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.

• A. $2\pi \sqrt{\frac{m}{k}}$
• B. $2\pi \sqrt{\frac{2m}{k}}$
• C. $2\pi \sqrt{\frac{k}{m}}$
• D. $2\pi \sqrt{\frac{m}{2k}}$

Answer 1

The correct option is D

$2\pi \sqrt{\frac{m}{2k}}$

The center 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'.

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

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

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

$\Rightarrow$ Time period $T=2\pi \sqrt{\frac{m}{2k}}$