Ring and Time Period

Consider the thread to be placed horizontally and the center of the thread is the origin. The bead can slide along the positive X-axis. Let the position of the bead be:

$\vec{r}_b = (x,0)$

The position of any point on the ring can be parameterised using polar coordinates as such:

$\vec{r}_p = (R\cos{\theta},R\sin{\theta})$

Let the charge per unit length on the ring be:

$\lambda = \frac{Q}{2 \pi R}$

The electrostatic force along the X direction due on the bead due to a ring element is:

$dF_e = -\frac{\lambda R q \ d\theta}{4 \pi}\left(\frac{R\cos{\theta}-x}{\left((R\cos{\theta}-x)^2 + R^2 \sin^2{\theta}\right)^{3/2}}\right)$ $\implies F_e = -\frac{ Q q \ d\theta}{8 \pi^2}\left(\frac{R\cos{\theta}-x}{\left(R^2 + x^2 -2Rx\cos{\theta}\right)^{3/2}}\right)$

Now, since we are considering small oscillations, then $x$ is small, therefore $x^2$ can be neglected. This makes the expression:

$\implies dF_e = -\frac{ Q q \ d\theta}{8 \pi^2}\left(\frac{R\cos{\theta}-x}{\left(R^2 -2Rx\cos{\theta}\right)^{3/2}}\right)$

By using binomial expansion approximation of the denominator (try this yourself), we get:

$\implies dF_e =-\frac{ Q q \ d\theta}{8 \pi^2 R^3}\left(R\cos{\theta}-x\right)\left(1 + \frac{3x\cos{\theta}}{R}\right)$

The total electrostatic force along X is therefore:

$F_e = \int_{0}^{2 \pi} -\frac{ Q q \ d\theta}{8 \pi^2 R^3}\left(R\cos{\theta}-x\right)\left(1 + \frac{3x\cos{\theta}}{R}\right)$ $\implies F_e = -\frac{ Q q x}{8 \pi R^3}$

Applying Newton's second law:

$\ddot{x} = -\frac{ Q q x}{8 \pi m R^3}$ $\implies \omega^2 = -\frac{ Q q}{8 \pi m R^3}$ $\implies T = 2 \pi \sqrt{\frac{8 \pi m R^3}{Qq}}$

Plugging in values:

$\implies \boxed{T = 2 \pi \sqrt{8 \pi}}$

I am very impressed by the analytical solution provided by @Karan Chatrath . I will provide a numerical solution. For each time step, the program numerically integrates to find the force. So it is a double integral, with one integral being spatial and the other integral being temporal. This is very computationally inefficient, but it is also very convenient for me. To ensure good results I ran three different cases (time step $\Delta t$ , number of ring partitions $N$ , and initial displacement from the ring center $x_0$ ).

Trial 1
$\Delta t = 10^{-3} \\ N = 10^4 \\ x_0 = 10^{-2} \\ \text{Calculated period } \approx 31.5$

Trial 2
$\Delta t = \frac{1}{2} \times 10^{-3} \\ N = 2 \times 10^4 \\ x_0 = 10^{-2} \\ \text{Calculated period } \approx 31.3$

Trial 3
$\Delta t = \frac{1}{2} \times 10^{-3} \\ N = 2 \times 10^4 \\ x_0 = 2 \times 10^{-2} \\ \text{Calculated period } \approx 31.4$

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import math

Num = 20000
dt = 0.0005

m = 2.0
Q = 2.0
R = 1.0
e0 = 1.0
q = 1.0

k = 1.0/(4.0*math.pi*e0)

dtheta = 2.0*math.pi/Num
dQ = Q*(dtheta/(2.0*math.pi))

###########################################

t = 0.0
count = 0

y = 0.0

x = 0.02
xd = 0.0
xdd = 0.0

while x >= 0.0:

    x = x + xd*dt
    xd = xd + xdd*dt

    Fx = 0.0
    theta = 0.0

    while theta <= 2.0*math.pi:

        xc = R*math.cos(theta)
        yc = R*math.sin(theta)

        Dx = x - xc
        Dy = y - yc

        D = math.hypot(Dx,Dy)

        ux = Dx/D
        uy = Dy/D

        dF = k*q*dQ/(D**2.0)
        dFx = dF*ux

        Fx = Fx + dFx

        theta = theta + dtheta

    xdd = Fx/m

    t = t + dt
    count = count + 1

    if count % 100 == 0:
        print t,x

print ""
print ""
print dt
print Num
print ""
print t
print (4.0*t)

#############################

#0.001    # x0 = 0.01
#10000

#7.876
#31.504

#############################


#0.0005   # x0 = 0.01
#20000

#7.825
#31.3

#############################

#0.0005    # x0 = 0.02
#20000

#7.8485
#31.394