Electric fields

Damped and Forced Oscillations, Resonance

Learning Objective

Understand how damping and external driving forces affect simple harmonic motion (SHM), and relate these concepts to electric fields in oscillatory circuits.

Key Concepts

• Simple Harmonic Motion (SHM) – ideal, undamped case.
• Damping – energy loss per cycle, characterised by a damping coefficient \$b\$.
• Forced (driven) oscillations – external periodic force \$F{\text{d}}(t)=F0\cos(\omega_{\text{d}}t)\$.
• Resonance – condition where the driving frequency matches the natural frequency of the system.
• Electric field analogy – RLC circuit as a mechanical oscillator.

Damped Oscillations

The equation of motion for a damped oscillator is

\$m\ddot{x}+b\dot{x}+kx=0\$

where \$m\$ is the mass, \$k\$ the spring constant, and \$b\$ the damping coefficient.

Three regimes are distinguished by the discriminant \$\Delta = b^{2}-4mk\$:

1. Underdamped (\$b^{2}<4mk\$): oscillatory motion with exponentially decaying amplitude.
2. Critically damped (\$b^{2}=4mk\$): fastest return to equilibrium without overshoot.
3. Overdamped (\$b^{2}>4mk\$): slow, non‑oscillatory return.

Forced Oscillations

When a periodic driving force \$F{\text{d}}(t)=F0\cos(\omega_{\text{d}}t)\$ acts on the system, the equation becomes

\$m\ddot{x}+b\dot{x}+kx = F0\cos(\omega{\text{d}}t)\$

The steady‑state solution has the form

\$x(t)=A(\omega{\text{d}})\cos\!\big(\omega{\text{d}}t-\phi\big)\$

with amplitude

\$A(\omega{\text{d}})=\frac{F0/m}{\sqrt{(\omega0^{2}-\omega{\text{d}}^{2})^{2}+(\gamma\omega_{\text{d}})^{2}}}\$

and phase lag

\$\tan\phi=\frac{\gamma\omega{\text{d}}}{\omega0^{2}-\omega_{\text{d}}^{2}}\$

where \$\omega_0=\sqrt{k/m}\$ is the natural angular frequency and \$\gamma=b/m\$ the damping constant.

Resonance

Resonance occurs when the driving frequency \$\omega{\text{d}}\$ maximises the amplitude \$A\$. Differentiating \$A(\omega{\text{d}})\$ gives the resonance condition

\$\omega{\text{r}}=\sqrt{\omega0^{2}-\frac{\gamma^{2}}{2}}\$

For light damping (\$\gamma\ll\omega0\$) this reduces to \$\omega{\text{r}}\approx\omega_0\$.

The quality factor \$Q\$ quantifies the sharpness of resonance:

\$Q=\frac{\omega_0}{\gamma}=\frac{1}{b}\sqrt{\frac{k}{m}}\$

A high \$Q\$ means a narrow resonance peak and large amplitude at \$\omega_{\text{r}}\$.

Electric Field Analogy – RLC Circuit

An RLC series circuit obeys the same differential equation as a damped driven mechanical oscillator, with the substitutions

The circuit equation is

\$L\frac{d^{2}q}{dt^{2}}+R\frac{dq}{dt}+\frac{q}{C}= \mathcal{E}0\cos(\omega{\text{d}}t)\$

The voltage across the capacitor \$VC = q/C\$ plays the role of the displacement \$x\$, and the electric field \$E\$ inside the capacitor is \$E = VC/d\$, linking the mechanical amplitude to an electric field amplitude.

Common Misconceptions

• Assuming resonance occurs at the natural frequency even for heavily damped systems – the resonance frequency shifts lower.
• Confusing the amplitude at resonance with the maximum possible amplitude; the latter also depends on the driving force magnitude \$F_0\$.
• Neglecting the phase relationship – at resonance the phase lag is \$90^\circ\$ (or \$\pi/2\$ rad).
• Thinking that higher \$Q\$ always means larger amplitude – \$Q\$ affects the width of the resonance peak, not the absolute amplitude without considering \$F_0\$.

Example Problem

Problem: A mass \$m=0.5\;\text{kg}\$ is attached to a spring with \$k=200\;\text{N m}^{-1}\$ and experiences a damping force \$b\dot{x}\$ with \$b=2\;\text{kg s}^{-1}\$. It is driven by a force \$F0=10\;\text{N}\$ at a frequency \$f{\text{d}}=2\;\text{Hz}\$. Determine the steady‑state amplitude and the phase lag.

1. Calculate \$\omega_0=\sqrt{k/m}= \sqrt{200/0.5}=20\;\text{rad s}^{-1}\$.
2. Find \$\gamma=b/m=2/0.5=4\;\text{s}^{-1}\$.
3. Convert driving frequency: \$\omega{\text{d}}=2\pi f{\text{d}}=2\pi(2)=4\pi\;\text{rad s}^{-1}\approx12.57\;\text{rad s}^{-1}\$.
4. Amplitude:

   \$\$A=\frac{F0/m}{\sqrt{(\omega0^{2}-\omega{\text{d}}^{2})^{2}+(\gamma\omega{\text{d}})^{2}}}

   =\frac{10/0.5}{\sqrt{(400-158)^{2}+(4\times12.57)^{2}}}\approx0.058\;\text{m}\$\$

5. Phase lag:

   \$\$\tan\phi=\frac{\gamma\omega{\text{d}}}{\omega0^{2}-\omega_{\text{d}}^{2}}

   =\frac{4\times12.57}{400-158}\approx0.21\;\Rightarrow\;\phi\approx12^{\circ}\$\$

Thus the mass oscillates with an amplitude of about \$5.8\;\text{cm}\$ and lags the driving force by roughly \$12^{\circ}\$.

Suggested Diagram