Physics · Oscillations

Free damped harmonic oscillation

Q: What does β describe?

$\beta=\frac b{2m}$ describes the exponential envelope decay rate of a free oscillator with linear viscous resistance. Its unit is s⁻¹, whereas resistance coefficient b has unit kg/s. After time 1/β the envelope is 1/e of its initial magnitude; after ln 2/β it halves. β alone does not determine whether oscillation occurs: compare it with $\omega_0=\sqrt{\frac km}$. Underdamping requires $\beta<\omega_0$. Equivalent peaks separated by Td give β = ln(|x₁/x₂|)/Td. Sensor offset and peak identification must be correct; adjacent opposite-sign extrema have only half that spacing and cannot be substituted without adjusting the time interval.

Q: Why is ω_d smaller than the undamped frequency?

The free linear damped oscillator has characteristic equation $r^2+2\beta r+\omega_0^2=0$, with complex roots $-\beta\pm i\sqrt{\omega_0^2-\beta^2}$. Their imaginary part is ωd. Positive underdamping therefore gives $\omega_d<\omega_0$; as $\beta\to0$ they coincide. For $\omega_0=5\,\mathrm{rad/s}$ and $\beta=1\,\mathrm{s^{-1}}$, $\omega_d=\sqrt{24}\,\mathrm{rad/s}\approx4.899\,\mathrm{rad/s}$. The decrease is small for weak damping but becomes substantial near the critical threshold. A steadily driven oscillator instead follows drive frequency Ω, and displacement resonance frequency must not automatically be identified with ωd. All three describe different aspects of the model. Therefore first identify whether your observation is a free decay, a steady driven motion, or the largest response amplitude in a frequency sweep.

Q: When does the system still oscillate?

A free linear viscous oscillator has an oscillatory component when $\beta 0$; $\beta=0$ is the undamped limit. At $\beta=\omega_0$ the critical solution is $(C+Dt)e^{-\omega_0t}$. Above that threshold it consists of two real decaying exponential components. Those regimes have no repeating free oscillation period, although suitable initial velocity can still produce one equilibrium crossing. For $m=0.80\,\mathrm{kg}$ and $k=20\,\mathrm{N/m}$, the critical coefficient is $2\sqrt{mk}=8.00\,\mathrm{kg/s}$. Periodic external forcing can produce a steady periodic response even outside the underdamped regime. Before computing a free period, check the motion equation and damping threshold; an observed drive period alone does not establish underdamped natural motion.

Q: How do I find the time for the amplitude to halve?

Set A(t)/A₀ = 1/2 for the exponential envelope $A(t)=A_0e^{-\beta t}$. Taking logarithms gives t = ln 2/β, provided $\beta>0$ and the linear decay model applies. With $\beta=0.40\,\mathrm{s^{-1}}$ this is about 1.733 s; substitution gives e⁻⁰·⁶⁹³¹⁴⁷ = 1/2. The quantity being halved is the envelope, not instantaneous displacement x(t), which can cross zero earlier. Equivalent peaks follow the same exponential ratio. Do not apply the calculation unchanged to varying damping, dry friction or sustained driving. Exact instantaneous energy decay also does not generally follow that same simple exponential form throughout a cycle.

Free underdamped motion with linear viscous resistance has an exponential envelope and damped natural angular frequency $\omega_d=\sqrt{\frac km-\beta^2}$ .

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Formula

Calculate

x (t) = A_{0} e^{- βt} cos (ω_{d} t + φ)

x and A₀ in m · β in s⁻¹ · ωd in rad/s · t in s · φ in rad (dimensionless) · b in kg/s · k in N/m

Variables & units

Variables & units
Symbol	Meaning	Unit
$x(t)$	Displacement at time t	m
$A₀$	Initial envelope amplitude	m
$β$	Damping rate b/(2m)	s⁻¹
$ω_{\mathrm{d}}$	Damped angular frequency	rad/s
$φ$	Phase angle	rad

LaTeX

x(t)=A_0 e^{-\beta t}\cos(\omega_d t+\varphi)

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Damped cosine between two exponential envelopes. A₀ denotes initial envelope magnitude; displacement and its actual extrema depend on phase. — Underdamped linear model with chosen zero phase. Envelope ±A₀e⁻ᵝᵗ decays exponentially; the depicted motion does not start from rest. Instantaneous total energy is not generally E₀e⁻²ᵝᵗ.

Damped oscillation calculator

Find the displacement at any time, the period, the amplitude half-life or the damping ratio of a spring oscillator whose friction is proportional to its speed.

Calculation model

Mass m

Mass of the oscillating body, e.g. 0.8 kg on a coil spring.

Spring stiffness k

Spring constant, the force per metre of stretch, e.g. 20 N/m for a soft spring.

N/m

Damping coefficient b

Damping coefficient b of the drag force F = −b · v, e.g. 1.6 kg/s; 0 means no damping.

