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Lesson 96 — Mechanical vibrations: mass-spring-damper

m x'' + c x' + k x = F(t). Natural frequency, damping, resonance. Underdamped, critically damped, overdamped.

Used in: Spécialité Maths (France, Terminale) · Leistungskurs alemão Klasse 12 · University Physics (global)

m\ddot x + c\dot x + kx = F(t) \quad\Longleftrightarrow\quad \ddot x + 2\zeta\omega_0\dot x + \omega_0^2 x = \frac{F(t)}{m}

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Rigorous notation, full derivation, hypotheses

Complete oscillator — four regimes

Characteristic equation and regimes

$\lambda^2 + 2\zeta\omega_0\lambda + \omega_0^2 = 0$ . Discriminant $\Delta = 4\omega_0^2(\zeta^2 - 1)$ .

Theorem· Four oscillator regimes

Undamped ( $c = 0$ , $\zeta = 0$ ): $\lambda = \pm i\omega_0$ . $x(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t)$

Underdamped ( $0 < \zeta < 1$ ): $\lambda = -\zeta\omega_0 \pm i\omega_d$ , with $\omega_d = \omega_0\sqrt{1-\zeta^2}$ . $x(t) = e^{-\zeta\omega_0 t}(A\cos\omega_d t + B\sin\omega_d t)$

Critically damped ( $\zeta = 1$ ): double root $\lambda = -\omega_0$ . $x(t) = (A + Bt)e^{-\omega_0 t}$

Overdamped ( $\zeta > 1$ ): $\lambda_{1,2} = \omega_0(-\zeta \pm \sqrt{\zeta^2-1}) < 0$ . $x(t) = Ae^{\lambda_1 t} + Be^{\lambda_2 t}$

"The most important case is $b^2 - 4km < 0$ , which occurs when the damping is small... In this case the solution oscillates with exponentially decaying amplitude." — Lebl, Notes on Diffy Qs, §2.4

Harmonic forced response

For $F(t) = F_0\cos\omega t$ : particular solution (steady-state)

$x_p(t) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4\zeta^2\omega_0^2\omega^2}}\cos(\omega t - \phi)$

where $\tan\phi = 2\zeta\omega_0\omega/(\omega_0^2 - \omega^2)$ .

Qualitative regime diagram

Free response ( $F = 0$ , $x(0) = x_0 > 0$ , $\dot x(0) = 0$ ): underdamped oscillates while decaying; critical and over converge monotonically.

Solved examples

Example— 1· Underdamped free oscillator with ICs

Problem. $m = 1\,\text{kg}$ , $k = 4\,\text{N/m}$ , $c = 2\,\text{N.s/m}$ . $x(0) = 1\,\text{m}$ , $\dot x(0) = 0$ . Find $x(t)$ .

Strategy. Calculate $\omega_0$ , $\zeta$ , identify the regime, solve the IVP.

Solution.

$\omega_0 = \sqrt{4/1} = 2$ rad/s. $\zeta = 2/(2\sqrt{1 \cdot 4}) = 2/4 = 0.5$ . Since $\zeta < 1$ : underdamped.

$\omega_d = 2\sqrt{1-0.25} = 2\sqrt{0.75} = \sqrt{3} \approx 1.732$ rad/s.

$x(t) = e^{-t}(A\cos\sqrt{3}\,t + B\sin\sqrt{3}\,t)$ .

IC: $x(0) = A = 1$ . $\dot x(0) = -A + \sqrt{3}\,B = 0 \Rightarrow B = 1/\sqrt{3}$ .

$x(t) = e^{-t}\!\left(\cos\sqrt{3}\,t + \frac{1}{\sqrt{3}}\sin\sqrt{3}\,t\right)$

Verification. The amplitude decays with $e^{-t}$ and oscillates with $\omega_d = \sqrt{3}$ . At $t = 0$ : $x = 1$ . $\dot x(0) = -1 + \sqrt{3}/\sqrt{3} = -1 + 1 = 0$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.4, Example 2.4.1 — CC-BY-SA.

