Question

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. $2\pi \sqrt{\frac{2M}{k}}$

Answer 1

Answer: (C)

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 $SHM,F=-kx...(ii)$
Comparing $(i)$ and $(ii)$, we get $k=4k$
$\therefore$ Time period, $T=2\pi \sqrt{\frac{M}{4k}}$