understand that a resistive force acting on an oscillating system causes damping
Damped and Forced Oscillations – Cambridge A‑Level Physics (9702) – Section 17.3
Learning Objectives
- Understand that a resistive (damping) force acting on an oscillating system causes the amplitude to decrease with time.
- Identify the three damping regimes required by the syllabus and be able to sketch their displacement‑time graphs.
- Explain how damping influences natural (free) oscillations.
- Describe forced (driven) oscillations, resonance and the effect of damping on the resonance curve.
- Relate the mechanical oscillator to its electrical analogue (RLC circuit).
- Apply the concepts to real‑world examples.
1. Simple Harmonic Motion – Recall
For an ideal (undamped) mass–spring system
\[
m\ddot{x}+kx=0
\]
Solution
\[
x(t)=A\cos(\omega_0 t+\phi),\qquad
\omega_0=\sqrt{\frac{k}{m}}
\]
2. Introducing a Resistive (Damping) Force
Real systems experience a force opposite to the motion, usually proportional to the velocity:
\[
F_{\text d}=-b\dot{x}\qquad (b>0)
\]
Including this term gives the damped equation of motion
\[
m\ddot{x}+b\dot{x}+kx=0
\]
where b (kg s⁻¹) is the damping coefficient.
3. Damping Regimes (Cambridge Syllabus Labels)
| Regime (syllabus term) | Text‑book name | Condition on | Qualitative behaviour |
|---|---|---|---|
| Light damping | Underdamped | \(b<2\sqrt{mk}\) | Oscillatory motion; amplitude decays exponentially. |
| Critical damping | Critically damped | \(b=2\sqrt{mk}\) | Fastest return to equilibrium without overshoot. |
| Heavy damping | Over‑damped | \(b>2\sqrt{mk}\) | Non‑oscillatory, slower return than the critical case. |
Displacement‑time sketches
- Light damping (underdamped) – oscillatory decay:
Key cue: peaks follow an exponential envelope; the motion crosses equilibrium many times.
- Critical damping – fastest monotonic return:
Key cue: no overshoot, return is as rapid as possible.
- Heavy damping (over‑damped) – slow monotonic return:
Key cue: no overshoot, return is slower than the critical case.
4. Free (Natural) Oscillations with Damping
4.1 Underdamped solution
Define the damping ratio and the damped angular frequency:
\[
\zeta=\frac{b}{2\sqrt{mk}},\qquad
\omegad=\omega0\sqrt{1-\zeta^{2}}\;( \zeta<1 )
\]
Displacement:
\[
x(t)=A\,e^{-\zeta\omega0 t}\cos\!\bigl(\omegad t+\phi\bigr)
\]
Amplitude decay:
\[
A(t)=A\,e^{-\zeta\omega_0 t}
\]
Time‑constant \(\tau=1/(\zeta\omega_0)\) governs how quickly the envelope shrinks.
4.2 Critical and Heavy damping
When \(\zeta\ge 1\) the solution is a sum of two non‑oscillatory exponentials:
\[
x(t)=C1e^{\lambda1 t}+C2e^{\lambda2 t},
\qquad
\lambda{1,2}= -\zeta\omega0\pm\omega_0\sqrt{\zeta^{2}-1}
\]
For \(\zeta=1\) (critical) the two roots coincide and the solution becomes
\[
x(t)=(C1+ C2 t)\,e^{-\omega_0 t}.
\]
4.3 Summary table – effect of damping on natural motion
| Regime | Behaviour | Frequency of motion |
|---|---|---|
| Underdamped (\(\zeta<1\)) | Oscillatory with exponential envelope | \(\omegad=\omega0\sqrt{1-\zeta^{2}}\) (slightly lower than \(\omega_0\)) |
| Critically damped (\(\zeta=1\)) | Monotonic return, fastest without overshoot | No periodic motion |
| Over‑damped (\(\zeta>1\)) | Monotonic return, slower than critical | No periodic motion |
5. Energy Loss due to Damping
Mechanical energy of the oscillator (potential + kinetic) decays as
\[
E(t)=\frac12 kA^{2}e^{-2\zeta\omega_0 t}.
\]
The instantaneous power removed by the damping force is
\[
\frac{dE}{dt}=-b\dot{x}^{2}\;<0,
\]
showing that the resistive force continuously does negative work, producing the exponential decay of amplitude.
6. Forced (Driven) Oscillations
6.1 Equation of motion
\[
m\ddot{x}+b\dot{x}+kx = F_{0}\cos(\omega t)
\]
6.2 General solution
The total solution is the sum of the homogeneous (transient) part and a particular (steady‑state) part.
- Transient term – the same as the free‑oscillation solution; it dies away after a few time‑constants \(\tau=1/(\zeta\omega_0)\).
- Steady‑state term – the motion that remains after the transient has vanished:
\[
x(t)=X(\omega)\cos\!\bigl(\omega t-\delta\bigr)
\]
6.3 Amplitude and phase of the steady‑state
\[
X(\omega)=\frac{F{0}/m}{\sqrt{\bigl(\omega0^{2}-\omega^{2}\bigr)^{2}+\bigl(2\zeta\omega_0\omega\bigr)^{2}}}
\]
\[
\tan\delta=\frac{2\zeta\omega0\omega}{\;\omega0^{2}-\omega^{2}\;}
\]
6.4 Resonance
- Resonance condition – the amplitude \(X(\omega)\) reaches a maximum when the driving frequency is close to the natural frequency. Damping shifts the exact peak slightly.
