at  the points A and B which are at distances a and b respectively from \[O\] and has velocity \[\upsilon \] when half way between them. Find the period of S.H.M.

Answer:

                Clearly, C is the mean position of S.H.M., as shown in Fig.                                 The amplitude of S.H.M. is                                 \[A=AC=CB=\frac{AB}{2}=\frac{b-a}{2}\]                                 The velocity at the mean position \[C\]will be                                 \[\upsilon =\omega A=\frac{2\pi }{T}.\frac{b-a}{2}\]                                 \[\therefore \]  \[T=\frac{\pi (b-a)}{\upsilon }.\]