Solution

The First Way

Force Way

The diagonal
$$d=l\times\sqrt(2)$$
$$d1=d-x$$
There are 2 ways to find out difference of force
first
$$\Delta F=F2-F1=k×q×q×(1/(d×d)-1(d1×d1))$$
The second way
$$F=k×q×q/(d1×d1)$$
$$dF/dx=-2k×q×q/(d-x)\wedge3$$
There is picture for dF

k' is stifness of spring

$$k'×dx=dF$$
$$k'=2×k×q×q×dx/d\wedge3$$
But the spring located between 2 object.We need period one of them

The period formula
$$T=2×\Pi×\sqrt(m/K)$$
But the spring is between 2 object
We need use half of spring
so
$$K=k'×2$$
The period is
$$T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q))$$

$$

T=2×\Pi×\sqrt(4×\Pi×\varepsilon×m×l×l/\sqrt(2)×q×q)
$$
The second way

Energy way

First We need to find out relationship between elongation and sohrtening.

The sollution

again

$$d=a×\sqrt2$$
then
$$a\wedge2=(d/2-x)\wedge2+(d/2+y)\wedge2$$

$$a×a=a×a+d×y-d×x$$
so
$$x=y$$
$$dx/dt=Vx$$
$$dy/dt=Vy$$

T=2×\Pi×\sqrt(m×d\wedge3/(4×k×q×q))

$$