The Central Idea
A standard forced mass-spring model is
\[ \boxed{ mx''+cx'+kx=F(t) }. \]Here:
- \(m\) is mass;
- \(c\) is the damping coefficient;
- \(k\) is the spring constant;
- \(x(t)\) is displacement from equilibrium;
- \(F(t)\) is an external force.
The equation combines inertia, damping, restoring force, and external input.
The Unforced Undamped Model
If there is no damping and no external force, then
\[ c=0 \]and
\[ F(t)=0. \]The equation becomes
\[ mx''+kx=0. \]Dividing by \(m\),
\[ x''+\frac{k}{m}x=0. \]Natural Angular Frequency
Define
\[ \boxed{ \omega_0=\sqrt{\frac{k}{m}} }. \]Then the undamped equation becomes
\[ x''+\omega_0^2x=0. \]The characteristic equation is
\[ r^2+\omega_0^2=0. \]Therefore,
\[ r=\pm i\omega_0. \]Natural Oscillation
The general solution is
\[ \boxed{ x(t)=C_1\cos(\omega_0t)+C_2\sin(\omega_0t) }. \]The system oscillates indefinitely in the ideal undamped model.
The angular frequency is
\[ \omega_0. \]Period and Frequency
If the angular frequency is \(\omega_0\), then the period is
\[ \boxed{ T=\frac{2\pi}{\omega_0} }. \]The ordinary frequency is
\[ \boxed{ f=\frac{1}{T}=\frac{\omega_0}{2\pi} }. \]Angular frequency is measured in radians per unit time, while \(f\) is measured in cycles per unit time.
Amplitude-Phase Form
The solution
\[ C_1\cos(\omega_0t)+C_2\sin(\omega_0t) \]can be rewritten as
\[ \boxed{ x(t)=A\cos(\omega_0t-\phi) }. \]The amplitude is
\[ A=\sqrt{C_1^2+C_2^2}. \]The phase angle \(\phi\) determines the horizontal shift of the oscillation.
Initial Position and Velocity
A mechanical initial-value problem typically specifies
\[ x(0)=x_0 \]and
\[ x'(0)=v_0. \]These determine the constants in the oscillatory solution.
Example: Undamped Oscillation
Solve
\[ x''+9x=0, \]with
\[ x(0)=2, \qquad x'(0)=0. \]The natural angular frequency is
\[ \omega_0=3. \]Therefore,
\[ x=C_1\cos3t+C_2\sin3t. \]Complete the Undamped Example
From
\[ x(0)=2, \]we obtain
\[ C_1=2. \]Differentiate:
\[ x'=-3C_1\sin3t+3C_2\cos3t. \]Since
\[ x'(0)=0, \]we get
\[ 3C_2=0. \]Thus,
\[ C_2=0. \]Therefore,
\[ \boxed{x(t)=2\cos3t}. \]Damping
Real systems often lose energy through friction, resistance, drag, or other dissipative effects.
A common model assumes damping proportional to velocity:
\[ F_d=-cx'. \]The unforced damped equation is
\[ \boxed{ mx''+cx'+kx=0 }. \]The Damped Characteristic Equation
For
\[ mx''+cx'+kx=0, \]the characteristic equation is
\[ \boxed{ mr^2+cr+k=0 }. \]Its discriminant is
\[ \boxed{ \Delta=c^2-4mk }. \]The sign of \(\Delta\) determines the damping regime.
Three Damping Regimes
The system is:
- overdamped when \(c^2>4mk\);
- critically damped when \(c^2=4mk\);
- underdamped when \(c^2<4mk\).
These three cases correspond to distinct real roots, a repeated real root, and complex-conjugate roots.
Overdamping
If
\[ c^2>4mk, \]then the characteristic roots are distinct and real.
In a stable physical system with \(m,c,k>0\), both roots are negative.
The solution has the form
\[ \boxed{ x=C_1e^{r_1t}+C_2e^{r_2t} }. \]The system returns toward equilibrium without oscillating.
Critical Damping
If
\[ c^2=4mk, \]the characteristic equation has a repeated root.
The solution is
\[ \boxed{ x=(C_1+C_2t)e^{rt} }. \]Critical damping marks the boundary between oscillatory and nonoscillatory return to equilibrium.
Underdamping
If
\[ c^2<4mk, \]the roots are complex:
\[ r=-\frac{c}{2m}\pm i\omega_d. \]The solution is
\[ \boxed{ x(t) = e^{-ct/(2m)} \left( C_1\cos\omega_dt+C_2\sin\omega_dt \right) }. \]The system oscillates while its amplitude decays.
