MA1035: Engineering Modeling using Dynamic Systems

Session 05: Mechanical Vibrations and RLC Circuits


Prof. Alejandro Ucan-Puc

Tecnológico de Monterrey • Departamento de Ciencias

Learning Outcomes

By the end of this session, you can:

  1. Translate weights, spring stretches, units, and initial data into a model.
  2. Solve free motion and identify its damping type, period, frequency, amplitude, and phase.
  3. Solve forced motion and distinguish transient from steady-state response.
  4. Translate a series RLC circuit into the same differential equation and solve it.

One Equation, Three Physical Readings

$$m x'' + c x' + k x = f(t)$$

TermMechanical meaningCircuit meaning
$m x''$inertia$Lq''$ (inductance)
$c x'$damping$Rq'$ (resistance)
$kx$spring restoring force$q/C$ (capacitor voltage)
$f(t)$applied force$E(t)$ (applied voltage)

Session Roadmap · 100 Minutes

0–10Modeling: Newton + Hooke10–28Undamped free motion
28–33Qualitative reasoning33–55Damped free motion
55–60Break / retrieval quiz60–78Forced motion and resonance
78–95Series RLC circuits95–100Wrap-up / exit ticket

Modeling · Choose the Sign Before the Math

Convention: measure displacement $x$ from equilibrium, with downward positive.

If the spring's static extension is $s$, Newton's law gives

$$m x'' = mg - k(s+x) - c x' + f(t).$$

At equilibrium, $mg=ks$. Cancel the weight and static spring force:

$$m x''+c x'+kx=f(t).$$

Modeling · Read the Data Carefully

  • Weight $W$ is a force; mass is $m=W/g$ (slugs when $W$ is in lb and $g=32$ ft/s$^2$).
  • Hooke's law: $k=W/\text{stretch}$, using consistent length units.
  • Convert inches to feet before calculating $k$.
  • Downward positive: 4 in below gives $x(0)=+1/3$ ft; 1 ft above gives $x(0)=-1$ ft.
  • An initial upward velocity is negative; a downward velocity is positive.

Undamped Free Motion · The Model

Set $c=0$ and $f(t)=0$. Then

$$x''+\omega^2x=0, \qquad \omega^2=\frac{k}{m}.$$

$$x(t)=c_1\cos(\omega t)+c_2\sin(\omega t)=A\sin(\omega t+\phi).$$

  • Angular frequency: $\omega$ rad/s
  • Period: $T=2\pi/\omega$ seconds
  • Frequency: $f=1/T$ cycles/s (Hz)

Worked Example · Undamped Spring

A 24-lb weight stretches a spring 4 in. It is pulled 4 in below equilibrium and released from rest. Find its motion.

  • Stretch: $4$ in $=1/3$ ft, so $k=24/(1/3)=72$ lb/ft.
  • Mass: $m=24/32=3/4$ slug.
  • Initial data: $x(0)=+1/3$ ft, $x'(0)=0$.

Worked Example · Solve and Interpret

$$\omega^2=\frac{k}{m}=96, \qquad \omega=4\sqrt{6}\ \text{rad/s}.$$

Applying the initial conditions gives

$$x(t)=\frac{1}{3}\cos(4\sqrt{6}\,t)\ \text{ft}.$$

Its amplitude is $1/3$ ft and its period is $T=\pi/(2\sqrt{6})$ s. With no damping, the motion continues indefinitely.

Your Turn · 5 Minutes

A mass of $1/2$ slug is attached to a spring with $k=36$ lb/ft. It is displaced 2 in below equilibrium and released from rest.

  1. Find the angular frequency and period.
  2. Write $x(t)$ using downward-positive coordinates.

Compare your sign and units with a partner before revealing the result.

