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
\[T=\,2\pi
\,\sqrt{\frac{x}{|a|}}\,=2\pi \,\sqrt{\frac{M}{4k}}\]
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