Question

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

Answer 1

Draw a rough diagram at equilibrium position.

Find equilibrium position.

Let in the equlibrium position, the spring has extended by an amount $x_0$.
Tension in the spring $=k{x}_0$

For equilibrium of the mass $M$,
$Mg=2k{x}_0$

Draw a rough diagram when block slightly displaced.



Find time period at equilibrium position.

Let the mass be pulled through a distance 𝑦 and then released. But string is inextensible, hence the spring alone will contribute the total extension $y+y=2y$,to lower the mass down by $y$from initial equilibrium mean position $x_0$ . So, netextension in the spring $(x_0+2y)$.

$2k(x_0+2y)-Mg=Ma$

$2k{x}_0+4ky-Mg=Ma\Rightarrow Ma=4ky$

$\overrightarrow{a}=-\left(\frac{4k}{M}\right)\overrightarrow{y}$

$k\text{and}M$ beign constant.

$a\alpha -x$. Hence, motion is SHM.

Comparing the above acceleration expression with standard SHM equation

$a=-{\omega }^{2}x,$ we get

${\omega }^2=\frac{4k}{M}\Rightarrow \omega =\sqrt{\frac{4k}{M}}$

Time period,

$T=\frac{2\pi }{\omega }$

$T=2\pi \sqrt{\frac{M}{4k}}$

Final Answer:
$T=2\pi \sqrt{\frac{M}{4k}}$