## College Physics (4th Edition)

(a) When the speed is a maximum, the extension of the spring is $0.392~m$ (b) The maximum speed is $1.96~m/s$
(a) The speed is a maximum at the equilibrium point, which is the point where the upward force of the spring is equal in magnitude to the body's weight. We can find the extension of the spring at the equilibrium point: $kd = mg$ $d = \frac{mg}{k}$ $d = \frac{(0.60~kg)(9.80~m/s^2)}{15~N/m}$ $d = 0.392~m$ When the speed is a maximum, the extension of the spring is $0.392~m$ (b) We can find the maximum speed: $v_m = A~\omega$ $v_m = A~\sqrt{\frac{k}{m}}$ $v_m = (0.392~m)~\sqrt{\frac{15~N/m}{0.60~kg}}$ $v_m = 1.96~m/s$ The maximum speed is $1.96~m/s$