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Q.
The time period of mass $M$ when displaced from its equilibrium position and then released for the system as shown in figure is
Oscillations
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
$F = Mg - 2k(2l + 2x) = - 4kx \quad... (i)$
$F\propto x$ and $-ve$ sign shows that it is directed towards mean position. Hence, the mass executes simple harmonic motion.
For $SHM, F = - kx \quad... (ii) $
Comparing $(i)$ and $(ii)$, we get $k = 4k$
$\therefore $ Time period, $T = 2\pi \sqrt{\frac{M}{4k}}$
