The motion is angular simple harmonic motion (SHM). The angular frequency $\omega_{ang}$ is related to the period $T$ by $\omega_{ang} = 2\pi/T$.

[Q1] The maximum angular speed $\omega_{max}$ occurs at zero angular displacement ($\theta=0$) and is given by:

$$\omega_{max} = \theta_m \omega_{ang} = \theta_m \frac{2\pi}{T}$$

[Q2] The angular speed $\omega$ at an arbitrary angular displacement $\theta$ is:

$$\omega = \omega_{ang}\sqrt{\theta_m^2 - \theta^2} = \frac{2\pi}{T}\sqrt{\theta_m^2 - \theta^2}$$

[Q3] The magnitude of the angular acceleration $|\alpha|$ at displacement $\theta$ is:

$$|\alpha| = \omega_{ang}^2 |\theta| = \left(\frac{2\pi}{T}\right)^2 |\theta|$$

Given $\theta_m = \pi$ rad and $T = 0.500$ s. First, calculate the angular frequency:

$\omega_{ang} = 2\pi / 0.500 \text{ s} = 4\pi$ rad/s.

[Q1] $\omega_{max} = (\pi \text{ rad})(4\pi \text{ rad/s}) = 4\pi^2$ rad/s [Q2] $\omega = (4\pi \text{ rad/s}) \sqrt{(\pi \text{ rad})^2 - (\pi/2 \text{ rad})^2} = 4\pi \sqrt{\frac{3\pi^2}{4}} = 2\sqrt{3}\pi^2$ rad/s [Q3] $|\alpha| = (4\pi \text{ rad/s})^2 (\pi/4 \text{ rad}) = 4\pi^3 \text{ rad/s}^2$