Initial displacement x₀

Displacement at the start t = 0 with sign, e.g. 4 cm; the side you pull towards counts as positive.

Initial velocity v₀

Velocity at the start with sign; 0 m/s if you release the mass from rest.

Time t

Time since the start, e.g. 2 s.

Filled with example values. Type your own and the result updates straight away.

Result: Displacement x

x = -0.5444 cm

With m = 0.8 kg, k = 20 N/m, b = 1.6 kg/s, x₀ = 4 cm, v₀ = 0 m/s and t = 2 s, x = -0.5444 cm.

Working · rearrange for x, substitute, evaluate

1. Rearrange for x
$x=e^{-\beta t}\left[x_0\cos(\omega_dt)+\frac{v_0+\beta x_0}{\omega_d}\sin(\omega_dt)\right]$
2. Substitute values (in SI units)
- x₀ = 4 cm = 0.04 m
$e^{-1\cdot 2}\left[0.04\cdot\cos(4.89898\cdot 2)+\frac{0+1\cdot 0.04}{4.89898}\cdot\sin(4.89898\cdot 2)\right]=-0.00544368$
3. Result
x = -0.5444 cm

Quick tips

Check the regime first: ζ = b / (2 · √(m · k)). Only for ζ < 1 does the body oscillate with a period; from ζ = 1 upwards there is no period.
Light damping barely changes the period: Td = T₀ / √(1 − ζ²). At ζ = 0.2 it is only about 2 % longer than without damping.

Assumptions and limits

Free linear viscous motion with ζ<1 and t≥0; x₀ and v₀ are signed values at t=0 relative to equilibrium. This mode assumes neither zero phase nor release from rest. It does not apply to critical or overdamped motion.

Derivation & background

Linear restoring force and viscous resistance give $m\ddot x+b\dot x+kx=0$ with $\beta=\frac b{2m}$ . For $\beta<\omega_0=\sqrt{\frac km}$ , the solution is $x=A_0e^{-\beta t}\cos(\omega_dt+\varphi)$ , with $\omega_d=\sqrt{\omega_0^2-\beta^2}$ . A₀ and φ must match x₀ and v₀; A₀ is the initial envelope magnitude, not necessarily initial displacement. Critical and overdamped solutions have no repeating oscillatory component, but may cross equilibrium once depending on initial velocity.

Exam blueprint

Validity range

Free one-dimensional linear motion about stable equilibrium with constant m, k and viscous b. The cosine form with real ωd applies only to $0<\beta<\omega_0$ ; $\beta=0$ is the undamped limit. Critical and overdamped regimes require other solutions. The model excludes external driving and dry or quadratic resistance.

Derivation steps

The characteristic equation of the free linear oscillator separates damping regimes and gives the exponential envelope.

1Measure x from equilibrium and assume linear forces −kx and −bẋ with constant m > 0, $k>0$ and b > 0.
2Newton gives $m\ddot x+b\dot x+kx=0$ . Define $\beta=\frac b{2m}$ and $\omega_0^2=\frac km$ .
3Substitute $x=Ce^{rt}$ : $r^2+2\beta r+\omega_0^2=0$ has $r=-\beta\pm\sqrt{\beta^2-\omega_0^2}$ .
4For $\beta<\omega_0$ roots are $-\beta\pm i\omega_d$ with $\omega_d=\sqrt{\omega_0^2-\beta^2}$ . Only this regime has a free oscillatory component.
5The real solution is $x=e^{-\beta t}(C\cos\omega_dt+D\sin\omega_dt)$ , with $C=x_0$ and $D=\frac{v_0+\beta x_0}{\omega_d}$ .
6With $A_0=\sqrt{C^2+D^2}$ and matching phase, $x=A_0e^{-\beta t}\cos(\omega_dt+\varphi)$ . $A(t)=A_0e^{-\beta t}$ is the envelope.
7At $\beta=\omega_0$ , $x=(C+Dt)e^{-\omega_0t}$ ; for $\beta>\omega_0$ two real decaying exponential components occur, without a free oscillation period.

Rearrangements

Envelope from its initial scale and decay rate

A(t)=A_0e^{-\beta t}

Underdamped motion of a linear mass-spring system with constant m, b and k.

Task variant

m=0.50 kg, b=1.0 kg/s and k=8.0 N/m. Find β and ω_d.

For m=0.50 kg, b=1.0 kg/s and k=8.0 N/m, $\beta=\frac{b}{2m}=1.0\,\mathrm{s^{-1}}$ and $\omega_d=\sqrt{16-1}\,\mathrm{rad\,s^{-1}}=\sqrt{15}\,\mathrm{rad\,s^{-1}}\approx3.87\,\mathrm{rad\,s^{-1}}$ .
At $t=\ln2\,\mathrm s$ , the envelope is $A_0e^{-\ln2}=\frac{A_0}{2}$ .

Check: $m\omega_d^2+\frac{b^2}{4m}=k$ gives $0.5\cdot15+0.5=8\,\mathrm{N\,m^{-1}}$ .

