Damped oscillations

Start with one release and three dampers

A 2 kg body is attached to a horizontal spring with stiffness k = 10 N/m. Choose a fixed coordinate increasing to the right, with equilibrium at x_eq = 1 m. Release the body at x = 1.2 m with zero velocity. What changes when the damping coefficient changes?

Supply a fixed inertial frame, constant mass, a massless linear spring and a fixed support. Weight and the support force cancel, giving zero vertical acceleration. There is no external drive or other horizontal force. These are an ideal model’s conditions, not measurements of a particular spring or damper.

Position, displacement, velocity, and acceleration and Newton’s laws and free-body diagrams prepare the signed quantities and force balance. The definitions needed here are also introduced locally. Work, kinetic energy, potential energy, and power is a published return path for the energy section.

A first reading can compare the three response curves and their table before checking the solution formulas. Geometry, trigonometry, and coordinate systems is a published return path for sine and cosine. Derivatives, Taylor series, and local approximation and Ordinary differential equations are optional return paths for the derivatives and equation.

Give each force its own rule

Let ξ = x − x_eq be signed displacement from equilibrium. The initial state is ξ(0) = 0.2 m and v(0) = 0 m/s. Since x_eq is fixed, v = dξ/dt; acceleration is a = dv/dt = d²ξ/dt². These are rates with respect to actual time t.

The spring force is F_s = −kξ. Supply a linear viscous damping force F_d = −cv, with a constant c ≥ 0 in kg/s. It opposes velocity and is zero when velocity is zero. This rule differs from a constant-magnitude sliding-friction rule, and is not a model of every kind of drag.

The net horizontal force gives ma = −kξ − cv, or mξ″ + cξ′ + kξ = 0. A prime here denotes a time derivative. Every term has units of newtons. There is no extra “force of motion” added to this equation.

Initially F_s = −2 N and F_d = 0 N, so a(0) = −1 m/s² for all three cases. Being released at rest does not make the displaced body an equilibrium state.

Identify three coefficient regimes

For positive constant m and k, the undamped frequency parameter is ω₀ = √(k/m). Define γ = c/(2m), in s⁻¹. The boundary coefficient is c_crit = 2√(mk). Here c_crit = 4√5 kg/s, approximately 8.944 kg/s.

If 0 < c < c_crit, a nonzero unforced response is underdamped: a decreasing exponential multiplies a sine-and-cosine combination. At c = c_crit it is critically damped, with a repeated exponential rate. At c > c_crit it is overdamped, with two distinct real exponential rates. The initial state selects their coefficients.

One way to check this classification is to try ξ = A e^(rt), where A has units of metres and r has units of s⁻¹. Substitution gives mr² + cr + k = 0. Its rates are r = −γ ± √(γ² − k/m). The repeated-rate case also needs the independent solution t e^(−γt), multiplied by a coefficient in m/s; repeating the same exponential twice would not use both initial data.

For c = 4 kg/s the rates are −1 ± 2i s⁻¹; for c = 4√5 kg/s the repeated rate is −√5 s⁻¹; for c = 12 kg/s they are −1 s⁻¹ and −5 s⁻¹. Here i² = −1. The complex rates are a compact mathematical representation; the displacement plotted below is real. Complex-number algebra is optional for this first comparison.

Compare responses with the same initial state

Use the dimensionless time τ = t/(1 s). The exponential arguments and the numerical radian phase 2τ are dimensionless. Both initial data select these exact model solutions; the three cases differ only in c.

Underdamped, c = 4 kg/s: xi_u(t) = (1/5 m)*exp(-tau)*(cos(2*tau) + (1/2)*sin(2*tau)). Critically damped, c = 4√5 kg/s: ξ_c(t) = (1/5 m)(1 + √5τ)e^(−√5τ). Overdamped, c = 12 kg/s: ξ_o(t) = (1/4 m)e^(−τ) − (1/20 m)e^(−5τ).

