1. Predict
At identical initial conditions and v₀ = 0, predict whether underdamped, critical and overdamped motion cross equilibrium.
Dynamics
Damped spring–mass system
Add a resistive force proportional to velocity. Compare underdamped, critically damped and overdamped motion with the same initial conditions.
Does increasing the damping always bring the system to equilibrium faster?
2. Run a controlled experiment
Set m = 1 kg, k = 4 N/m, x₀ = 0.8 m, v₀ = 0. For ζ = 0.25, record x, K, U and E at three times.
3. Compare and measure
Compare ζ = 0.25, 1 and 2. Estimate the earliest time after which |x| stays below 0.05 m until the end of observation; note the finite observation limit.
4. Examine the relationship
Relate energy loss to velocity: dE/dt = −bv². What is the loss rate at a turning point? Where is mechanical energy transferred?
5. Conclusions and limits
Test your initial interpretation with a different v₀. State the linear viscous-drag assumption and explain why damped motion does not repeat with the same amplitude.
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Spring–mass motion
Ready02 / LIVE MEASUREMENTS
Position x—m
Velocity v—m/s
Acceleration a—m/s²
Net force—N
Damping coefficient b—kg/s
Damping regime
Position versus time
Change the parameters
Kinetic, potential and total energy
Kinetic energy KPotential energy UMechanical energy E
Kinetic energy K—J
Potential energy U—J
Mechanical energy E—J
Transferred energy E₀ − E—J
K: solid blue line; U: dashed orange line; E = K + U: dotted lavender line. Graphs are drawn up to the selected instant. E₀ − E is the energy transferred to the surroundings by the resistive force.
Compare three damping regimes: position–time
Mass, stiffness, initial position and initial velocity are identical here; only damping ratio differs (ζ = 0.25, 1, 2). Curves build up to the selected time. The experiment above uses the selected ζ; this comparison uses three fixed ζ values.
Underdamped (ζ = 0.25)Critically damped (ζ = 1)Overdamped (ζ = 2)
Ask. Test. Explain.
Ask a question
Does stronger damping always mean faster settling?
Test it
Set v₀ = 0; inspect the three damping curves and the energy legend.
Interpret the result
Is absence of oscillation the same as conservation of mechanical energy?
Grid spacing: 0.25 m. The position scale is fixed; equilibrium is at x = 0.
The F arrow shows the direction of the net force; its length is not a quantitative force scale.
Equation of motion and its solution
Equation of motion and its solution
For an ideal horizontal spring, the restoring force is Fₛ = −k x. The position x is measured relative to equilibrium. The initial conditions are x(0) = x₀ and v(0) = v₀. The natural angular frequency is ω₀ = √(k/m), and the period is T₀ = 2π/ω₀.
In this experiment, friction is a linear viscous resistance of the form Fᵈ = −b v, not dry surface friction. We define b ≥ 0, β = b/(2m), ω₀ = √(k/m) and ζ = β/ω₀. The equation of motion is m ẍ + b ẋ + k x = 0.
Underdamped
0 < ζ < 1: the oscillation amplitude decreases exponentially. The damped angular frequency is ω_d = √(ω₀² − β²).
Critically damped
ζ = 1: the two characteristic roots coincide. The solution is the product of an exponential factor and a factor linear in time. There is no periodic oscillation in this regime.
Overdamped
ζ > 1: there are two distinct negative real roots. The motion is not periodic; at long times, the more slowly decaying exponential term dominates if its coefficient is nonzero. Depending on the initial velocity, the mass can cross the equilibrium position once.
In the overdamped solution, q = √(β² − ω₀²), r₊ = −β + q, r₋ = −β − q; C₊ = (v₀ − r₋x₀)/(r₊ − r₋), C₋ = (r₊x₀ − v₀)/(r₊ − r₋).
We use the standard characteristic-root solution for the linearly damped oscillator. Thornton and Marion, Classical Dynamics of Particles and Systems, 5th edition, Chapter 3, is a further-reading reference on this topic. The expressions here were derived independently; they have not been checked against a specific numbered equation in the book.
Dots denote derivatives with respect to time. In the expressions below, t = 0 is the selected initial instant.
Kinetic, potential and total energy
K = m v²/2, U = k x²/2 and E = K + U. Since the power delivered by the resistive force is −b v², mechanical energy cannot increase. Energy is not destroyed; it is transferred from the mechanical system described by the model to the surroundings.
Virial theorem: time averages
For stable, bound, conservative motion, an appropriate time average gives 2⟨K⟩ = ⟨x dU/dx⟩. For a harmonic potential, averaging over a full period gives ⟨K⟩ = ⟨U⟩ = E/2. This does not mean the energies are equal at every instant; we do not apply the same equality directly to damped transient motion.
Model and assumptions
One-dimensional horizontal motion, constant mass, a massless ideal Hookean spring and constant coefficients. Weight and the normal force balance vertically. There is no external driving force or measurement noise. The damped page uses only resistance proportional to velocity. Changing a parameter starts the experiment with new initial conditions; the transition during the parameter change is not modelled.
Symbols and derivatives: how to read them
| Notation | Meaning and unit |
|---|---|
| Time; s denotes seconds. The light clock uses ns = 10⁻⁹ s. | |
| Position relative to equilibrium (m); x₀ is initial position. Its sign indicates the side of equilibrium. | |
| First time derivative of position: velocity (m/s). One dot means one derivative with respect to time. | |
| Second time derivative of position: acceleration (m/s²), the rate of change of velocity. Two dots mean two time derivatives. | |
| Mass m (kg), spring stiffness k (N/m), viscous damping coefficient b (kg/s). Drag is −bv and spring force is −kx. | |
| ω₀ is natural angular frequency (rad/s); β = b/(2m) is decay rate (s⁻¹); ζ = β/ω₀ is dimensionless damping ratio. This β is not the light clock’s v/c. | |
| Damped angular frequency (rad/s) in the underdamped regime; used as an oscillation frequency only for ζ < 1. | |
| v₀ is initial velocity (m/s). A, B and C± are solution coefficients (m) fixed by initial position and velocity, not independent parameters. | |
| Two exponential decay rates (s⁻¹) for overdamping. C₊ + C₋ = x₀ and r₊C₊ + r₋C₋ = v₀ enforce the initial conditions. | |
| K is kinetic, U is spring potential and E is total mechanical energy (J). With damping, lost mechanical energy is transferred to the environment as thermal energy. | |
| Angle brackets indicate time averages, here over a complete period or an integer number of periods of undamped periodic motion. |
References and further reading
The books below offer further study of the relevant concepts. The simulation’s assumptions are stated separately above.
- Stephen T. Thornton, Jerry B. Marion. Classical Dynamics of Particles and Systems. Edition 5 · Chapter(s) 3. Cengage. View source ↗
- David Halliday, Robert Resnick, Jearl Walker. Fundamentals of Physics. Edition 12 · Chapter(s) 15. Wiley. View source ↗
- Hugh D. Young, Roger A. Freedman. University Physics with Modern Physics. Edition 15 · Chapter(s) 14. Pearson. View source ↗