Suspended from a nail in a wall, a closed circular wire experiences...

The time period of a physical pendulum is given by:

$T = 2\pi \sqrt{\dfrac{I}{Mgd}}$

For a closed circular wire (ring) of mass $M$ and radius $R$ oscillating in its own plane about a point on its circumference, the moment of inertia is:

$I = I_{cm} + MR^2 = MR^2 + MR^2 = 2MR^2$

The distance of the center of mass from the suspension point is $d = R$. Substituting these into the time period formula:

$T = 2\pi \sqrt{\dfrac{2MR^2}{MgR}} = 2\pi \sqrt{\dfrac{2R}{g}}$

Given $T = 2 \text{ s}$ and $g = \pi^2 \text{ m/s}^2$:

$2 = 2\pi \sqrt{\dfrac{2R}{\pi^2}} \Rightarrow 1 = \sqrt{2R} \Rightarrow 2R = 1 \text{ m}$

The farthest point on the wire from the point of suspension is diametrically opposite, at a distance $L = 2R = 1 \text{ m}$.

The angular frequency of oscillation is:

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{2} = \pi \text{ rad/s}$

The angular amplitude in radians is:

\theta_0 = 2^{}^{{}^{\circ}} = 2 \times \dfrac{\pi}{180} = \dfrac{\pi}{90} \text{ rad}

The maximum angular velocity of the pendulum as it passes through the mean position is:

\Omega_\max = \omega \theta_0 = \pi \times \dfrac{\pi}{90} = \dfrac{\pi^2}{90} \text{ rad/s}

The maximum linear speed of the farthest point is:

v_\max = L \Omega_\max = 1 \times \dfrac{\pi^2}{90} = \dfrac{\pi^2}{90} \text{ m/s}

Converting to cm/s:

v_\max = \dfrac{\pi^2}{90} \times 100 = \dfrac{10\pi^2}{9} \text{ cm/s}

Using $\pi^2 \approx 9.87$:

v_\max \approx \dfrac{10 \times 9.87}{9} \approx 10.96 \text{ cm/s} \approx 11 \text{ cm/s}

Answer: $11 \text{ cm/s}$