Each starts at ξ = 0.2 m with v = 0 m/s and a = −1 m/s². The underdamped curve crosses equilibrium; the other two selected curves stay positive and approach it. That last observation depends on the supplied initial data, not on damping labels alone.

The diagram uses one common time and displacement scale for the three responses. In its lower panel, the same underdamped displacement is compared with its positive and negative exponential bounds. Some late markers are close enough to overlap; the tables preserve the separate values. All decimal entries below are rounded evaluations of exact model formulas, not measured precision.

Separate the bound, crossings and turns

For the underdamped solution, v(t) = −(1/2 m/s)e^(−τ) sin(2τ). A zero of displacement is generally not a zero of velocity. At the first equilibrium crossing the spring force is zero, but velocity is negative and the damping force points right.

The coefficients inside its trigonometric combination are C = 1/5 m and D = 1/10 m. Their combined magnitude is R = √(C² + D²) = √5/10 m, approximately 0.223607 m. Thus |ξ_u(t)| ≤ R e^(−τ). The dashed curves are bounds, not the body’s displacement or curves obtained by connecting its actual peaks. In particular, R is larger than the initial displacement 0.2 m.

Define ψ = arctan(2), the numerical radian angle between 0 and π/2 whose tangent is 2. The two displayed positive displacement zeros occur at t = ((π − ψ)/2) s and ((2π − ψ)/2) s, approximately 1.017222 s and 2.588018 s. Their velocities have opposite, nonzero signs.

The first positive local minimum occurs instead at t = (π/2) s, approximately 1.570796 s. There ξ = −(1/5 m)e^(−π/2) and v = 0 m/s. Damping force is zero at that instant, but the spring force points right. The table’s exact zeros use these analytic conditions; machine sine roundoff does not turn them into small physical residuals.

Account for mechanical energy

For the chosen mass and spring, mechanical energy is E = (1/2)mv² + (1/2)kξ². Initially it is 0.2 J: all spring energy and zero kinetic energy. The two forms can exchange while their total decreases.

Differentiating within this constant-m, constant-k, unforced model gives dE/dt = v(ma + kξ) = −cv² ≤ 0. The damping force’s power is F_d v = −cv², in watts. This account tracks the body–spring mechanical energy. A description of the damper and surroundings would also need their variables and energy transfers.

At a turning point v = 0, the instantaneous energy derivative is zero. This does not make E constant over an interval. Nor is the exact instantaneous E(t) generally just E(0)e^(−2γt): kinetic and potential terms have different phase factors.

The nonzero initial state approaches ξ = 0 and v = 0 as t grows. It does not reach that whole state at a finite time in this linear model: uniqueness would otherwise make it the constant equilibrium solution. Crossing ξ = 0 with nonzero velocity is a different event.

Test overdamping with a different release

Keep the overdamped parameters m = 2 kg, k = 10 N/m and c = 12 kg/s, but supply ξ(0) = 0.2 m and v(0) = −2 m/s. The larger initial velocity points toward equilibrium. This is a new initial-value problem, not a hidden change to the plotted cases.

Write ξ(t) = A e^(−τ) + B e^(−5τ). The initial conditions give A + B = 1/5 m and −A − 5B = −2 m, hence A = −1/4 m and B = 9/20 m. The selected displacement is ξ(t) = −(1/4 m)e^(−τ) + (9/20 m)e^(−5τ).

It crosses equilibrium at t = (ln(9/5)/4) s, approximately 0.146947 s. At that crossing v = −(1 m/s)e^(−τ), which is nonzero. Overdamping prevents repeated oscillatory crossing in this unforced two-real-rate model, but it does not prohibit a single crossing for every initial state.

Keep frequency and validity conditions visible

In the underdamped case the sine-and-cosine factor has angular frequency 2 rad/s and repeats after π s. The complete displacement satisfies ξ_u(t + π s) = e^(−π)ξ_u(t), so it is not a periodic motion. A repeat interval of a carrier factor is different from a period of the complete decaying response.

