$A$ $5\; kg$ collar is attached to a spring of spring constant $500\; N m^{-1}$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10.0\; cm$ and released. Calculate
$(a)$ the period of oscillation.
$(b)$ the maximum speed and
$(c)$ maximum acceleration of the collar.

(N/A) The period of oscillation is given by $T = 2 \pi \sqrt{\frac{m}{k}}$.
Substituting the values,$T = 2 \pi \sqrt{\frac{5.0\; kg}{500\; N m^{-1}}} = 2 \pi \sqrt{0.01} = 2 \pi \times 0.1 = 0.2 \pi \; s \approx 0.63\; s$.
$(b)$ The maximum speed is given by $v_{max} = A \omega$,where $\omega = \sqrt{\frac{k}{m}}$.
$v_{max} = 0.1\; m \times \sqrt{\frac{500\; N m^{-1}}{5\; kg}} = 0.1 \times \sqrt{100} = 0.1 \times 10 = 1.0\; m s^{-1}$.
$(c)$ The maximum acceleration is given by $a_{max} = \omega^2 A$.
$a_{max} = \left(\frac{k}{m}\right) A = \left(\frac{500\; N m^{-1}}{5\; kg}\right) \times 0.1\; m = 100 \times 0.1 = 10\; m s^{-2}$.