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:

\[ \zeta=\frac{b}{2\sqrt{mk}},\qquad \omega_d=\omega_0\sqrt{1-\zeta^{2}}\;( \zeta<1 ) \]

Displacement:

\[ x(t)=A\,e^{-\zeta\omega_0 t}\cos\!\bigl(\omega_d t+\phi\bigr) \]

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:

\[ x(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}, \qquad \lambda_{1,2}= -\zeta\omega_0\pm\omega_0\sqrt{\zeta^{2}-1} \]

For \(\zeta=1\) (critical) the two roots coincide and the solution becomes

\[ x(t)=(C_1+ C_2 t)\,e^{-\omega_0 t}. \]

4.3 Summary table – effect of damping on natural motion

Regime	Behaviour	Frequency of motion
Underdamped (\(\zeta<1\))	Oscillatory with exponential envelope	\(\omega_d=\omega_0\sqrt{1-\zeta^{2}}\) (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

\[ X(\omega)=\frac{F_{0}/m}{\sqrt{\bigl(\omega_0^{2}-\omega^{2}\bigr)^{2}+\bigl(2\zeta\omega_0\omega\bigr)^{2}}} \] \[ \tan\delta=\frac{2\zeta\omega_0\omega}{\;\omega_0^{2}-\omega^{2}\;} \]

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, \(\zeta\ll1\)) \[ \omega_{r}\approx\omega_0\sqrt{1-2\zeta^{2}} \]
• 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)

1. 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.
2. 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”.
3. 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 \(\omega_r\) and the quality factor \(Q\); compare with the mechanical expressions \(\omega_r\) and \(1/(2\zeta)\).

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 \(A(t)=A e^{-\zeta\omega_0 t}\); energy decays as \(E(t)=\tfrac12kA^{2}e^{-2\zeta\omega_0 t}\).
• 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 \(\omega_0\); 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 ( \(m\leftrightarrow L,\; b\leftrightarrow R,\; k\leftrightarrow 1/C\) ).