If c = 0 with the same mass, spring and initial state, the model gives ξ(t) = (1/5 m)cos(√5τ), constant mechanical energy 0.2 J and a nonconstant-motion period (2π/√5) s. If both initial displacement and velocity are zero, any of the three damping cases gives the constant equilibrium solution.

The formulas are mathematical solutions for t ≥ 0 s; the drawing displays only 0 ≤ t ≤ 4 s. A real application must check the spring’s linear range, whether a velocity-proportional damper is suitable, other forces and the inertial-frame approximation. A general claim about the “fastest return” also needs a specified initial state and a return criterion or tolerance; the coefficient label alone supplies neither.

Worked examples

Try it yourself

1. For the plotted release, does increasing c change the net force at exactly t = 0 s? Find that force and acceleration.

   Show the answer and reasoning

   No. F_net(0) = −2 N and a(0) = −1 m/s² for every plotted c.

   The stipulated v(0) = 0 makes −cv zero at release. The common displacement and stiffness give −kξ = −2 N. Later velocities and damping forces can differ.

2. Keep m = 2 kg and k = 10 N/m, but choose c = 8 kg/s. Which regime is it, and what is the numerical radian carrier frequency?

   Show the answer and reasoning

   Underdamped, with γ = 2 s⁻¹ and carrier angular frequency 1 rad/s.

   The boundary is 4√5 kg/s, and 8 < 4√5. Compute γ = c/(2m) = 2 s⁻¹; k/m − γ² = 5 s⁻² − 4 s⁻² = 1 s⁻². This frequency belongs to the oscillatory factor, not a period of the full damped motion.

3. For the plotted underdamped release, may the bound coefficient R be replaced by the initial displacement 0.2 m without checking the sine coefficient? Find the correct R.

   Show the answer and reasoning

   No. R = √5/10 m, approximately 0.223607 m.

   The same initial velocity condition supplies D = 1/10 m because differentiating the exponential factor contributes −C/(1 s). Thus R² = (1/5 m)² + (1/10 m)². Using only C omits the second part of the state.

4. At the first underdamped equilibrium crossing, which of F_s and F_d is zero, and which way does the nonzero force point?

   Show the answer and reasoning

   F_s = 0 N. F_d is nonzero and points right.

   The crossing has ξ = 0 and v < 0. Hence −kξ = 0, while −cv > 0 for c = 4 kg/s. At the later displacement minimum the roles reverse: v = 0, with a positive spring force.

5. At the first underdamped minimum, dE/dt = 0. Does this mean that E is constant over the whole interval 0 ≤ t ≤ 4 s?

   Show the answer and reasoning

   No. A zero instantaneous derivative at the turn does not give constant energy on the interval.

   The identity dE/dt = −cv² holds throughout the unforced model. Velocity is nonzero on portions of that interval and c = 4 kg/s, so energy decreases there. The initial E is 0.2 J.

6. For the plotted underdamped case, compare ξ_u(0) and ξ_u(π s). Is π s a period of the complete motion?

   Show the answer and reasoning

   ξ_u(0) = 0.2 m; ξ_u(π s) = (1/5 m)e^(−π), approximately 0.008643 m. No.

   The carrier trigonometric factor repeats, but the exponential factor multiplies displacement by e^(−π). Unequal endpoint values already rule out that proposed period; a nonzero decaying motion cannot have a positive period.

What to carry forward

• A damping force needs a stated rule; linear viscous damping opposes velocity and differs from a constant-magnitude friction law.
• Damping regimes describe the equation’s rates, while both initial data select the response and its possible crossing.
• An exponential bound, a displacement zero and a turning point are three different objects.
• For this unforced ideal model mechanical energy has derivative −cv², and its loss does not imply arrival at the whole equilibrium state in finite time.
• Use explicit units, common axes, a finite displayed domain and the force model’s validity conditions when comparing responses.