foundation28 min studyLesson 7 in Physics

Harmonic motion and the oscillator model

Derive oscillation from a restoring force, connect phase to initial data, and distinguish free motion from damping and driving.

Before you begin

On this page

The question for today

What you’ll understand

Explain why a linear restoring force produces sinusoidal motion.
Relate frequency, phase, energy, and initial conditions.
Interpret damping as an explicit change to the ideal model.

An oscillator is more than a mass on a spring. Near a stable equilibrium, many smooth forces are approximately proportional to displacement, making the same equation describe a pendulum, a vibrating molecule, or one mode of a field. Understanding this shared model now will make the quantum harmonic oscillator and the mode expansion of fields much less mysterious.

A stable equilibrium supplies the restoring tendency

Choose displacement zero at a stable equilibrium. For an ideal spring, a positive displacement produces a negative force, and a negative displacement produces a positive force. The force points toward equilibrium, but the mass usually passes through equilibrium because it still has velocity there. Position and velocity together specify the state; position alone does not tell you which way motion continues.

The harmonic approximation comes from keeping the leading nonzero curvature of a smooth potential near its minimum. It can fail for large excursions, slack springs, or contact with a stop. A pendulum is approximately harmonic only when its angular excursion is small enough to replace sine of the angle by the angle in radians. The approximation has a range, not a universal guarantee.

m\ddot x=-kx,\qquad \omega_0=\sqrt{\frac{k}{m}},\qquad x(t)=A\cos(\omega_0t+\phi)

For an ideal, undriven, undamped spring with k > 0 and constant mass m, ω₀ is angular frequency in rad/s. A is the nonnegative amplitude, and phase φ is determined with A by the initial position and velocity.↔ Scroll wide equations horizontally.

Phase connects the graph to the initial state

Amplitude sets the largest displacement; angular frequency sets how quickly phase advances. The period is the time required for one complete cycle, not the time from one turning point to the opposite one. Frequency in hertz counts cycles per second and therefore differs from angular frequency by a factor of two pi. Write the units next to each to prevent mixing them.

Differentiating the displacement shifts the phase of velocity. Speed is greatest at equilibrium, where spring energy is smallest, and vanishes at turning points, where spring energy is largest. Acceleration is opposite displacement and is greatest in magnitude at the turning points. These relationships are a consistency check on a sketch, not just separate formulas to memorize.

T=\frac{2\pi}{\omega_0},\qquad E=\frac12m\dot x^2+\frac12kx^2=\frac12kA^2

The energy equality is constant only for the undamped, undriven ideal oscillator. Energy has units joules, and amplitude must be expressed in length units consistent with k. The period is independent of amplitude only within this harmonic model.↔ Scroll wide equations horizontally.

Damping and driving change the question

A resistive force proportional to velocity adds damping. In the underdamped regime, oscillations continue with a decaying envelope and a frequency below the undamped natural frequency. Critical damping separates oscillatory decay from overdamped return. Overdamping does not mean faster return in every useful sense; a strongly overdamped system can approach equilibrium very slowly.

An external periodic drive can supply energy that damping removes. Resonance concerns the response of this driven system and depends on damping and on which response is measured. A free decaying oscillator is not itself a resonance demonstration. Later, coupled oscillators form normal modes; quantizing those modes supplies the ladder of excitations used for free quantum fields, while nonlinear interactions couple them.

\ddot x+2\zeta\omega_0\dot x+\omega_0^2x=0,\qquad \omega_d=\omega_0\sqrt{1-\zeta^2}

This free damped model uses dimensionless damping ratio ζ. The displayed oscillation frequency ωd applies only for 0 ≤ ζ < 1; critical and overdamped solutions have different forms. There is no external driving term in this equation.↔ Scroll wide equations horizontally.

Work it through

Predict motion from a release

A $0.50\,\mathrm{kg}$ mass attached to an ideal spring of stiffness $8.0\,\mathrm{N\,m^{-1}}$ is released from rest at $x=0.10\,\mathrm{m}$ . Find angular frequency, period, maximum speed, and total energy.

Compute $\omega_0=\sqrt{k/m}=\sqrt{8.0/0.50}=4.0\,\mathrm{rad\,s^{-1}}$ . Release from positive maximum displacement sets $A=0.10\,\mathrm{m}$ and permits phase $\phi=0$ .
The period is $T=2\pi/4.0\approx1.57\,\mathrm{s}$ . Differentiating $x=A\cos(4t)$ gives $v=-4A\sin(4t)$ , so the largest speed is $A\omega_0=0.40\,\mathrm{m\,s^{-1}}$ .
At release all mechanical energy is spring energy: $E=kA^2/2=(8.0)(0.10)^2/2=0.040\,\mathrm{J}$ . At equilibrium, $mv_{\max}^2/2$ gives the same value.

The result

The model predicts $\omega_0=4.0\,\mathrm{rad\,s^{-1}}$ , $T\approx1.57\,\mathrm{s}$ , $v_{\max}=0.40\,\mathrm{m\,s^{-1}}$ , and $E=0.040\,\mathrm{J}$ . Damping would invalidate the constant-energy conclusion.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Explore three kinds of return

Change damping ratio ζ and natural frequency ω₀ in the analytic unit-mass model. Every reset starts at x = 1 with velocity zero. Compare underdamped, critical, and overdamped returns; there is no driving force or collision in this lab.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe same initial displacement can return by oscillation or by monotonic decay, depending on damping.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Total energy is fixed. At equilibrium the spring potential is minimum, so kinetic energy and speed are largest. At both turning points, velocity is momentarily zero even though restoring acceleration is nonzero.

Your turn

Try it yourself

Keep the spring stiffness fixed but increase the attached mass by a factor of four. How do angular frequency and period change? If the release amplitude stays the same, how does total energy change?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Inspect the mass dependence of ω₀ and the amplitude dependence of spring energy separately.

Compare with the solution

Since $\omega_0=\sqrt{k/m}$ , quadrupling mass halves angular frequency and doubles period. At release from rest with the same spring and amplitude, $E=kA^2/2$ is unchanged. Maximum speed halves so that the larger mass still has the same maximum kinetic energy. This comparison assumes both oscillators remain ideal and undamped.

Keep the idea

Take it with you

Linear restoring forces near stable equilibria produce harmonic motion.
Initial position and velocity determine amplitude and phase together.
Damping, driving, and large amplitudes must be modeled rather than hidden inside the ideal oscillator.

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