A particle of mass 0.25 kg moves along a line under the force F = -(49 N/m) x…

(a) $\omega = \sqrt{\dfrac{49\ \text{N/m}}{0.25\ \text{kg}}} = 14\ \text{rad/s}$, so $T = \dfrac{2\pi}{14\ \text{rad/s}} = 0.449\ \text{s}$.

(b) The force is largest at the ends: $kA = 49\ \text{N/m} \times 0.050\ \text{m} = 2.45\ \text{N}$.

(c) $a = -\dfrac{k}{m}x = -\dfrac{49\ \text{N/m}}{0.25\ \text{kg}} \times 0.030\ \text{m} = -5.88\ \text{m/s}^2$.