Question

Consider the situation shown in figure (12-E10). 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

The centre of mass of the system should not change during the motion.

So, if the block 'm' on the left moves towards right a distance 'x', the block on the right moves towards left a distance 'x'. So, total compression of the spring is 2x.

By energy method,

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

$\Rightarrow m{v}^2+2k{x}^2=C$

Taking derivative of both sides with respect to 't'.

$m2v\frac{dv}{dt}+2k\times 2x\frac{dx}{dt}=0$

$\therefore ma+2kx=0$

$[becausev=\frac{dx}{dt}anda=\frac{dv}{dt}]$

$\Rightarrow -\frac{a}{x}=\frac{2k}{m}{\omega }^2$

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

$\Rightarrow$ Time period = $T=2\pi \sqrt{\left(\frac{m}{2k}\right)}$