question_answer

The time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure is

A)         \[2\pi \sqrt{\frac{M}{k}}\]                      

B) \[2\pi \sqrt{\frac{M}{2k}}\]

C) \[2\pi \sqrt{\frac{M}{4k}}\]

D)                    \[2\pi \sqrt{\frac{2M}{k}}\]

Correct Answer: C

Solution :

When mass M is

suspended from the given system as shown in figure, let \[l\] be the length through which mass M moves down before it comes to rest.

In this situation, both the spring and string will be stretched by length l. Since string is inextensible, so spring is stretched by length\[2l\]. The tension along the string and spring is the same.

In equilibrium, \[Mg=2\,(k2l)\]

If mass M is pulled down through small distance x, then

\[F=Mg-2k(2l+2x)=-4kx\]           ...(i)

\[F\propto x\] and \[-ve\] sign shows that it is directed towards mean position. Hence, the mass executes simple harmonic motion.

For SHM, \[F=-k'x\] b                  ...(ii)

Comparing (i) and (ii), we get \[k'=4k\]

\[\therefore \] Time period, \[T=2\pi \sqrt{\frac{M}{4k}}\]