Damped Angular Frequency
For the underdamped case,
\[ \boxed{ \omega_d = \sqrt{ \frac{k}{m} - \frac{c^2}{4m^2} } }. \]Since damping reduces the oscillation rate,
\[ \omega_d<\omega_0 \]when \(c>0\).
The Decaying Envelope
In
\[ x(t) = e^{-ct/(2m)} \left( C_1\cos\omega_dt+C_2\sin\omega_dt \right), \]the factor
\[ e^{-ct/(2m)} \]controls the decay of the amplitude.
The trigonometric part produces the oscillation.
Example: Classify a Damped System
Consider
\[ x''+4x'+13x=0. \]Here,
\[ m=1,\qquad c=4,\qquad k=13. \]Compute
\[ c^2-4mk = 16-52 = -36. \]Since the discriminant is negative, the system is underdamped.
Complete the Damped Example
The characteristic equation is
\[ r^2+4r+13=0. \]Therefore,
\[ r=-2\pm3i. \]The solution is
\[ \boxed{ x=e^{-2t} \left( C_1\cos3t+C_2\sin3t \right) }. \]The oscillation frequency is \(3\), while the amplitude decays according to \(e^{-2t}\).
External Forcing
If an external force acts on the system, the equation becomes
\[ \boxed{ mx''+cx'+kx=F(t) }. \]The solution has the form
\[ \boxed{ x=x_h+x_p }. \]The homogeneous solution \(x_h\) describes the system's natural response.
The particular solution \(x_p\) describes the response to the forcing.
Transient and Steady-State Response
In many stable damped systems,
\[ x_h(t)\to0 \]as \(t\to\infty\).
The homogeneous response is then called the transient response.
The persistent forced response is called the steady-state response.
Sinusoidal Forcing
A common forcing function is
\[ F(t)=F_0\cos(\omega t). \]The differential equation becomes
\[ \boxed{ mx''+cx'+kx = F_0\cos(\omega t) }. \]Here \(\omega\) is the forcing angular frequency.
A Trial Particular Solution
For sinusoidal forcing, a useful trial form is
\[ x_p = A\cos(\omega t)+B\sin(\omega t). \]Its derivatives have the same sine-cosine structure, so substitution reduces the differential equation to algebraic equations for \(A\) and \(B\).
Example: Simple Forced Equation
Consider
\[ x''+x=2\cos2t. \]The homogeneous equation is
\[ x''+x=0. \]Therefore,
\[ x_h=C_1\cos t+C_2\sin t. \]Try
\[ x_p=A\cos2t+B\sin2t. \]Complete the Forced Example
Differentiate twice:
\[ x_p'' = -4A\cos2t-4B\sin2t. \]Then
\[ x_p''+x_p = -3A\cos2t-3B\sin2t. \]Match with
\[ 2\cos2t. \]Therefore,
\[ -3A=2, \qquad -3B=0. \]Thus,
\[ A=-\frac23, \qquad B=0. \]The full solution is
\[ \boxed{ x=C_1\cos t+C_2\sin t-\frac23\cos2t }. \]Forcing Frequency versus Natural Frequency
A forced oscillator involves two important frequencies:
- the natural frequency of the system;
- the frequency of the external forcing.
Their relationship can strongly affect the response amplitude.
Resonance
In an ideal undamped system, resonance occurs when the forcing frequency matches the natural frequency.
Consider
\[ x''+\omega_0^2x = F_0\cos(\omega_0t). \]The ordinary trial function
\[ A\cos(\omega_0t)+B\sin(\omega_0t) \]overlaps with the homogeneous solution.
Resonant Trial Form
Because the usual trial duplicates the homogeneous solution, multiply by \(t\).
A resonant trial may take the form
\[ x_p = t \left( A\cos\omega_0t+B\sin\omega_0t \right). \]The factor of \(t\) allows the forced response amplitude to grow with time in the ideal undamped model.
Example: Ideal Resonance
Consider
\[ x''+x=\cos t. \]The natural frequency is
\[ \omega_0=1. \]The forcing frequency is also
\[ \omega=1. \]Therefore, the system is forced at its natural frequency.
Resonant Particular Solution
A suitable particular solution is
\[ x_p=\frac12t\sin t. \]Thus,
\[ \boxed{ x=C_1\cos t+C_2\sin t+\frac12t\sin t }. \]The factor \(t\) causes the oscillation amplitude to grow without bound in this ideal model.
Damping Changes Resonance
In a damped system,
\[ mx''+cx'+kx=F_0\cos(\omega t), \]the steady-state amplitude remains finite when \(c>0\).
The frequency that maximizes the steady-state amplitude is generally close to, but not always exactly equal to, the undamped natural frequency.