Check · Problem 2

$$\omega=\sqrt{36/(1/2)}=6\sqrt{2}\ \text{rad/s}, \qquad T=\frac{\pi}{3\sqrt{2}}\ \text{s}.$$

The initial displacement is $2$ in $=1/6$ ft, so

$$x(t)=\frac{1}{6}\cos(6\sqrt{2}\,t)\ \text{ft}.$$

Qualitative Analysis · Read Before Solving

Suppose the motion satisfies

$$x''+\frac{1}{4}e^{-0.1t}x=0.$$

As $t\to\infty$, the restoring-force coefficient tends to zero.

Predict: Why should we not expect sustained oscillation? What equation describes the long-time trend?

Qualitative Analysis · What the Limit Says

For large $t$, $e^{-0.1t}/4\approx0$, so the equation approaches $x''=0$.

$$x(t)\approx C_1t+C_2.$$

The restoring force weakens away; it cannot keep pulling the mass back strongly enough for continuing oscillations. The solution may keep stretching rather than settling into a periodic motion.

Habit to build: inspect the coefficients and limiting equation before reaching for a solution method.

Damped Free Motion · Classify First

$$m x''+\beta x'+kx=0, \qquad mr^2+\beta r+k=0.$$

DiscriminantMotionResponse shape
$\beta^2-4mk>0$OverdampedTwo real roots; no oscillation
$\beta^2-4mk=0$Critically dampedRepeated root; fastest non-oscillatory return
$\beta^2-4mk<0$UnderdampedDecaying oscillation

Damped Free Motion · Solution Forms

  • Overdamped: $x=c_1e^{r_1t}+c_2e^{r_2t}$ for distinct real roots.
  • Critical: $x=(c_1+c_2t)e^{rt}$ for a repeated root.
  • Underdamped: $x=e^{\alpha t}(c_1\cos\omega t+c_2\sin\omega t)$, with $\alpha<0$.

Sketch each case: mark equilibrium, whether it crosses, and how the amplitude changes over time.

Worked Example · Critical Damping

A 4-lb weight stretches a spring 2 ft. The damping constant is $\beta=1$. The mass is pulled 1 ft above equilibrium and launched downward at 12 ft/s.

  • $m=4/32=1/8$ slug; $k=4/2=2$ lb/ft.
  • $$\frac18x''+x'+2x=0 \quad\Longrightarrow\quad x''+8x'+16x=0.$$
  • Double root $r=-4$; $x(0)=-1$, $x'(0)=+12$.

Worked Example · Apply the Initial Data

With a repeated root, $x=(c_1+c_2t)e^{-4t}$. From $x(0)=-1$, $c_1=-1$.

$$x'(0)=c_2-4c_1=12 \quad\Longrightarrow\quad c_2=8.$$

$$x(t)=(-1+8t)e^{-4t}\ \text{ft}.$$

This is critically damped: it crosses equilibrium once, then returns toward equilibrium without oscillating.

Read the Clock Reference

Equilibrium occurs when $x=0$:

$$-1+8t=0 \quad\Longrightarrow\quad t=\frac18\ \text{s after release}.$$

Velocity is $x'=(12-32t)e^{-4t}$. Its zero is at $t=3/8$ s, where

$$x(3/8)=2e^{-3/2}\ \text{ft}.$$

$3/8$ s is measured from release. The time after passing equilibrium is $3/8-1/8=1/4$ s.

Break · Retrieval Check

Without notes, write down:

  1. The discriminant test for all three damping cases.
  2. One sign convention you must state before modeling.
  3. The relation between angular frequency, period, and frequency.

Take a short break, then compare answers with a partner.

Forced Motion · Transient + Steady State

For sinusoidal forcing,

$$m x''+c x'+kx=F_0\cos(\gamma t), \qquad x=x_c+x_p.$$

  • $x_c$: complementary (transient) response; it decays when the damped system is stable and $c>0$.
  • $x_p$: particular (steady-state) response; it follows the ongoing input.
  • Use undetermined coefficients for sinusoidal inputs. If the trial duplicates a homogeneous term, multiply by $t$.

