Two blocks rest on a smooth horizontal surface. They are connected by a spring of force constant k. If the system is set into oscillation, find its time period.

Let x₁ and x₂ be displacement of the two blocks at any point of time.
Extension in the spring is x = x₂ – x₁
For motion of m₁; ${\mathrm{n}}_1\frac{{\mathrm{d}}^2{\mathrm{x}}_1}{{\mathrm{dt}}^2}=\mathrm{kx}\Rightarrow \frac{{\mathrm{d}}^2{\mathrm{x}}_1}{{\mathrm{dt}}^2}=\frac{\mathrm{k}}{{\mathrm{m}}_1}\mathrm{x}....\left(\mathrm{i}\right)$
For motion of m₂; ${\mathrm{m}}_2\frac{{\mathrm{d}}^2{\mathrm{x}}_2}{{\mathrm{dt}}^2}=-\mathrm{kx}\Rightarrow \frac{{\mathrm{d}}^2{\mathrm{x}}_2}{{\mathrm{dt}}^2}=-\frac{\mathrm{k}}{{\mathrm{m}}_2}\mathrm{x}.....\left(\mathrm{ii}\right)$
Equation (ii) – (i)
$\frac{{\mathrm{d}}^2\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}{{\mathrm{dt}}^2}=-\mathrm{k}\left(\frac{1}{{\mathrm{m}}_1}+\frac{1}{{\mathrm{m}}_2}\right)\mathrm{x}$
$\Rightarrow \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\frac{\mathrm{k}}{\mathrm{\mu }}\mathrm{x}$ $\left[\mathrm{where} \mathrm{\mu }=\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right]$
$\therefore \mathrm{\omega }=\sqrt{\frac{\mathrm{k}}{\mathrm{\mu }}}\Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{\mathrm{\mu }}{\mathrm{k}}}$