Example— 2· Critical damping

Problem. $m = 1\,\text{kg}$ , $k = 9\,\text{N/m}$ , $c$ chosen critically. $x(0) = 2\,\text{m}$ , $\dot x(0) = -6\,\text{m/s}$ .

Strategy. $c_c = 2\sqrt{mk} = 6$ . $\zeta = 1$ . $\lambda = -\omega_0 = -3$ .

Solution.

$x(t) = (A + Bt)e^{-3t}$ .

$x(0) = A = 2$ . $\dot x(0) = B - 3A = -6 \Rightarrow B = -6 + 6 = 0$ .

$x(t) = 2e^{-3t}$

The system returns to equilibrium exponentially, without oscillation.

Verification. $\dot x = -6e^{-3t}$ , $\ddot x = 18e^{-3t}$ . $\ddot x + 6\dot x + 9x = 18 - 36 + 18 = 0$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.4, Example 2.4.2 — CC-BY-SA.

Example— 3· Sinusoidal forcing and steady-state

Problem. $\ddot x + 2\dot x + 5x = 10\cos 2t$ . Find the complete solution with $x(0) = 0$ , $\dot x(0) = 0$ .

Strategy. $y_h$ with roots $-1 \pm 2i$ ; $y_p$ with coefficients to be determined.

Solution.

$y_h = e^{-t}(C_1\cos 2t + C_2\sin 2t)$ .

Trial: $x_p = A\cos 2t + B\sin 2t$ .

$x_p'' = -4A\cos 2t - 4B\sin 2t$ , $x_p' = -2A\sin 2t + 2B\cos 2t$ .

Substituting: $(5A - 4A + 4B)\cos 2t + (5B - 4B - 4A)\sin 2t = 10\cos 2t$ .

$(A + 4B)\cos 2t + (B - 4A)\sin 2t = 10\cos 2t$ .

System: $A + 4B = 10$ , $B - 4A = 0 \Rightarrow B = 4A$ . Then $A + 16A = 10 \Rightarrow A = 10/17$ , $B = 40/17$ .

$x = e^{-t}(C_1\cos 2t + C_2\sin 2t) + \frac{10}{17}\cos 2t + \frac{40}{17}\sin 2t$ .

IC $x(0) = 0$ : $C_1 + 10/17 = 0 \Rightarrow C_1 = -10/17$ .

$\dot x(0) = 0$ : $-C_1 + 2C_2 + 80/17 = 0 \Rightarrow C_2 = -50/17$ .

$x(t) = \frac{e^{-t}}{17}(-10\cos 2t - 50\sin 2t) + \frac{10\cos 2t + 40\sin 2t}{17}$

Verification. As $t \to \infty$ : $x \to x_p = (10\cos 2t + 40\sin 2t)/17$ — the steady-state. Amplitude = $10\sqrt{1+16}/17 = 10\sqrt{17}/17 = 10/\sqrt{17} \approx 2.43$ m.

Source. Lebl, Notes on Diffy Qs, §2.6, Example 2.6.1 — CC-BY-SA.

Example— 4· Pure resonance

Problem. $\ddot x + \omega_0^2 x = F_0\cos\omega_0 t$ (no damping, forcing at natural frequency). Show that the amplitude grows without bound.

Strategy. $\omega = \omega_0$ is the resonance frequency; modification rule.

Solution.

$x_h = C_1\cos\omega_0 t + C_2\sin\omega_0 t$ .

Since $\cos\omega_0 t$ is in $x_h$ , the trial $x_p = t(A\cos\omega_0 t + B\sin\omega_0 t)$ .

$x_p'' = \omega_0^2 x_p \cdot (-) + (-2\omega_0 A\sin\omega_0 t + 2\omega_0 B\cos\omega_0 t)$ .