Common mistakes

Confusing the undamped natural frequency ω₀ with ω_d.

Damping lowers the angular frequency: $\omega_d=\sqrt{\omega_0^2-\beta^2}$ , while amplitude falls as $e^{-\beta t}$ .

Exam context

Shock absorption, decay curves, resonance and real mechanical oscillators.

Formula cluster

Real oscillations

The envelope describes energy loss while the cosine component carries the remaining periodic motion.

T = 2π·√(m/k)

Spring Pendulum (Oscillation Period)

T = 2π·√(l/g)

Simple Pendulum (Period)

T = 2π·√(L·C)

Thomson formula for an LC circuit

Worked example

For m = 0.50 kg, b = 1.0 kg/s and k = 8.0 N/m, $\beta=\frac b{2m}=1.0\,\mathrm{s^{-1}}$ and $\omega_d=\sqrt{\frac km-\beta^2}=\sqrt{15}\,\mathrm{rad\,s^{-1}}\approx3.87298\,\mathrm{rad\,s^{-1}}$ . The period is $T_d=\frac{2\pi}{\omega_d}\approx1.62231\,\mathrm s$ . At $t=\frac{\ln2}{\beta}\approx0.693147\,\mathrm s$ , the envelope halves. Check: $m\omega_d^2+\frac{b^2}{4m}=0.5\cdot15+0.5=8\,\mathrm{N\,m^{-1}}$ .

Applications

Mass-spring systems in fluids, shock absorbers, decay measurements and estimating damping parameters.

How do you calculate with Damped harmonic oscillation?

Here is how to work through a typical Damped harmonic oscillation (x(t) = A₀ e^(−βt) cos(ωdt+φ)) task step by step:

1
Task
m=0.50 kg, b=1.0 kg/s and k=8.0 N/m. Find β and ω_d.
Solution path
1. For m=0.50 kg, b=1.0 kg/s and k=8.0 N/m, $\beta=\frac{b}{2m}=1.0\,\mathrm{s^{-1}}$ and $\omega_d=\sqrt{16-1}\,\mathrm{rad\,s^{-1}}=\sqrt{15}\,\mathrm{rad\,s^{-1}}\approx3.87\,\mathrm{rad\,s^{-1}}$ .
2. At $t=\ln2\,\mathrm s$ , the envelope is $A_0e^{-\ln2}=\frac{A_0}{2}$ .
Check: $m\omega_d^2+\frac{b^2}{4m}=k$ gives $0.5\cdot15+0.5=8\,\mathrm{N\,m^{-1}}$ .

Understand and apply the equation

Damping reduces free excursions and can prevent oscillations altogether. The linear spring–mass model distinguishes underdamping, critical damping, and overdamping. Initial values and measured peaks determine the applicable time function and period.

Why an oscillation fades

A mass on a spring may cross its equilibrium repeatedly while its excursions become smaller. Friction removes mechanical energy. Here we use one specific friction model: the resisting force is proportional to instantaneous velocity and points opposite to it. This model gives a time function that can be checked.[1]

You need the spring force, Newton’s law, and the distinction between displacement, velocity, and acceleration. Damping first decides whether free oscillations exist at all. A damped period is meaningful only in the oscillatory regime. An externally driven oscillation follows a different equation of motion.[1]

Which equation describes a free damped oscillation?

x measures displacement from stable equilibrium. A linear restoring force −kx and viscous resistance −bẋ act on constant mass m. Assume m > 0, $k>0$ and b > 0, with no external drive. Resistance opposes instantaneous velocity, not necessarily x. If v is negative, resistance is positive. A constant weight can be absorbed by shifting equilibrium in a vertical linear spring arrangement. The spring–mass oscillator without friction is this model’s zero-damping limit.[1]

\begin{aligned}m\ddot x+b\dot x+kx&=0\\{} \beta&=\frac b{2m}\\{} \omega_0&=\sqrt{\frac{k}{m}}\end{aligned}

β has unit s⁻¹; b has $\mathrm{kg\,s^{-1}}=\mathrm{N\,s\,m^{-1}}$ . ω₀ is the undamped natural angular frequency.

When do underdamping, critical damping and overdamping occur?

After division by m, the equation is $\ddot x+2\beta\dot x+\omega_0^2x=0$ . Substitution $x=Ce^{rt}$ gives $r^2+2\beta r+\omega_0^2=0$ . Roots $r=-\beta\pm\sqrt{\beta^2-\omega_0^2}$ determine the motion’s form. Dimensionless damping ratio $\zeta=\frac{\beta}{\omega_0}$ compares resistance with the natural time scale.

\begin{aligned}\beta&<\omega_0\\x&=e^{-\beta t}(C\cos\omega_dt+D\sin\omega_dt)\\\omega_d&=\sqrt{\omega_0^2-\beta^2}\end{aligned}

Underdamping has an oscillatory component with a decaying envelope.[1]

\beta=\omega_0:\quad x=(C+Dt)e^{-\omega_0t}

Critical damping does not produce repeated sinusoidal oscillation.

\begin{aligned}\beta&>\omega_0\\x&=Ce^{r_+t}+De^{r_-t}\\r_\pm&=-\beta\pm\sqrt{\beta^2-\omega_0^2}\end{aligned}

Overdamping has two negative real decay rates.