Steady-State Amplitude
For
\[ mx''+cx'+kx=F_0\cos(\omega t), \]the steady-state amplitude is
\[ \boxed{ A(\omega) = \frac{F_0}{ \sqrt{(k-m\omega^2)^2+c^2\omega^2} } }. \]This formula shows how forcing frequency and damping control response size.
Phase Lag
The steady-state response can be written
\[ x_{ss}(t) = A(\omega)\cos(\omega t-\delta). \]The phase lag \(\delta\) satisfies
\[ \boxed{ \tan\delta = \frac{c\omega}{k-m\omega^2} }. \]The response does not necessarily reach its maximum at the same instant as the applied force.
Transient Decay in a Forced Damped System
Suppose the homogeneous roots have negative real parts.
Then
\[ x_h(t)\to0. \]Eventually,
\[ x(t)\approx x_p(t). \]The long-term behavior is therefore controlled primarily by the forcing.
Mechanical Interpretation of the Terms
In
\[ mx''+cx'+kx=F(t), \]each term has a distinct role:
- \(mx''\): inertia;
- \(cx'\): damping or resistance;
- \(kx\): restoring force;
- \(F(t)\): external input.
The differential equation balances these competing effects.
Units as a Model Check
Every term in
\[ mx''+cx'+kx=F(t) \]must have units of force.
If \(x\) is length and \(t\) is time, then:
- \(m\) has units of mass;
- \(c\) has units of force divided by velocity;
- \(k\) has units of force divided by displacement.
Dimensional consistency provides a fast modeling check.
Electrical Analogy
Oscillation equations also appear in electrical circuits.
A series RLC circuit may lead to an equation such as
\[ Lq''+Rq'+\frac{1}{C}q=E(t), \]where \(q(t)\) is charge.
The structure parallels
\[ mx''+cx'+kx=F(t). \]This is an example of the same mathematics governing different physical systems.
Common Error: Confusing Natural and Forcing Frequencies
The natural frequency is determined by the system parameters.
The forcing frequency is determined by the external input.
They are conceptually different even when they happen to have the same numerical value.
Common Error: Forgetting the Homogeneous Solution
For a forced equation,
\[ mx''+cx'+kx=F(t), \]a particular solution alone is not the complete general solution.
The full form is
\[ \boxed{x=x_h+x_p}. \]Common Error: Using the Ordinary Trial at Resonance
If the forcing term duplicates a homogeneous solution form, the usual trial function fails.
Multiply the trial by \(t\) enough times to make it linearly independent of the homogeneous solution.
Common Error: Assuming Every Oscillation Has Constant Amplitude
Undamped free oscillation can have constant amplitude.
Damped oscillation has a decreasing amplitude.
Resonant forcing in an ideal undamped model can produce increasing amplitude.
The amplitude behavior depends on the system and forcing.
Common Error: Calling Every Large Response Resonance
Resonance refers to frequency-dependent amplification associated with the relationship between forcing and natural system behavior.
A large response caused simply by a large forcing amplitude is not, by itself, evidence of resonance.
Common Error: Ignoring Damping in Long-Term Behavior
In a stable damped system, the transient solution often decays.
Therefore, long-term behavior may differ substantially from the initial motion.
The steady-state forced response can dominate after the transient disappears.
A Reliable Oscillation Workflow
- Identify \(m\), \(c\), \(k\), and \(F(t)\).
- Write the equation in standard form.
- Solve the associated homogeneous equation.
- Use the discriminant to classify damping.
- Identify the natural frequency when appropriate.
- Find a suitable particular solution for the forcing.
- Modify the trial form if resonance causes overlap.
- Combine \(x_h\) and \(x_p\).
- Apply initial conditions.
- Separate transient and steady-state behavior.
- Interpret amplitude, frequency, and phase.
- Check units and physical plausibility.
Example: Full Damped-Forced Workflow
Consider
\[ x''+2x'+5x=10\cos2t. \]First solve the homogeneous equation:
\[ r^2+2r+5=0. \]Therefore,
\[ r=-1\pm2i. \]Thus,
\[ x_h = e^{-t} \left( C_1\cos2t+C_2\sin2t \right). \]Complete the Forced Workflow
Try
\[ x_p=A\cos2t+B\sin2t. \]Then
\[ x_p' = -2A\sin2t+2B\cos2t \]and
\[ x_p'' = -4A\cos2t-4B\sin2t. \]Substitute into
\[ x''+2x'+5x=10\cos2t. \]Matching coefficients gives
\[ A+4B=10, \] \[ B-4A=0. \]From
\[ B=4A, \]we obtain
\[ 17A=10. \]Therefore,
\[ A=\frac{10}{17}, \qquad B=\frac{40}{17}. \]The full solution is
\[ \boxed{ x(t) = e^{-t} \left( C_1\cos2t+C_2\sin2t \right) + \frac{10}{17}\cos2t + \frac{40}{17}\sin2t }. \]Interpret the Full Forced Solution
The term
\[ e^{-t} \left( C_1\cos2t+C_2\sin2t \right) \]is the transient response.
