For one block system:
Let the extension l is produced when a force F is applied to the free end of the spring. For the maximum extension,
F = kl
Where k is the spring constant.
Hence, the maximum extension, $\mathrm{l}=\frac{\mathrm{F}}{\mathrm{k}}$
For the two-block system:
The displacement (x) produced is:
$\mathrm{x}=\frac{\mathrm{l}}{2}$
$\mathrm{Net} \mathrm{force}, \mathrm{F}=+2\mathrm{kx}=2\mathrm{k}\frac{\mathrm{l}}{2}$
$\therefore \mathrm{l}=\frac{\mathrm{F}}{\mathrm{k}}$
$\mathrm{For} \mathrm{one} \mathrm{block} \mathrm{system}:$
$\mathrm{For} \mathrm{mass} \left(\mathrm{m}\right) \mathrm{of} \mathrm{the} \mathrm{block},$
$\mathrm{F}=\mathrm{ma}=\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}$
$\mathrm{where} \mathrm{x} \mathrm{is} \mathrm{the} \mathrm{displacement} \mathrm{of} \mathrm{the} \mathrm{block} \mathrm{in} \mathrm{time} \mathrm{t}.$
$\therefore \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\mathrm{kx}$
$\mathrm{Negative} \mathrm{sign} \mathrm{shows} \mathrm{that} \mathrm{the} \mathrm{direction} \mathrm{of} \mathrm{elastic} \mathrm{force} \mathrm{is} \mathrm{opposite} \mathrm{to} \mathrm{the} \mathrm{direction} \mathrm{of} \mathrm{displacement}.$
$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\left(\frac{\mathrm{k}}{\mathrm{m}}\right)\mathrm{x}=-{\mathrm{\omega }}^2\mathrm{x}$
$\mathrm{where} {\mathrm{\omega }}^2=\frac{\mathrm{k}}{\mathrm{m}}$
$\mathrm{\omega }=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
$\therefore \mathrm{Time} \mathrm{period} \mathrm{of} \mathrm{the} \mathrm{oscillation}, \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{\sqrt{{\displaystyle \frac{\mathrm{k}}{\mathrm{m}}}}}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
$\mathrm{For} \mathrm{the} \mathrm{two} \mathrm{block} \mathrm{system}:$
$\mathrm{F}=\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}$
$\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-2\mathrm{kx}$
$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=-\left[\frac{2\mathrm{k}}{\mathrm{m}}\right]\mathrm{x}=-{\mathrm{\omega }}^2\mathrm{x}$
$\mathrm{where},$
$\mathrm{Angular} \mathrm{frequency}, \mathrm{\omega }=\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}$
$\therefore \mathrm{Time} \mathrm{period}, \mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}}$