Substituting into $\ddot x + \omega_0^2 x = F_0\cos\omega_0 t$ :

$-2\omega_0 A\sin\omega_0 t + 2\omega_0 B\cos\omega_0 t = F_0\cos\omega_0 t$ .

Thus $A = 0$ , $B = F_0/(2\omega_0)$ .

$x_p(t) = \frac{F_0}{2\omega_0} t\sin(\omega_0 t)$

The amplitude $\frac{F_0}{2\omega_0} t$ grows linearly — pure resonance.

Verification. $x_p = t\sin(\omega_0 t) \cdot F_0/(2\omega_0)$ . Differentiating and substituting confirms the equation.

Source. Lebl, Notes on Diffy Qs, §2.6, Example 2.6.2 — CC-BY-SA.

Example— 5· Overdamped: settling time

Problem. $m = 2\,\text{kg}$ , $k = 8\,\text{N/m}$ , $c = 12\,\text{N.s/m}$ . $x(0) = 0.1\,\text{m}$ , $\dot x(0) = 0$ . Calculate $\zeta$ , the solution $x(t)$ , and the time for $|x| < 0.01\,\text{m}$ .

Strategy. $\omega_0 = 2$ , $\zeta = 12/(2\sqrt{16}) = 12/8 = 1.5$ : overdamped.

Solution.

$\lambda_{1,2} = 2(-1.5 \pm \sqrt{2.25-1}) = 2(-1.5 \pm \sqrt{1.25}) = -3 \pm \sqrt{5}$ .

$\lambda_1 \approx -3 + 2.236 = -0.764$ ; $\lambda_2 \approx -5.236$ .

$x(t) = Ae^{-0.764t} + Be^{-5.236t}$ .

IC: $A + B = 0.1$ ; $-0.764A - 5.236B = 0$ .

Solving: $B = -0.764A/5.236$ . Then $A(1 + 0.764/5.236) = 0.1 \Rightarrow A \approx 0.0838$ , $B \approx 0.0162$ .

For $|x| < 0.01$ : dominated by $Ae^{-0.764t} < 0.01 \Rightarrow t > \ln(A/0.01)/0.764 \approx \ln(8.38)/0.764 \approx 2.75$ s.

Verification. $x(0) = A + B \approx 0.1$ . $\dot x(0) = -0.764A - 5.236B \approx -0.064 + 0.085 \approx 0$ . Correct.

Source. Lebl, Notes on Diffy Qs, §2.4, Exercise 2.4.4 — CC-BY-SA.

Exercise list

24 exercises · 6 with worked solution (25%)