No free oscillation period is defined for critical or overdamped motion. Depending on initial velocity, a single zero crossing can still occur; no oscillation does not mean no overshoot for every initial condition. At very large β the slow root approaches $-\frac{\omega_0^2}{2\beta}$ : more resistance can slow return. Claims about fast critical return require a specified comparison of initial conditions.

Initial conditions instead of an unsuitable phase

\begin{aligned}C&=x_0\\D&=\frac{v_0+\beta x_0}{\omega_d}\\A_0&=\sqrt{C^2+D^2}\end{aligned}

This applies in the underdamped regime and satisfies $x(0)=x_0$ and $\dot x(0)=v_0$ .

\begin{aligned}x&=A_0e^{-\beta t}\cos(\omega_dt+\varphi)\\C&=A_0\cos\varphi\\D&=-A_0\sin\varphi\end{aligned}

Phase must match C and D; A₀ is the initial envelope magnitude.

Using $x=A_0e^{-\beta t}\cos(\omega_dt)$ without phase fixes $v_0=-\beta A_0$ . That is not release from rest. With $v_0=0$ and $x_0>0$ , D is positive and phase negative. A₀ can exceed x₀ even though release occurs at an actual displacement maximum. This distinction prevents an envelope label being mistaken for measured initial displacement.

Release from rest inside the envelope

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For the first example,

v_0=0

requires a sine term too. The envelope starts at

A_0\approx40.825\,\mathrm{mm}

, above initial displacement. It is a bound, not a sequence of all actual extrema.[1]

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How quickly does the envelope halve and what do the peaks measure?

\begin{aligned}A(t)&=A_0e^{-\beta t}\\T_d&=\frac{2\pi}{\omega_d}\\\tan(\omega_dt+\varphi)&=-\frac\beta{\omega_d}\end{aligned}

The last equation follows from $\dot x=0$ and locates actual extrema.

The curve stays between ±A(t), but generally does not meet those bounds at displacement maxima because the envelope itself moves. Equivalent peaks on the same side are separated by Td. Since amplitude decreases, x(t + Td) differs from x(t); Td is phase recurrence time, not strict periodicity of the full motion.

\begin{aligned}\delta&=\ln\frac{|x_n|}{|x_{n+1}|}=\beta T_d\\{} t_{1/2}&=\frac{\ln2}{\beta}\end{aligned}

xn and xn+1 are successive equivalent peaks; adjacent opposite-sign extrema are only Td/2 apart.

Over N full peak intervals use $\beta=\frac{\ln\!\left|\frac{x_1}{x_{N+1}}\right|}{NT_d}$ . Evaluate peaks relative to the same equilibrium. An offset near the noise floor makes ratios unstable; changing damping or Coulomb friction does not produce constant exponential decay. The logarithmic decrement expresses the ratio of equivalent peaks as a dimensionless decrease per complete phase cycle.

Energy loss without a false exponential equality

\begin{aligned}E&=\frac12m\dot x^2+\frac12kx^2\\\frac{\mathrm dE}{\mathrm dt}&=\dot x(m\ddot x+kx)=-b\dot x^2\le0\end{aligned}

Dissipated power is bẋ²; mechanical energy cannot increase without a drive.

Exact instantaneous energy is generally not simply $E(0)e^{-2\beta t}$ . Velocity and displacement contain different phase contributions because differentiating introduces an additional term. At the same phase after Td, however, both x and v scale by $e^{-\beta T_d}$ , so $E(t+T_d)=E(t)e^{-2\beta T_d}$ exactly. Under weak damping, a cycle-averaged exponential energy curve approximates the slow evolution.

Example: damped release from rest

Calculating the matching phase and envelope

Given: m = 0.800 kg, k = 20.0 N/m, $b=1.60\,\mathrm{kg/s}$ , x₀ = +0.0400 m, $v_0=0$ .
Find: Find regime, Td, the complete solution and envelope half-life.

