Unchanged oscillations

Let us first see why the initial set up oscillates in the first place.

when we pull the mass M down , since the angular momentum of the disc is conserved angular velocity will increase on reduction in radius of its path due to pulling down the hanging block. The increment in angular velocity and decrement in the radius are so related that the tension force will not be sufficient to provide the sufficient acceleration thus causing the disc to increase the radius of its path by simultaneously pulling the hanging M, setting up oscillations in the system in such a way that the hanging mass M starts up and down oscillations and the mass m radial oscillations.

let the initial angular velocity of the mass m be $\omega_{0}$

$\omega_{0}r^{2}=\omega_{1}(r-x)^{2}$

$m\omega_{1}^{2}(r-x)-m\omega_{0}^{2}r)=(m+M)a$

we get angular frequency of oscillations as $\omega=\omega_{0} \sqrt{\frac{3m}{m+M}}$

Now comes the tricky part how can we find the force due to a charged disc. let the radius of the disc be a with charge b

let us consider a cylindrical gaussian surface of small radius x and length 2z symmetrical to the centre of the cylinder.

for this small radius we can consider that the vertical component of electric field near the axis doesn't vary with x

the flux can be given by $\frac{bx^{2}}{\varepsilon_{0}a^{2}}=2\frac{2kb}{a^{2}}(1-\frac{z}{\sqrt{z^{2}+a^{2}}})(\pi x^{2})+2\int_{0}^{z}E_{radial}2\pi xdz$

by diffrenciating and applying newtons libnitz formula we obtain

$E_{radial}=\frac{kbx}{(z^{2}+a^{2})^{3/2}}$ here as a is very large we can neglect x

the main aim of me giving this question was to find this approximation.

for the time period or frequency to be the same it is necessary that no other external forces restore the system.

$\Delta E_{vertical}Q+\Delta E_{radial}q=0$

$\frac{2kQb}{a^{3}}x=-\frac{kqb}{a^{3}}$

$\frac{q}{Q}=-2$