Application 12Understanding 2Modeling 5Challenge 3Proof 2

Ex. 96.1Application
Undamped spring: $m = 1\,\text{kg}$ , $k = 4\,\text{N/m}$ . Calculate $\omega_0$ , period, and write the general solution.
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Ex. 96.2Application
$m = 1\,\text{kg}$ , $k = 16\,\text{N/m}$ . Classify the regime for: (a) $c = 2$ , (b) $c = 8$ , (c) $c = 10$ .
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Ex. 96.3ApplicationAnswer key
$\ddot x + 6\dot x + 9x = 0$ , $x(0) = 1$ , $\dot x(0) = -1$ . Classify and solve.
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Ex. 96.4Application
$\ddot x + 4\dot x + 8x = 0$ , $x(0) = 3$ , $\dot x(0) = 0$ . Solve and calculate $\omega_d$ .
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Ex. 96.5Application
$\ddot x + 6\dot x + 5x = 0$ , $x(0) = 2$ , $\dot x(0) = -4$ . Overdamped — solve.
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Ex. 96.6Application
$\ddot x + 4x = \cos t$ (no damping). Calculate the steady-state amplitude.
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Ex. 96.7Application
Solve $\ddot x + 4x = 2\cos 3t$ .
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Ex. 96.8ApplicationAnswer key
Pure resonance: solve $\ddot x + 4x = \cos 2t$ . What happens to the amplitude?
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Ex. 96.9Application
Solve $\ddot x + 2\dot x + 5x = 5\cos t$ .
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Ex. 96.10ApplicationAnswer key
In a vibration test, two consecutive peaks measure $x_1 = 1.20$ m and $x_2 = 0.89$ m. Calculate the logarithmic decrement and the damping ratio $\zeta$ .
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Ex. 96.11Modeling
Automotive suspension: $m = 100\,\text{kg}$ , $k = 50000\,\text{N/m}$ , $c = 2000\,\text{N.s/m}$ . Calculate $\omega_0$ , $c_c$ , and $\zeta$ . Is it under or overdamped?
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Ex. 96.12Modeling
$m = 1\,\text{kg}$ , $k = 100\,\text{N/m}$ , $c = 2$ N.s/m. Calculate $\omega_0$ , $\zeta$ , peak frequency, and amplification factor.
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Ex. 96.13Modeling
Pendulum of length $L = 1\,\text{m}$ . Calculate $\omega_0$ and the period $T$ . (Use $g = 9.8\,\text{m/s}^2$ .)
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Ex. 96.14Modeling
Vibration isolation: to isolate a machine from 4 Hz vibration (from the floor), what should be the maximum natural frequency of the support?
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Ex. 96.15Modeling
Series RLC circuit: $L = 0.01\,\text{H}$ , $C = 100\,\mu\text{F}$ , $R = 10\,\Omega$ . Calculate $\omega_0$ and $Q$ .
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Ex. 96.16Understanding
How does the damped frequency $\omega_d$ compare to the natural frequency $\omega_0$ in the underdamped regime?
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Ex. 96.17Understanding
In control design, when is critical damping preferred versus underdamped?
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Ex. 96.18Application
Two springs with $k_1 = 200\,\text{N/m}$ and $k_2 = 300\,\text{N/m}$ connected in series with mass $m = 5\,\text{kg}$ . Calculate $k_{\text{eq}}$ and $\omega_0$ .
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Ex. 96.19Application
For the damped oscillator with $F = F_0\cos\omega t$ , write the formula for the steady-state amplitude and phase.
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Ex. 96.20ChallengeAnswer key
Compare the response at $\omega = \omega_0$ for (a) $\zeta = 0$ and (b) $\zeta = 0.05$ . What is the maximum amplitude in each case?
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Ex. 96.21ChallengeAnswer key
Beats: $\ddot x + 4x = \cos 2.1t$ , $x(0) = 0$ , $\dot x(0) = 0$ . Calculate the beat frequency and sketch the solution qualitatively.
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Ex. 96.22Challenge
Apply variation of parameters to the underdamped oscillator $\ddot x + 2\dot x + 5x = e^{-t}\sin t$ .
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Ex. 96.23ProofAnswer key
Demonstrate that the total energy $E = \frac{1}{2}m\dot x^2 + \frac{1}{2}kx^2$ of the damped oscillator ( $c > 0$ ) is strictly decreasing.
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Ex. 96.24Proof
Use Abel's theorem to show that the Wronskian of $y_1 = e^{-\zeta\omega_0 t}\cos\omega_d t$ and $y_2 = e^{-\zeta\omega_0 t}\sin\omega_d t$ is always non-zero ( $0 < \zeta < 1$ ).
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Sources

Notes on Diffy Qs — Jiří Lebl · v6.6 · §2.4–2.6 · EN · CC-BY-SA. Primary source.
University Physics Volume 1 — OpenStax · §15.4–15.6 · EN · CC-BY.
Elementary Differential Equations — William F. Trench · §6.1–6.2 · EN · open.

Lesson 96 — Mechanical vibrations: mass-spring-damper

Complete oscillator — four regimes

Equation of motion

Characteristic equation and regimes

Harmonic forced response

Resonance

Qualitative regime diagram

Solved examples

Exercise list

Sources