Resonance · Match the Trial to the Input

Solve the particular-response part of

$$x''+4x=\cos(2t).$$

The natural frequency is $2$, the same as the forcing frequency. The usual cosine/sine trial duplicates a homogeneous solution, so include a factor of $t$:

$$x_p(t)=\frac{t}{4}\sin(2t).$$

Resonance · Interpret the Response

The resonant particular solution has an amplitude envelope proportional to $t$: in the ideal undamped model, the amplitude grows without bound.

With damping, the long-time amplitude stays bounded; it reaches a peak near the natural frequency.

Check: In a forced problem, name which part eventually fades and which part remains while the input continues.

Series RLC · The Same Structure

Let $q(t)$ be charge and $i(t)=q'(t)$ current. Kirchhoff's voltage law gives

$$Lq''+Rq'+\frac{1}{C}q=E(t).$$

MechanicalElectrical
mass $m$inductance $L$
damping $c$resistance $R$
spring constant $k$$1/C$
displacement $x$charge $q$
velocity $x'$current $i=q'$
force $f(t)$voltage $E(t)$

Worked Example · RLC Circuit

Take $L=1$ H, $R=2\ \Omega$, $C=1/5$ F, and a constant source $E=10$ V, with $q(0)=0$ and $i(0)=0$.

$$q''+2q'+5q=10.$$

The characteristic roots are $-1\pm2i$ and the constant particular solution is $q_p=2$ C. Thus

$$q(t)=2-e^{-t}(2\cos2t+\sin2t)\ \text{C}.$$

RLC · Current and Steady State

Differentiate charge to obtain

$$i(t)=q'(t)=5e^{-t}\sin2t\ \text{A}.$$

  • $R^2-4L/C=4-20<0$: the circuit is underdamped.
  • As the transient decays, $q(t)\to2$ C and $i(t)\to0$ A.
  • The capacitor's steady voltage is $q/C=2/(1/5)=10$ V, matching the source.

Common Modeling and Solution Errors

  • Using weight as mass: compute $m=W/g$.
  • Reversing the sign for “above/below” or “upward/downward”: draw the positive axis.
  • Leaving a stretch in inches while using $g$ in ft/s$^2$: convert first.
  • Confusing $\omega$ (rad/s) with $T$ (s): $T=2\pi/\omega$.
  • Forgetting the $t$ multiplier when a forcing term resonates with a homogeneous solution.
  • Reporting elapsed time from release when the question starts the clock at an equilibrium crossing.

Practice · Free and Qualitative Motion

  1. Solve $x''+9x=0$, $x(0)=2$, $x'(0)=-3$. Find amplitude, period, and phase.
  2. Solve $x''+16x=0$, $x(0)=0$, $x'(0)=8$. Find its period and frequency.
  3. Classify and solve each IVP: (a) $x''+5x'+4x=0$, $x(0)=1$, $x'(0)=0$; (b) $x''+4x'+4x=0$, $x(0)=0$, $x'(0)=1$; (c) $x''+2x'+5x=0$, $x(0)=1$, $x'(0)=0$.
  4. Describe the long-time behavior of $x''+\frac14e^{-0.1t}x=0$ without solving it.

Practice · Forced Motion and Circuits

  1. For $x''+2x'+5x=10\cos t$, find a solution and identify its transient and steady-state parts.
  2. For a series circuit with $L=1$ H, $R=2\ \Omega$, $C=1/5$ F and $E(t)=10\cos t$ V, write the IVP for $q(t)$ when $q(0)=i(0)=0$; classify its damping.
  3. Sketch one time series or phase-plane portrait for each damping case. Mark equilibrium crossings and explain your classification.

Before calculating, vote: over-, critically, or underdamped?

Exit Ticket

  1. State the three damping cases and the discriminant test for each.
  2. For an RLC circuit, what electrical quantity corresponds to mass, damping, and spring stiffness?
  3. In one sentence, distinguish transient from steady-state response.

One model, three readings: mechanical free motion, mechanical forced motion, and electrical charge response.