Approach: Check the regime first, then apply both initial conditions.
$\begin{aligned}\beta&=\frac b{2m},&\omega_0&=\sqrt{\frac km}\\\omega_d&=\sqrt{\omega_0^2-\beta^2}\\C&=x_0,&D&=\frac{v_0+\beta x_0}{\omega_d}\end{aligned}$
First put force balance into standard form; substitute the numerical values in the next step.
$\begin{aligned}m\ddot x+b\dot x+kx&=0\\\beta&=\frac{b}{2m}\\\omega_0&=\sqrt{\frac km}\\\omega_d&=\sqrt{\omega_0^2-\beta^2}\quad(\beta<\omega_0)\end{aligned}$
Determine parameters and regime.
$\begin{aligned}\beta&=1.00\,\mathrm{s^{-1}}\\{} \omega_0&=5.00\,\mathrm{rad/s}\\{} \omega_d&=\sqrt{24}=4.89898\,\mathrm{rad/s}\end{aligned}$
Insert initial values into C and D.
$\begin{aligned}C&=0.0400\,\mathrm m\\{} D&=\frac{0.0400}{\sqrt{24}}=0.00816497\,\mathrm m\end{aligned}$
Determine envelope magnitude and phase.
$\begin{aligned}A_0&=0.0408248\,\mathrm m\\{} \varphi&=-\arctan(\frac{D}{C})=-0.201358\,\mathrm{rad}\end{aligned}$
Calculate recurrence time and half-life.
$\begin{aligned}T_d&=\frac{2\pi}{\sqrt{24}}=1.28255\,\mathrm s\\{} t_{1/2}&=\frac{\ln2}{\beta}=0.693147\,\mathrm s\end{aligned}$

Result: The motion is underdamped. With time in seconds, the solution is $x(t)=e^{-t}\left(0.0400\cos(\sqrt{24}\,t)+0.00816497\sin(\sqrt{24}\,t)\right)\,\mathrm m$ .

Check: At t = 0, x = 0.0400 m and $v=-\beta C+\omega_dD=0$ . A₀ is about 2.06% greater than x₀. Energy derivative is zero at release from rest, already ruling out an everywhere exact curve $E(0)e^{-2\beta t}$ .

Example: parameters from equivalent peaks

Reconstructing a decay measurement

Given: Successive positive maxima relative to equilibrium: 15.0 mm and 9.00 mm; separation 1.20 s. Moving mass 0.600 kg. Linear viscous model.
Find: Find β, ωd, ω₀, b and k, then predict the next peak.

Approach: The two same-side peaks are one damped period apart.
$\begin{aligned}\delta&=\ln\frac{X_1}{X_2}\\\beta&=\frac\delta{T_d}\\\omega_d&=\frac{2\pi}{T_d}\\b&=2m\beta,&k&=m(\omega_d^2+\beta^2)\end{aligned}$
Substitute the given values with matching units.
$\begin{aligned}\delta&=\ln\frac{x_n}{x_{n+1}}\\\beta&=\frac{\delta}{T_d}\\\omega_d&=\frac{2\pi}{T_d}\\\omega_0^2&=\omega_d^2+\beta^2\\b&=2m\beta\\k&=m\omega_0^2\end{aligned}$
Obtain decrement and decay rate from the ratio.
$\begin{aligned}\delta&=\ln(\frac{15}{9})=0.510826\\{} \beta&=\frac{\delta}{1.20}=0.425688\,\mathrm{s^{-1}}\end{aligned}$
Calculate angular frequencies from peak spacing.
$\begin{aligned}\omega_d&=\frac{2\pi}{1.20}=5.23599\,\mathrm{rad/s}\\{} \omega_0&=\sqrt{\omega_d^2+\beta^2}=5.25326\,\mathrm{rad/s}\end{aligned}$
Use known mass to recover damping and stiffness.
$\begin{aligned}b&=2m\beta=0.510826\,\mathrm{kg/s}\\{} k&=m\omega_0^2=16.5581\,\mathrm{N/m}\end{aligned}$

Result: β ≈ 0.426 s⁻¹, b ≈ 0.511 kg/s and k ≈ 16.6 N/m; the next positive peak is 5.40 mm.

Check: $e^{-\beta T_d}=0.600$ reproduces $\frac9{15}$ and gives $9\cdot0.600=5.40\,\mathrm{mm}$ . $\zeta\approx0.0810<1$ confirms underdamping. Without known mass, peak data supply only $\frac bm$ and $\frac km$ , not b and k separately.

Free motion, driving and model limits

Explore further: driving and resonance

With an external sinusoidal force F₀cos(Ωt), that force appears on the equation’s right side. After the free transient decays, the linear steady response oscillates at Ω. Thus ω₀, ωd and Ω are three different quantities. Displacement-amplitude resonance additionally occurs at $\Omega_R=\sqrt{\omega_0^2-2\beta^2}$ if $\beta<\frac{\omega_0}{\sqrt2}$ ; stronger damping gives no peak at positive frequency. Other measured quantities can have different resonance conditions.[2]

Explore further: derive the resonance condition

For $\Omega>0$ , minimise the squared denominator $D(\Omega)=(\omega_0^2-\Omega^2)^2+4\beta^2\Omega^2$ . Its derivative is $D\prime(\Omega)=4\Omega(\Omega^2-\omega_0^2+2\beta^2)$ . An interior amplitude maximum therefore requires $\Omega^2=\omega_0^2-2\beta^2>0$ ; otherwise the largest amplitude measure occurs at the static limit $\Omega=0$ . This precise peak condition is derived here by differentiating the cited amplitude formula.

