Sidney P. answered • 11/08/21

Astronomy, Physics, Chemistry, and Math Tutor

a) When weight balances spring force, mg = kx, x = mg/k = (0.500 kg) (9.81 m/s^{2}) / (50.0 N/m) = 0.0951 m = 9.51 cm below spring's equilibrium.

b) 2π f = √ (k/m) = √ (50.0 kg/s^{2}) / (0.500 kg) = 10.0 rad/s and f = 10.0/2π = 1.59 Hz.

c) Starting at 10 cm below equilibrium, x = -10.0 cos(ωt), where ω = 2πf = 10.0 rad/s. At 1/3 period, ωt = 2π/3 radians and cos(2π/3) = -1/2. Then x = (-10.0) (-1/2) = +5.0 cm above equilibrium.

d) Dimensions of 2π √(m/k): √[kg ÷ (N/m = kg/s^{2})] = √(s^{2}) = s.