Question

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached. It is then displaced downward an additional 5.0 cm and released. Its position as a function of time is approximately?

Answer 1

The spring constant is
k = 5.0kg*9.8 m/s^2/0.10 m = 490 N/m
The oscillation angular frequency is
w = sqrt (k/M) = 9.90 rad/s

The mass vibrates with an amplitude of 5 cm about the equilibrium position (10 cm stretch). Measured about that position, the deflection (positive downward) is
Y = 5.0 cm * cos (wt)= 5 cos (9.90 t)