Spring Constant & Oscillatory Motion

Question

A 500.0 gram mass is hung from a spring having a spring constant of 50.0 N/m.

a) How much does the spring stretch when the mass is at rest and it is not oscillating? This is called the equilibrium position of the hanging mass.

b) Suppose now that the hanging mass is stretched an additional 10.0 cm beyond its equilibrium position and it is released from rest at t= 0 seconds. The mass now begins to oscillate with an amplitude of 10.0 cm. What is the frequency of the oscillating mass in Hertz?

c) What is the position of the mass after one third of a period. Take x=0 to be its equilibrium position (when the mass is at rest). Remember the mass begins at x=10.0 cm.

d) Show by dimensional analysis that the period has units of seconds in the equation 𝑻= 𝟐𝝅√(𝒎/𝒌)

Answer 1

a) When weight balances spring force, mg = kx, x = mg/k = (0.500 kg) (9.81 m/s²) / (50.0 N/m) = 0.0951 m = 9.51 cm below spring's equilibrium.

b) 2π f = √ (k/m) = √ (50.0 kg/s²) / (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/s²)] = √(s²) = s.