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

For a closed circular wire (ring) of radius $R$ suspended from a point on its circumference and oscillating in its own plane, the moment of inertia about the point of suspension is $I = mR^2 + mR^2 = 2mR^2$.

The distance from the point of suspension to the center of mass is $d = R$.

The time period of this physical pendulum is given by:

$T = 2\pi \sqrt{\dfrac{I}{mgd}} = 2\pi \sqrt{\dfrac{2mR^2}{mgR}} = 2\pi \sqrt{\dfrac{2R}{g}}$

Given $T = 2 \text{ s}$ and $g = \pi^2 \text{ m/s}^2$, substituting these values yields:

$2 = 2\pi \sqrt{\dfrac{2R}{\pi^2}}$

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

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

The angular frequency of oscillation is $\omega = \dfrac{2\pi}{T} = \pi \text{ rad/s}$.

The angular amplitude is \theta_0 = 2^{}^{{}^{\circ}} = 2 \times \dfrac{\pi}{180} = \dfrac{\pi}{90} \text{ rad}.

At the extreme position, the angular velocity is zero, so the radial (centripetal) acceleration is zero. The acceleration is purely tangential.

The maximum angular acceleration at the extreme position is $\alpha = \omega^2 \theta_0$.

The linear acceleration of the farthest point is:

$a = L \alpha = L \omega^2 \theta_0$

Substituting the values:

$a = 1 \times \pi^2 \times \dfrac{\pi}{90} = \dfrac{\pi^3}{90} \text{ m/s}^2$

Converting to $\text{cm/s}^2$:

$a = \dfrac{100 \pi^3}{90} \text{ cm/s}^2 = \dfrac{10 \pi^3}{9} \text{ cm/s}^2$

Using $\pi \approx 3.14$, $\pi^3 \approx 31.0$:

$a \approx \dfrac{10 \times 31.0}{9} \approx 34.4 \text{ cm/s}^2 \approx 34 \text{ cm/s}^2$

In simple harmonic motion, the acceleration is always directed towards the mean position.

Answer: $34 \text{ cm/s}^2$ towards the mean position