A particle executes SHM in a line 4 cm long. Its velocity when passing through the centre of line is 12 cm/s. The period will be

Length of the line = Distance between extreme positions of oscillation = 4 cm  
So, Amplitude  $\mathrm{a}=2\mathrm{cm}.$
also  ${\mathrm{v}}_{\max }=12\mathrm{cm}/\mathrm{s}.$
 $\because {\mathrm{v}}_{\max }=\mathrm{\omega a}=\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{a}$
   $\Rightarrow \mathrm{T}=\frac{2\mathrm{\pi a}}{{\mathrm{v}}_{\max }}$ $=\frac{2\times 3.14\times 2}{12}$ $=1.047\sec$