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 \[\upsilon =\omega A=\frac{2\pi }{T}.\frac{b-a}{2}\] \[\therefore \] \[T=\frac{\pi (b-a)}{\upsilon }.\]
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