Velocity of particle

I am getting upto this stage :

v_{app y.} = Fm∫x√(l²-x²) dt

What is the answer ???

i m not able to post the whole soln since i m not able to upload the image in which i have done......
in this question firstly find acceleration at any arbitary separation r.......
that acceleration of mass m comes out to be.....a(r)=(F/2m)r/2((l/2)^2 -(r/2)^2))^ .5
relative accel will be 2a(r)......
v(r) be rel velo at any general separation r
now since a=vdv/dx
so...2a(r)=v(r)dv(r)/dr

2a(r)dr=v(r)dv(r)

integrating lhs wrt r and rhs wrt v...with limits from 0 to x and 0 to v....wher v is rel velo when sep is x

we get the answer

@aveek: In the problem, we have to find the relative velocity of approach of the particles.

Well after trying the solution I have got upto this point:
If we find the work done by the force on any of the particles for a small displacement dx and then integrate it from l to x,we would find out the total work done on any one of the particles.

Now,simply applying work-energy theorem:Change in k=work done by the apllied force(f here)

Thus we can find the individual velocy of each particle and hence find their relative velocity.

hey is there any prob with my method????plzz tell....

Hint:
Take a reference frame which is fixed to the mid point of the thread. It is a non-inertial one (as the entire system is accelerating in the direction of F with an acceleration a₀=F2m.)
Now, in this reference frame, we have a "effective gravity" of a₀ in the opposite direction of a₀.

@Kaymant Sir and saggy....can't we proceed like this ....

vdvdx = F2mx√(l²-x²)

hence relative velocity = 2v