Two small blocks of mass m and 4m are connected to two springs as shown in fig. Both springs have stiffness K and they are in their natural length when the blocks are at point O. Both the blocks are pushed so that both the springs get compressed by a distance a. First the block of mass m is released 

and after it travels through a distance $\left(1-\frac{\sqrt{3}}{2}\right)\mathrm{a}$ the second block is also released.
AIt is observed that at a distance s  from point O the two blocks collide and
the time the two blocks need to collide after the block of mass 4m is released is $t_0. then$

Till the point of collision, motion of both the blocks remain simple harmonic. For spring block system of mass m, the time period of SHM will be $2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=\mathrm{t}$ (say)
Angular frequency $\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
For spring block system of mass 4m, the corresponding time period is 2T [and angular frequency = $\frac{\mathrm{\omega }}{2}$]
If a particle P performs uniform circular motion in x-y plane with constant angular speed $\mathrm{\omega }$, foot of perpendicular drawn from it on the X axis [or Y axis] performs SHM with angular frequency $\mathrm{\omega }$.

Let P₁" and P₂ be two points rotating on a circle of radius a in clockwise sense with angular speed $\mathrm{\omega } \mathrm{and} \frac{\mathrm{\omega }}{2}$ respectively. Motion of foot of perpendicular from P₁ on X axis represents the motion of mass m and the motion of
foot of perpendicular from P₂ on the X axis represents the motion of mass 4m.
According to the question, when P₁ reaches P₁' (i.e., block of mass m moves to Q₁ getting displaced by( $\mathrm{a}-\frac{\sqrt{3}}{2}\mathrm{a}$) then P2 starts rotating. Q2 is foot of perpendicular of P₂ representing the motion of 4m. The blocks collide when P₁ reaches P₁'' and P₂ reaches P'₂ (see fig). Angle rotated by P₁ is twice that covered by P₂ due to its double angular speed.
$\therefore \mathrm{If} <{\mathrm{P}}_2{\mathrm{OP}}_2^{\mathrm{\prime }}=\mathrm{\varphi } \mathrm{then} <{\mathrm{P}}_1^{\mathrm{\prime }}{\mathrm{OP}}_1^{\prime \prime }=2\mathrm{\varphi }$
Then $<{\mathrm{P}}_1^{\prime \prime }{\mathrm{OP}}_2=<{\mathrm{P}}_2{\mathrm{OP}}_2^{\mathrm{\prime }}=\mathrm{\varphi }$
But $<{\mathrm{P}}_1^{\prime \prime }{\mathrm{OP}}_2=<{\mathrm{P}}_2{\mathrm{OP}}_2^{\mathrm{\prime }}=\mathrm{\varphi }$
$\begin{array}{l}\therefore \frac{\mathrm{\pi }}{6}+2\mathrm{\varphi }+\mathrm{\varphi }=\mathrm{\pi }\\ \Rightarrow \mathrm{\varphi }=\frac{5\mathrm{\pi }}{18}\end{array}$
$\therefore$ OQ₁ = distance from O where collision takes place $=\mathrm{acos}\mathrm{\varphi }=\mathrm{acos}\left(\frac{5\mathrm{\pi }}{18}\right)=\mathrm{acos}{50}^{\circ }$
Time required for collision = time required by P₂ to rotate through $\frac{5\mathrm{\pi }}{18}$
$=\frac{2\mathrm{T}}{2\mathrm{\pi }}\cdot \frac{5\mathrm{\pi }}{18}=\frac{5\mathrm{T}}{18}=\frac{5\mathrm{\pi }}{9}\sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$