1. Dynamics Inside the Field (Turnaround):

When the rod enters a field region by a distance $x$, the charge inside is $q_{in} = \lambda x = \frac{q}{l}x$. The restoring force is:

$$ F = -q_{in} E = -\left( \frac{qE}{l} \right) x $$

This is the equation for SHM ($F = -m\omega^2 x$) with angular frequency:

$$ \omega = \sqrt{\frac{qE}{ml}} $$

The rod enters with velocity $v_0$, stops at max distance $x_m$, and exits. This corresponds to exactly half a period of SHM (from equilibrium to amplitude and back). However, in time domain of SHM cycle, going from $x=0$ to $x_{max}$ takes $T_{SHM}/4$. Returning takes another $T_{SHM}/4$.

Total time spent in one field interaction:

$$ t_{field} = \frac{T_{SHM}}{2} = \frac{1}{2} \left( \frac{2\pi}{\omega} \right) = \frac{\pi}{\omega} $$

2. Dynamics in the Gap (Coasting):

Between the fields, the rod travels a distance equal to the free space available. Since the rod has length $l$ and the gap is $L$, the center of mass travels a distance $L-l$ from the point it leaves the left field to the point it hits the right field.

Velocity in the gap is constant at $v_0$. From SHM relations, maximum velocity $v_0$ relates to amplitude $x_m$ by $v_0 = \omega x_m$.

Time for one gap traversal:

$$ t_{gap} = \frac{Distance}{Speed} = \frac{L-l}{v_0} = \frac{L-l}{\omega x_m} $$

3. Total Period Calculation:

One full oscillation cycle involves: Right Field Interaction $\to$ Coast Left $\to$ Left Field Interaction $\to$ Coast Right.

$$ T_{total} = 2 \times t_{field} + 2 \times t_{gap} $$ $$ T_{total} = 2 \left( \frac{\pi}{\omega} \right) + 2 \left( \frac{L-l}{\omega x_m} \right) $$

Factoring out $\frac{2}{\omega}$:

$$ T_{total} = \frac{2}{\omega} \left( \pi + \frac{L-l}{x_m} \right) $$

Substituting $\omega = \sqrt{\frac{qE}{ml}}$:

$$ T_{total} = 2 \sqrt{\frac{ml}{qE}} \left( \pi + \frac{L-l}{x_m} \right) $$