- Peak (resonant) frequency (light damping, \(\zeta\ll1\))
\[
\omega{r}\approx\omega0\sqrt{1-2\zeta^{2}}
\]
- Maximum steady‑state amplitude
\[
X{\max}= \frac{F{0}}{2m\omega_0\zeta}
\]
Thus weaker damping (smaller \(b\) or \(\zeta\)) gives a larger resonant response.
- Bandwidth (full width at half‑maximum)
\[
\Delta\omega\approx2\zeta\omega_0
\]
A larger \(\zeta\) produces a broader, less sharp resonance curve.
6.5 Sketch of the resonance curve
Key features to label: peak \(\omega{r}\), maximum amplitude \(X{\max}\), half‑maximum frequencies \(\omega_{r}\pm\Delta\omega/2\).
7. Mechanical ↔ Electrical Analogue
The differential equation for a series RLC circuit is
\[
L\ddot{q}+R\dot{q}+\frac{q}{C}=V_{0}\cos(\omega t)
\]
Comparing with the mechanical equation shows the following correspondence:
| Mechanical quantity | Electrical analogue |
|---|---|
| Mass \(m\) | Inductance \(L\) |
| Damping coefficient \(b\) | Resistance \(R\) |
| Spring constant \(k\) | Reciprocal capacitance \(1/C\) |
| Displacement \(x\) | Charge \(q\) |
| Force \(F\) | Voltage \(V\) |
Consequently the resonant angular frequency for a lightly damped RLC circuit is
\[
\omega_{r}\approx\frac{1}{\sqrt{LC}},
\]
and the quality factor \(Q=1/(2\zeta)=\omega_0L/R\) mirrors the mechanical damping ratio.
8. Practical Examples (with quantitative hints)
- Engineered structures – tuned‑mass damper
- A skyscraper of effective mass \(m=5\times10^{7}\,\text{kg}\) has a natural frequency \(\omega_0=0.2\;\text{rad s}^{-1}\). A tuned‑mass damper provides an additional damping coefficient \(b=8\times10^{5}\,\text{kg s}^{-1}\).
Exam‑style question: Calculate the damping ratio \(\zeta\) and state which regime the system lies in.
- A skyscraper of effective mass \(m=5\times10^{7}\,\text{kg}\) has a natural frequency \(\omega_0=0.2\;\text{rad s}^{-1}\). A tuned‑mass damper provides an additional damping coefficient \(b=8\times10^{5}\,\text{kg s}^{-1}\).
- Musical instrument – violin string
- A string of tension \(T=80\;\text{N}\) and linear density \(\mu=5\times10^{-4}\,\text{kg m}^{-1}\) has \(\omega_0\approx 400\;\text{rad s}^{-1}\). Air resistance and internal friction give \(b\approx0.03\;\text{kg s}^{-1}\).
Exam‑style question: Estimate \(\zeta\) and comment on whether the decay of the note is “lightly damped”.
- A string of tension \(T=80\;\text{N}\) and linear density \(\mu=5\times10^{-4}\,\text{kg m}^{-1}\) has \(\omega_0\approx 400\;\text{rad s}^{-1}\). Air resistance and internal friction give \(b\approx0.03\;\text{kg s}^{-1}\).
- Electrical analogue – series RLC circuit
- Take \(L=10\;\text{mH}\), \(C=100\;\mu\text{F}\) and \(R=5\;\Omega\).
Exam‑style question: Find the resonant frequency \(\omegar\) and the quality factor \(Q\); compare with the mechanical expressions \(\omegar\) and \(1/(2\zeta)\).
- Take \(L=10\;\text{mH}\), \(C=100\;\mu\text{F}\) and \(R=5\;\Omega\).
9. Summary Checklist (Cambridge Syllabus)
- Resistive force \(F_{\text d}=-b\dot{x}\) adds the term \(b\dot{x}\) to the equation of motion.
- Three damping regimes:
- Light damping (underdamped) – \(b<2\sqrt{mk}\) – oscillatory decay at \(\omega_d\).
- Critical damping – \(b=2\sqrt{mk}\) – fastest non‑overshooting return.
- Heavy damping (over‑damped) – \(b>2\sqrt{mk}\) – slow monotonic return.
- Underdamped amplitude falls as \(A(t)=A e^{-\zeta\omega0 t}\); energy decays as \(E(t)=\tfrac12kA^{2}e^{-2\zeta\omega0 t}\).
- For a driven system the transient (homogeneous) term disappears after a few \(\tau=1/(\zeta\omega_0)\), leaving the steady‑state amplitude \(X(\omega)\) and phase lag \(\delta\).
- Resonance occurs near \(\omega0\); the peak amplitude \(X{\max}\propto 1/\zeta\) and the bandwidth \(\Delta\omega\approx2\zeta\omega_0\).
- Mechanical quantities map directly onto an RLC circuit ( \(m\leftrightarrow L,\; b\leftrightarrow R,\; k\leftrightarrow 1/C\) ).
