Solution:

Using \( v_{\text{max}} = A \omega = 5 \) and \( a_{\text{max}} = A \omega^2 = 50\pi \), we get \( \omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{50\pi}{5} = 10\pi \, \text{rad/s} \). The time period is \( T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = \frac{1}{5} \, \text{s} \).