This system is a physical pendulum, whose period for small oscillations is given by $T = 2\pi\sqrt{I/(M_{tot}gd_{CM})}$. We calculate the required quantities: Total mass: $M_{tot} = m + m = 2m$. Moment of inertia about the pivot: $I = m x^2 + m l^2 = m(x^2+l^2)$. Distance from pivot to center of mass (CM): $d_{CM} = \frac{m \cdot x + m \cdot l}{m+m} = \frac{x+l}{2}$.

[Q1] Substituting these into the period formula:

$$T = 2\pi \sqrt{\frac{m(x^2+l^2)}{(2m) g \frac{x+l}{2}}} = 2\pi \sqrt{\frac{m(x^2+l^2)}{mg(x+l)}}$$ $$T = 2\pi \sqrt{\frac{x^2+l^2}{g(x+l)}}$$

[Q2] To minimize $T$, we must minimize the function $f(x) = \frac{x^2+l^2}{x+l}$ over the domain $x \in [0, l]$. We find the minimum by setting the derivative with respect to $x$ to zero:

$$\frac{df}{dx} = \frac{2x(x+l) - (x^2+l^2)(1)}{(x+l)^2} = 0$$

This requires the numerator to be zero:

$$2x^2 + 2xl - x^2 - l^2 = 0 \implies x^2 + 2xl - l^2 = 0$$

Using the quadratic formula for $x$:

$$x = \frac{-2l \pm \sqrt{(2l)^2 - 4(1)(-l^2)}}{2} = \frac{-2l \pm \sqrt{8l^2}}{2} = l(-1 \pm \sqrt{2})$$

Since $x$ must be a positive length, we choose the positive root:

$$x = l(\sqrt{2}-1)$$

This value is within the domain $[0,l]$ and corresponds to a minimum.