Question

Consider the situation shown in figure. Show that if the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. Calculate the time period.

Answer 1

Consider that both the blocks are displaced by x each then, by energy method

$\frac{1}{2}k(2x{)}^2+\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2=$ constant

$m{v}^2+2k{x}^2=$ constant

Taking derivative of both sides wrt t

$m\times 2v\frac{dN}{dt}+2k\times 2x\frac{dx}{dt}=0$

$ma+2kx=0$ $[\frac{dv}{dt}=a;\frac{dx}{dt}=v]$

$\Rightarrow a=-\left(\frac{2k}{m}\right)x$ $[a=-{\omega }^{2}x]$ i.e., it is SHM

$\Rightarrow \omega =\sqrt{\frac{2k}{m}}$

Thus, time period $\left(T\right)=2\pi \sqrt{\frac{m}{2k}}$

$[T=\frac{2\pi }{\omega }]$.