View equation: steady driven response

A_{\mathrm{steady}}(\Omega)=\frac{\frac{F_0}{m}}{\sqrt{(\omega_0^2-\Omega^2)^2+4\beta^2\Omega^2}}

Steady displacement amplitude under harmonic forcing; not the free decay envelope.[2]

The decay model requires constant linear stiffness and resistance proportional to v. Dry friction, quadratic drag, stops and nonlinear springs can generate other envelopes and amplitude-dependent frequencies. Check sensor offset, peak spacing, support motion and possible superposed modes in measurements. An apparently exponential segment alone proves neither a single mode nor constant parameters.

Returning to equilibrium: critical or overdamped?

More damping can make overdamped return slower. This comparison uses the same spring, mass, and release from rest so that damping alone accounts for the different motion.[1]

Returning to equilibrium: critical or overdamped?

Given: Two free linear oscillators each have m=1.0 kg, k=4.0 N/m, x₀=0.030 m, and v₀=0. Only b differs: 4.0 kg/s or 5.0 kg/s. Displacement is measured from each equilibrium; neither is driven.
Find: Determine regimes and time functions; compare displacements after 1.0 s.

Approach: the characteristic equation and both initial conditions determine the constants.
$\begin{aligned}\omega_0&=\sqrt{\frac km}=2\,\mathrm{s^{-1}}\\b_{\mathrm{crit}}&=2\sqrt{mk}=4\,\mathrm{kg/s}\\x_{\mathrm{crit}}(t)&=(C+Dt)e^{-\omega_0t}\end{aligned}$
At critical damping $C=x_0$ and $D=2x_0$ give zero initial velocity. With b=5 the roots are −1 and −4 per second.
$\begin{aligned}x_{\mathrm{crit}}(t)&=(0.030\,\mathrm m)\left(1+2\,\mathrm{s^{-1}}t\right)e^{-(2\,\mathrm{s^{-1}})t}\\x_{\mathrm{over}}(t)&=(0.040\,\mathrm m)e^{-(1\,\mathrm{s^{-1}})t}\\&\quad-(0.010\,\mathrm m)e^{-(4\,\mathrm{s^{-1}})t}\\x_{\mathrm{crit}}(1\,\mathrm s)&\approx0.0121802\,\mathrm m\\x_{\mathrm{over}}(1\,\mathrm s)&\approx0.0145320\,\mathrm m\end{aligned}$

Result: For this release from rest, the critical case is closer to equilibrium after 1 s. Both motions remain positive and do not oscillate. Increasing damping here does not mean a faster return.

Check: Both satisfy $x(0)=0.030\,\mathrm m$ and $v(0)=0$ . For overdamping, $v(t)=(0.040\,\mathrm{m/s})[e^{-(4\,\mathrm{s^{-1}})t}-e^{-(1\,\mathrm{s^{-1}})t}]$ , negative at positive times. The comparison applies to this release from rest; other initial velocities require their own analysis.

Five solved exercises

Calculate · Foundation

m = 0.50 kg, k = 8.0 N/m and b = 1.0 kg/s. Find regime and ωd.

Show hint

Compare β with ω₀ before evaluating the square root.

Learning goal: Check damping regime before calculating frequency.

Show solution and calculation

$\beta=1\,\mathrm{s^{-1}}$ and $\omega_0=4\,\mathrm{rad/s}$ .
Therefore underdamped, with $\omega_d=\sqrt{15}\,\mathrm{rad/s}\approx3.87298\,\mathrm{rad/s}$ and $T_d\approx1.62231\,\mathrm s$ .

\begin{aligned}\beta&=\frac b{2m}=\frac{1.0}{1.0}\,\mathrm{s^{-1}}=1\,\mathrm{s^{-1}}\\\omega_0&=\sqrt{\frac km}=4\,\mathrm{rad/s}\\\omega_d&=\sqrt{\omega_0^2-\beta^2}=\sqrt{15}\,\mathrm{rad/s}\\T_d&=\frac{2\pi}{\omega_d}\approx1.62231\,\mathrm s\end{aligned}

Rearrange · Application

$\beta=0.40\,\mathrm{s^{-1}}$ . How long until the envelope reaches 10%?

Show hint

Solve $e^{-\beta t}=0.10$ .

Learning goal: Find envelope decay time using logarithms.

Show solution and calculation

$t=\frac{\ln10}{0.40}\,\mathrm s\approx5.75646\,\mathrm s$ .
Substitution gives $e^{-2.302585}=0.10$ .
It does not require $x=0.10x_0$ at that instant.

\begin{aligned}\frac{A(t)}{A_0}&=e^{-\beta t}=0.10\\t&=-\frac{\ln(0.10)}\beta=\frac{\ln10}{0.40}\,\mathrm s\\&\approx5.75646\,\mathrm s\end{aligned}

Rearrange · Application

$m=0.80\,\mathrm{kg}$ and $k=20\,\mathrm{N/m}$ . What b threshold separates underdamping and overdamping?

