Question

A spring stress 0.150 m when a 0.300 kg mass is gently suspended from it as fig below. The spring is then set up horizontally with 0.300 kg mass resting on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point, and released from rest. Determine: a) the spring stiffness constant K; b) the amplitude of the horizontal oscillation A; c) the magnitude of the maximum velocity Vmax? d) the magnitude of the velocity V when the mass is 0.075 m from the equilibrium; e) the total energy

Answer 1

a F=kn = wg = kx = ing. 0.150 • 67 k 0-300 x9.8 Nm a 19.6 - spring stiffness constant k = 19.6./m Amplitude of horizontal oserllation. A= 0.100 m [maximum stretched position] Maximum velocity be max. a) - Yomon = KA2 = 196 x 0to? 0.300 19.6 x 0.1002 V 0-300 - 0.80829 m/s - Yox = 0.80899 wys. velocity at a distance a bom equilibrium a d) Now n=0.075 m. v 19.6 . 0.300 0.076272 = 0.534633 Tv = 0.534633831 m/s. e) Total energy = KA² = + x 196 x 0.100²] > 0.098 J.