question_answer 54)

                    Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in Fig.

Answer:

                        Let k = spring constant of the spring.                 When mass M is displaced through a distance I in the downward direction, then both the string and spring will be stretched by \[l\]. Since string is inextensible, so spring is stretched by 21. As the tension in string and spring is equal, so \[F=-2k\times 2l=-4\,kl\]                 This force is known as restoring force and it is balanced by the weight (Mg)                 \[\therefore \]  \[Mg=4\,kl\]                 Let the mass M is pulled down further through a distance x, then restoring force P is given by                 \[F'=-2k\,(2l+2x)\]                 \[\therefore \] Force acting on mass M is given by                 \[f=F'-F=-2k\,(2l+2x)+4kl\]                 \[=-4\,kx\]                          Hence, acceleration of mass M,                 \[a=\,\frac{f}{m}\,=\,-\left( \frac{4k}{M} \right)\,x\]                       Thus, motion of mass M is S.H.M.                 Time period of oscillation of mass M is given by                 \[T=\,2\pi \,\sqrt{\frac{x}{|a|}}\,=2\pi \,\sqrt{\frac{M}{4k}}\]