Show hint

Critical damping means $\beta=\omega_0$ .

Learning goal: Find the critical damping coefficient.

Show solution and calculation

$b_{\mathrm{crit}}=2\sqrt{mk}=8.00\,\mathrm{kg/s}$ .
Below this the free component oscillates; equality is critical damping and above it is overdamping.
For $b=1.60\,\mathrm{kg/s}$ , $\zeta=0.20$ .

\begin{aligned}\beta_{\mathrm{crit}}&=\omega_0\\\frac{b_{\mathrm{crit}}}{2m}&=\sqrt{\frac km}\\b_{\mathrm{crit}}&=2\sqrt{mk}=2\sqrt{0.80\cdot20}\,\mathrm{kg/s}\\&=8.00\,\mathrm{kg/s}\end{aligned}

Interpret · Application

A free linear spring-mass model has m=1 kg, k=4 N/m, b=6 kg/s and no drive. Find β, ω₀ and both characteristic roots. What does negative ωd² imply for an attempted free-period measurement?

Show hint

Check the regime first; negative ωd² means no real oscillation frequency here.

Learning goal: Interpret real decay rates without a free oscillation period.

Show solution and calculation

Approach: $\beta=3\,\mathrm{s^{-1}}$ exceeds $\omega_0=2\,\mathrm{rad/s}$ , giving overdamping.
Roots $(-3\pm\sqrt5)\,\mathrm{s^{-1}}$ are both negative; the free solution is a sum of two exponentials.
No free oscillation period is defined.

Check: both roots satisfy $r^2+6r+4=0$ . Particular initial conditions may allow one zero crossing, which does not establish periodic oscillation.

\begin{aligned}r_+&=(-3+\sqrt5)\,\mathrm{s^{-1}}\approx-0.763932\,\mathrm{s^{-1}}\\r_-&=(-3-\sqrt5)\,\mathrm{s^{-1}}\approx-5.236068\,\mathrm{s^{-1}}\end{aligned}

[1]

Error analysis · Application

For $\beta=1\,\mathrm{s^{-1}}$ , $\omega_d=2\,\mathrm{rad/s}$ , $x_0=0.020\,\mathrm m$ , v₀=0, a proposed solution is $x(t)=(0.020\,\mathrm m)e^{-\beta t}\cos(\omega_dt)$ . Check v(0) and correct the sine coefficient of the free solution.

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The exponential envelope derivative contributes to v(0) too.

Learning goal: Check initial velocity in a damped cosine solution.

Show solution and calculation

Approach: for $x=e^{-\beta t}[C\cos(\omega_dt)+D\sin(\omega_dt)]$ , $v(0)=-\beta C+\omega_dD$ .
Proposed $D=0$ gives −0.020 m/s, not zero.
With $C=0.020\,\mathrm m$ , D must be $0.010\,\mathrm m$ .

Check: both initial conditions then hold. Envelope amplitude $\sqrt{C^2+D^2}\approx0.0223607\,\mathrm m$ is not simply x₀; initial displacement and envelope parameter are different quantities.

\begin{aligned}D&=\frac{v_0+\beta x_0}{\omega_d}=0.010\,\mathrm m\\v(0)&=-\beta C+\omega_dD=-0.020+0.020=0\,\mathrm{m\,s^{-1}}\end{aligned}

[1]

In brief

Check the regime before calculating a period. Both initial values determine free motion. The exponential envelope bounds displacement but is not itself the motion. Energy decay and displacement decay are different quantities; driving adds a frequency and force to the model.[1]

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Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Refined question headings for free motion, damping regimes and the envelope; added paragraph connections to the undamped spring–mass oscillator and logarithmic decrement.
October 1, 2026Explained regimes and initial values step by step; compared critical and overdamped rest releases and graphed motion with envelopes. Preserved both previous examples and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Damped harmonic oscillation. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/gedaempfte-harmonische-schwingung

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Frequently asked questions about Damped harmonic oscillation

What does β describe?+

$\beta=\frac b{2m}$ describes the exponential envelope decay rate of a free oscillator with linear viscous resistance. Its unit is s⁻¹, whereas resistance coefficient b has unit kg/s. After time 1/β the envelope is 1/e of its initial magnitude; after ln 2/β it halves. β alone does not determine whether oscillation occurs: compare it with $\omega_0=\sqrt{\frac km}$ . Underdamping requires $\beta<\omega_0$ . Equivalent peaks separated by Td give β = ln(|x₁/x₂|)/Td. Sensor offset and peak identification must be correct; adjacent opposite-sign extrema have only half that spacing and cannot be substituted without adjusting the time interval.

