Let the charge q₀ be given a small displacement y towards the positive x axis. Let the instantaneous position of q₀ be x(t) and x(0) = y. Also, the initial velocity of q₀ is 0.
Let K = (1/4πε₀)
The net force acting on q₀ will be in the negative x direction, as the force exerted by the charge q at x = a exerts higher force.
Net Force acting on q₀ = - K q q₀ [ 1/(a-x)² - 1/(a+x)² ]
F = m d² x(t)/ dt² = - K q q₀ [ 4 a x /(a² - x²)² ]
d² x(t) / dt² = - 4 a k q q₀/m x / [ a⁴ * (1 - x²/a²)² ]
Let ω² = 4 k q q₀ / (m d³) = q q₀ /(π ε₀ m d³)
for x , y << a,
d² x(t) / dt² ≈ - ω² x [ 1 + 2 x²/a² ] ≈ -ω² x(t)
This is a simple harmonic motion. SHM
The angular frequency of oscillation = ω = √[ q q₀ / (π ε₀ m d³) ]
T = time period = 2π/ω = 2π √[ (π ε₀ m d³) / (q q₀) ]
the instantaneous displacement is : x(t) = y Cos (ω t)
as x = y , at t = 0 sec.
The amplitude of oscillation is y.
we can calculate the velocity, acceleration, and energy in the oscillations.
