## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $k = 24.5~N/m$ (b) $T = 0.898~s$ (c) $v_{max} = 0.700~m/s$ The maximum speed occurs as the mass passes through the equilibrium position.
(a) We can find the spring constant as: $kx = mg$ $k = \frac{mg}{x}$ $k = \frac{(0.500~kg)(9.80~m/s^2)}{0.20~m}$ $k = 24.5~N/m$ (b) We can find the period of oscillation as: $T = 2\pi~\sqrt{\frac{m}{k}}$ $T = 2\pi~\sqrt{\frac{0.500~kg}{24.5~N/m}}$ $T = 0.898~s$ (c) We can find the maximum speed as: $v_{max} = A~\omega$ $v_{max} = A~\sqrt{\frac{k}{m}}$ $v_{max} = (0.100~m)~\sqrt{\frac{24.5~N/m}{0.500~kg}}$ $v_{max} = 0.700~m/s$ The maximum speed occurs as the mass passes through the equilibrium position.