Why is ω_d smaller than the undamped frequency?+

The free linear damped oscillator has characteristic equation $r^2+2\beta r+\omega_0^2=0$ , with complex roots $-\beta\pm i\sqrt{\omega_0^2-\beta^2}$ . Their imaginary part is ωd. Positive underdamping therefore gives $\omega_d<\omega_0$ ; as $\beta\to0$ they coincide. For $\omega_0=5\,\mathrm{rad/s}$ and $\beta=1\,\mathrm{s^{-1}}$ , $\omega_d=\sqrt{24}\,\mathrm{rad/s}\approx4.899\,\mathrm{rad/s}$ . The decrease is small for weak damping but becomes substantial near the critical threshold. A steadily driven oscillator instead follows drive frequency Ω, and displacement resonance frequency must not automatically be identified with ωd. All three describe different aspects of the model. Therefore first identify whether your observation is a free decay, a steady driven motion, or the largest response amplitude in a frequency sweep.

When does the system still oscillate?+

A free linear viscous oscillator has an oscillatory component when $\beta<\omega_0$ , equivalently $b^2<4mk$ . Actual damped oscillation additionally needs $\beta>0$ ; $\beta=0$ is the undamped limit. At $\beta=\omega_0$ the critical solution is $(C+Dt)e^{-\omega_0t}$ . Above that threshold it consists of two real decaying exponential components. Those regimes have no repeating free oscillation period, although suitable initial velocity can still produce one equilibrium crossing. For $m=0.80\,\mathrm{kg}$ and $k=20\,\mathrm{N/m}$ , the critical coefficient is $2\sqrt{mk}=8.00\,\mathrm{kg/s}$ . Periodic external forcing can produce a steady periodic response even outside the underdamped regime. Before computing a free period, check the motion equation and damping threshold; an observed drive period alone does not establish underdamped natural motion.

How do I find the time for the amplitude to halve?+

Set A(t)/A₀ = 1/2 for the exponential envelope $A(t)=A_0e^{-\beta t}$ . Taking logarithms gives t = ln 2/β, provided $\beta>0$ and the linear decay model applies. With $\beta=0.40\,\mathrm{s^{-1}}$ this is about 1.733 s; substitution gives e⁻⁰·⁶⁹³¹⁴⁷ = 1/2. The quantity being halved is the envelope, not instantaneous displacement x(t), which can cross zero earlier. Equivalent peaks follow the same exponential ratio. Do not apply the calculation unchanged to varying damping, dry friction or sustained driving. Exact instantaneous energy decay also does not generally follow that same simple exponential form throughout a cycle.

What is the difference between amplitude and displacement?+

Displacement x(t) is signed position relative to equilibrium and can be positive, negative or zero. The nonnegative envelope $A(t)=A_0e^{-\beta t}$ bounds its magnitude, while the cosine describes rapid phase motion. Under damping, $|x(t)|\le A(t)$ . Actual extrema satisfy $\tan(\omega_dt+\varphi)=-\frac\beta{\omega_d}$ and therefore generally differ from envelope contact points. A₀ also need not equal x₀: release from rest requires a matching negative phase. With $x_0=0.0400\,\mathrm m$ , $\beta=1\,\mathrm{s^{-1}}$ and $\omega_d=\sqrt{24}\,\mathrm{rad/s}$ , the envelope starts at $A_0\approx0.040825\,\mathrm m$ even though initial displacement is only 0.0400 m. When comparing two maxima, also choose the same phase; the envelope and instantaneous position answer different measurement questions.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 1, §15.5: Damped Oscillations

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How do you calculate with Damped harmonic oscillation?

Why an oscillation fades

Which equation describes a free damped oscillation?

When do underdamping, critical damping and overdamping occur?

Initial conditions instead of an unsuitable phase

Release from rest inside the envelope

How quickly does the envelope halve and what do the peaks measure?

Energy loss without a false exponential equality

Example: damped release from rest

Calculating the matching phase and envelope

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Reconstructing a decay measurement

Free motion, driving and model limits

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Returning to equilibrium: critical or overdamped?

Five solved exercises

m = 0.50 kg, k = 8.0 N/m and b = 1.0 kg/s. Find regime and ωd.

β=0.40 s−1\beta=0.40\,\mathrm{s^{-1}}β=0.40s−1. How long until the envelope reaches 10%?

m=0.80 kgm=0.80\,\mathrm{kg}m=0.80kg and k=20 N/mk=20\,\mathrm{N/m}k=20N/m. What b threshold separates underdamping and overdamping?

A free linear spring-mass model has m=1 kg, k=4 N/m, b=6 kg/s and no drive. Find β, ω0 and both characteristic roots. What does negative ωd2 imply for an attempted free-period measurement?

In brief

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Frequently asked questions about Damped harmonic oscillation

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Retain Damped harmonic oscillation for exams

$\beta=0.40\,\mathrm{s^{-1}}$ . How long until the envelope reaches 10%?

$m=0.80\,\mathrm{kg}$ and $k=20\,\mathrm{N/m}$ . What b threshold separates underdamping and overdamping?

A free linear spring-mass model has m=1 kg, k=4 N/m, b=6 kg/s and no drive. Find β, ω₀ and both characteristic roots. What does negative ωd² imply for an attempted free-period measurement?