Two masses m₁ and m₂ are suspended together by a massless spring of constant k .
When the masses are in equilibrium, m₁ is removed without disturbing the system; the
amplitude of vibration is:

With mass m₂ alone, the extension of the spring $\ell$ is given as ${\mathrm{m}}_2\mathrm{g}=k\ell$ ..........(i) With mass $\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)$, the extension $\ell$' is given by $\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{g}=\mathrm{k\ell }\text{'}=\mathrm{k}\left(\ell +\mathrm{\Delta \ell }\right)$ .........(ii)

Hence $\mathrm{\Delta \ell }$ is the amplitude of vibration. Subtracting eq. (i) from Eq. (ii), we get

${\mathrm{m}}_1\mathrm{g}=\mathrm{k\Delta }\ell \text{ or }\mathrm{\Delta \ell }=\frac{{\mathrm{m}}_1\mathrm{g}}{\mathrm{k}}$