a small object of mass 1.0kg is hanging from an ideal spring with spring constant a at 0.1m above the equilibrium position and it’s velocity is -0.52m/s. What is the period of oscillation?

Answer 1

Hi Jennyfer,

I think you are asking about the inherent oscillation frequency of a spring-mass system. This is an inherent property of the system only determined by mass m and spring constant k and has nothing to do with the state of the mass (its position and velocity). Shortly speaking, the answer is ω = √(k/m).

The derivation is below:

We first choose the mass as our objective and set the vertical down direction as +x while the equilibrium is the origin. By analyzing the forces applied to the mass, we know that the mass has its gravity mg point to the down direction while the force from the spring is -kx (because the spring force always points to the equilibrium). By Newton's law:

mg - kx = mx''

This is a second-order linear inhomogeneous ordinary differential equation. We first solve the homogeneous version by omitting the mg part. We can use a trial solution e^st. By substituting it into the equation, we have:

s² + k/m = 0

Which means:

s = ± i√(k/m)

So the solution should be a linear combination of e^it√(k/m) and e^-it√(l/m). Remember e^iωt is a periodic function, so the frequency is ω = √(k/m).