Question

An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spring stretches 7.7 cm before it reaches its equilibrium position. If this system is allowed to oscillate, what will be its frequency? (Give your answer to the nearest 0.1 Hz)

Answer 1

In equilibrium position,
mg = k * ΔL
k = (mg) / (ΔL)
k = (8 * 9.81) / (0.077) N/m
k = 1019.22 N/m
frequency, f
f = (1/2π) √(k/m)
f = (1/2π) √(1019.22/8)
f = 1.7973 Hz