First, recall that spring force is F = -kx
The top charge exerts a force on the negative charge (ke*qQ) / ((d/2)^2 + x^2) which is directed upwards and left, at an angle theta. Now notice that the y-components of the charges canel each other out so there is no y-component in the final result.
So procceed in finding the x-component of the 2 forces. Rather than doing this twice, because of the symmetry you just need to mulitply the answer by 2.
The total force is: -2(ke*qQ) / ((d/2)^2 + x^2)^(1/2) * cos(theta) i
Now you must solve for cos(theta) in terms of the geometry of the figure.
Now to find the period. Recall that the formula
acceleration = angular frequency^2 * x
Cos = adjacent/ hypotenuse, so:
cos = x / ((d/2)^2 + x^2)^(1/2)
Now plug back into the equation for total force to get :
[-2(ke*qQ) / ((d/2)^2 + x^2)^(1/2)] * [x / ((d/2)^2 + x^2)^(1/2)] i = simply to get:
[-2(ke*qQ) * x / ((d/2)^2 + x^2)^(3/2)] i
In the question it states that x is small, so we can safely remove x from the denominator and simplify resulting in:
[-2(ke*qQ) * x / ((d/2)^2 )^(3/2)] i
Notice how the first part of this is a constant? easier to see when you rewrite it like so:
F = - [2(ke*qQ) / ((d/2)^3] * x,
(the first part in brackets is the k-value or the spring constant.)
now recall that F = ma, and hence to get acceleration just divide by m, and so:
a = - [2(ke*qQ) / m*((d/2)^3] * x
Now use the formula: a = - (angular frequency) ^2 * x
and solve for angular frequency.
Now recall that the Period is simply: T = 2pi / (angular frequency)
T = (pi/2) [(m*d^3)/(ke*qQ)]^(1/2)
and vmax is just angular frequency times acceleration:
so vmax = 4a * [(ke*qQ) / (m*d^3)] ^ (1/2)