A particle of charge $Q$ and mass M = 2m is tied to two identical particles, each having mass m and charge $q$. The strings are of equal length, $l$ each, and they are inextensible. The system is held at rest on a smooth horizontal surface with strings taut in a position where the strings make an angle of 60⁰ between them. From this position the system is released.

In original position, the COM of the system is located at centre of line AD (i.e., at a distance of $AG=\frac{\sqrt{3}}{4}l$  from A). The particle move such that the COM does not get displaced. When all three come to rest again, particle of mass 2m will be at D' [This particle experiences net force along AD and moves on this line itself] and the other two will be symmetrically placed at B’ and C’.

$\therefore$   A and D are extreme position of Oscillation of particle of mass 2m.
$\therefore$ Amplitude = $\frac{AD}{2}=\frac{\sqrt{3}}{4}l$ ;  

Speed will be max when particles are along same line.
 

By conservation of momentum of the system  :

At mean position, let v be velocity of B and C masses and v' be velocity of 2m mass. 

$2\mathrm{mv}+2\mathrm{mv}\text{'}=0\Rightarrow \mathrm{v}\text{'}=\mathrm{v}$

By conservation of energy : 

Net energy initially is 

$TE=\frac{k{q}^2}{l}+\frac{kQq}{l}$

Net energy in mean position is 

$TE=2\frac{kQq}{l}+\frac{k{q}^2}{2l}+\frac{1}{2}m{v}^2+\frac{1}{2}\left(2m\right)v^2$

Equating the above two  equations

$v=\frac{q}{4}\sqrt{\frac{1}{\pi {\in }_{0}ml}}$

Tension in the string in the mean position is 

$T=\frac{k{q}^2}{{\left(2l\right)}^2}+\frac{kQq}{l^2}+\frac{4m{v}^2}{l}$

$\Rightarrow T=\frac{q[5q+4Q]}{16\pi {\in }_0{l}^2}$