The Eternal Ring!

Mine is a different approach from that of @Steven Chase

Consider the left end of the wire to be the origin. Say the ring has moved a distance $x$ from the left end. At this instant, the potential energy of the system can be computed as such:

$V = -\int_{0}^{L}\frac{KQ \ \lambda \ dp}{\sqrt{R^2 + (p-x)^2}}$

This integral transforms to:

$V = -KQ \ \lambda \int_{-x}^{L-x}\frac{dz}{\sqrt{R^2 + z^2}} = -KQ \ \lambda \left(\sinh^{-1}\left(\frac{L-x}{R}\right) - \sinh^{-1}\left(\frac{-x}{R}\right)\right)$

At the instant when the ring is released from rest, the potential energy is:

$V_o = -KQ \ \lambda \int_{0}^{L}\frac{dz}{\sqrt{R^2 + z^2}}=-KQ \ \lambda \sinh^{-1}\left(\frac{L}{R}\right)$

At a general time $t$ , applying the law of conservation of energy leads to:

$\frac{M\dot{x}^2}{2} + V = V_o$ $\implies \frac{M\dot{x}^2}{2} = KQ \ \lambda \left(\sinh^{-1}\left(\frac{L-x}{R}\right) - \sinh^{-1}\left(\frac{-x}{R}\right)- \sinh^{-1}\left(\frac{L}{R}\right) \right)$ $\implies \dot{x} = \sqrt{\frac{2KQ \ \lambda}{M} \left(\sinh^{-1}\left(\frac{L-x}{R}\right) - \sinh^{-1}\left(\frac{-x}{R}\right)- \sinh^{-1}\left(\frac{L}{R}\right) \right)}$

Separating the variables and integrating from $0$ to $L$ gives half the time period.

$\implies T=2\int_{0}^{L} \frac{dx}{\sqrt{\frac{2KQ \ \lambda}{M} \left(\sinh^{-1}\left(\frac{L-x}{R}\right) - \sinh^{-1}\left(\frac{-x}{R}\right)- \sinh^{-1}\left(\frac{L}{R}\right) \right)}} \approx 6.3$

This last step was solved numerically.

Fun problem. I modeled this using numerical integration over time. On every time step, the program integrates over the length of the rod to get the total force. Due to symmetry, the ring can be modeled as an equivalent point charge a distance $R$ from the rod axis, and we consider only the force parallel to the rod axis. My highest resolution simulation uses a time step of $10^{-4}$ seconds and a spatial step (on the rod) of $10^{-7}$ meters. The convergence is quite good, and the time period comes out to $T \approx 6.304$ . Note that this simulation does not require or use series expansions.

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import math

# Constants

lam = 10.0**(-7.0)
L = 10.0**(-3.0)
Q = -10.0**(-6.0)
R = 10.0**(-2.0)
m = 0.9
k = 9.0 * (10.0**9.0)

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

# Spatial and temporal resolution parameters

dt0 = 10.0**(-3.0)
dxr0 = L/1000.0

scale = 10.0

dt = dt0/scale
dxr = dxr0/scale

dq = dxr*lam

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

# Numerical integration
# Ring is modeled as an equivalent point at (x,R)
# Infinitesimal length of rod is at (xr,0)

t = 0.0

y = R
yr = 0.0

x = 0.0
xd = 0.0
xdd = 0.0

while xd >= 0.0:

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

    Fx = 0.0

    xr = 0.0

    while xr <= L:

        Dx = x - xr
        Dy = y - yr

        D = math.hypot(Dx,Dy)

        ux = Dx/D

        dF = k*Q*dq/(D**2.0)

        dFx = dF*ux

        Fx = Fx + dFx

        xr = xr + dxr

    xdd = Fx/m

    t = t + dt

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


print dt
print dxr
print scale
print ""
print t
print (2.0*t)
print ""
print (x/L)

#>>> 
#0.001
#1e-06
#1.0

#3.153
#6.306

#0.999783603967
#>>> ================================ RESTART ================================
#>>> 
#0.0005
#5e-07
#2.0

#3.152
#6.304

#1.00039213401
#>>> ================================ RESTART ================================
#>>> 
#0.0002
#2e-07
#5.0

#3.1522
#6.3044

#0.999956744911
#>>>
#>>> ================================ RESTART ================================
#>>> 
#0.0001
#1e-07
#10.0

#3.1519
#6.3038

#1.00007838511
#>>> 

Force on ring by rod can be given by $F = \frac{1}{4\pi\epsilon_o}\frac{\lambda Q}{R}\bigg(\frac{R}{\sqrt{R^2+x^2}}-\frac{R}{\sqrt{R^2+(L-x)^2}}\bigg)$ $F = \frac{1}{4\pi\epsilon_o}\frac{\lambda Q}{R}\bigg(\frac{1}{\sqrt{1+(\frac{x}{R})^2}}-\frac{1}{\sqrt{1+(\frac{L-x}{R})^2}}\bigg)$

As maximum value of $(\frac{x}{R})^2 = 0.01$ , in binomial expansion of $\bigg(1-\frac{x}{R}\bigg)^{-\frac12}$ , we can ignore powers of $x$ greater than 2. Same is true for $(\frac{L-x}{R})$ term.

$F = \frac{1}{4\pi\epsilon_o}\frac{\lambda Q}{R}\bigg(\bigg(1-\frac12\bigg(\frac{x}{R}\bigg)^2\bigg)-\bigg(1-\frac12\bigg(\frac{L-x}{R}\bigg)^2\bigg)\bigg)$ $\boxed{F = \frac{1}{4\pi\epsilon_o}\frac{\lambda QL}{R^3}\bigg(\frac{L}2-x\bigg)}$

So, above motion represents SHM about $x=L/2$ . Force constant can be given by

$k = \frac{1}{4\pi\epsilon_o}\frac{\lambda QL}{R^3}$

$\boxed{k = 0.9}$

So, time period $T$ can be given as

$T = 2\pi\sqrt{\frac{m}{k}}$

$T = 2\pi \text{ sec}$

$\color{#3D99F6}{\boxed{T = 6.283 \text{ sec}}}$