Because of the factor \(e^{-t}\), it decays.
The terms
\[ \frac{10}{17}\cos2t + \frac{40}{17}\sin2t \]form the steady-state response.
At long times, the steady-state motion dominates.
Practice 1: Natural Frequency
Find the natural angular frequency of
\[ x''+25x=0. \]Practice 2: Period
Find the period of
\[ x''+16x=0. \]Practice 3: Classify Damping
Classify
\[ x''+6x'+9x=0. \]Practice 4: Classify Damping
Classify
\[ x''+2x'+10x=0. \]Practice 5: Solve an Undamped System
Solve
\[ x''+4x=0. \]Practice 6: Solve an Underdamped System
Solve
\[ x''+4x'+8x=0. \]Practice 7: Forced Motion
Find a particular solution of
\[ x''+4x=3\cos t. \]Practice 8: Identify Resonance
Does
\[ x''+9x=5\cos3t \]represent ideal resonance?
Practice 9: Transient or Steady State?
In
\[ x(t) = e^{-2t}\cos3t+4\cos t, \]identify the transient and steady-state terms.
Practice 10: Mechanical Interpretation
In
\[ mx''+cx'+kx=F(t), \]identify the roles of the four terms.
Practice Answers
Practice 1
Compare
\[ x''+\omega_0^2x=0 \]with
\[ x''+25x=0. \]Therefore,
\[ \boxed{\omega_0=5}. \]Practice 2
Here,
\[ \omega_0=4. \]Therefore,
\[ T=\frac{2\pi}{4}. \]Thus,
\[ \boxed{T=\frac{\pi}{2}}. \]Practice 3
The characteristic equation is
\[ r^2+6r+9=0. \]Thus,
\[ (r+3)^2=0. \]The repeated root means the system is
\[ \boxed{\text{critically damped}}. \]Practice 4
The characteristic equation is
\[ r^2+2r+10=0. \]Its discriminant is
\[ 4-40=-36. \]Therefore, the system is
\[ \boxed{\text{underdamped}}. \]Practice 5
The characteristic equation is
\[ r^2+4=0. \]Thus,
\[ r=\pm2i. \]Therefore,
\[ \boxed{ x=C_1\cos2t+C_2\sin2t }. \]Practice 6
The characteristic equation is
\[ r^2+4r+8=0. \]Therefore,
\[ r=-2\pm2i. \]Thus,
\[ \boxed{ x=e^{-2t} \left( C_1\cos2t+C_2\sin2t \right) }. \]Practice 7
Try
\[ x_p=A\cos t+B\sin t. \]Then
\[ x_p''=-A\cos t-B\sin t. \]Therefore,
\[ x_p''+4x_p = 3A\cos t+3B\sin t. \]Match with
\[ 3\cos t. \]Thus,
\[ A=1, \qquad B=0. \]Therefore,
\[ \boxed{x_p=\cos t}. \]Practice 8
The natural frequency is
\[ \omega_0=3. \]The forcing frequency is also
\[ \omega=3. \]Since there is no damping, this is
\[ \boxed{\text{ideal resonance}}. \]Practice 9
The decaying term
\[ \boxed{e^{-2t}\cos3t} \]is the transient response.
The persistent term
\[ \boxed{4\cos t} \]is the steady-state response.
Practice 10
In
\[ mx''+cx'+kx=F(t), \]- \(mx''\) represents inertia;
- \(cx'\) represents damping;
- \(kx\) represents restoring force;
- \(F(t)\) represents external forcing.
What to Remember
The standard damped, forced second-order model is
\[ \boxed{ mx''+cx'+kx=F(t) }. \]- without damping or forcing, the system oscillates at its natural frequency;
- damping changes the characteristic roots and causes energy loss;
- overdamped systems return without oscillating;
- critically damped systems lie at the boundary between oscillatory and nonoscillatory response;
- underdamped systems oscillate with a decaying amplitude;
- external forcing creates a particular response;
- the homogeneous response is often transient;
- the forced response may persist as a steady state;
- resonance occurs when forcing strongly aligns with natural system dynamics.
The central structural idea is
\[ \boxed{ \text{response} = \text{natural behavior} + \text{forced behavior} }. \]Where We Go Next
A single second-order equation can describe one state with acceleration, but many systems involve several interacting state variables whose rates depend on one another.
The next lesson, Systems of Differential Equations, develops coupled first-order equations, vector form, matrix representations, equilibrium states, eigenvalue methods, and interacting dynamic systems.