AP® Physics C: Mechanics · Unit 7 · Topic 7.2
The Mass-Spring Oscillator
A mass on a spring obeys m x'' = -kx, whose solution oscillates with angular frequency omega = √(k/m), so the period T = 2 π √(m/k) depends only on mass and stiffness, never on amplitude.
Newton's second law for a mass on a spring is m x'' = -kx: the restoring force is always proportional to displacement and points back toward equilibrium. Dividing by m gives x'' = -(k/m) x, the defining equation of simple harmonic motion. Its solution is a sine or cosine whose angular frequency is omega = √(k/m) — you can check this by plugging x = A cos(omega t) back into the equation and seeing that it only works when omega2 = k/m. From omega you get the period directly: T = 2 π / omega = 2 π √(m/k).
This tells you exactly why a heavier mass swings more slowly: more mass means more inertia resisting the spring's pull, so it takes longer to complete each swing even though the force law is identical. A stiffer spring (larger k) does the opposite — it yanks the mass back harder, shortening the period. Neither the amplitude nor the initial speed shows up anywhere in omega or T, which is the single most surprising fact in this topic.
This result sets up everything else in the unit. The same omega = √(k/m) reappears when you compute energy, since the oscillator trades kinetic and potential energy back and forth at a rate fixed by this same frequency. It also reappears, in disguise, in the pendulum equation and in driven oscillations, where matching an external driving frequency to this natural omega produces resonance.
Key terms
- Angular frequency (omega)
- The rate of oscillation in radians per second, equal to √(k/m) for a mass-spring system.
- Period (T)
- The time for one complete oscillation, equal to 2 π √(m/k).
- Spring constant (k)
- A measure of spring stiffness, defined by the force law F = -kx.
- Simple harmonic motion
- Motion where acceleration is proportional to and opposite the displacement from equilibrium.
Watch it explained
- How it worksHow does solving m times the second derivative of x equals -kx give omega as the root of k over m?
- Worked exampleFind the period of a 0.4 kg mass oscillating on a spring with k = 100 N/m.
- Worked exampleFind the spring constant needed to make a 2 kg mass oscillate with a period of 0.5 s.
- Common mistakeWhy doesn't pulling the mass back farther make each cycle take longer?
Check yourself
End of topic.