Introduction to Computational Physics – Oscillations & Harmonic Motion

Periodic Motion

• Definition: A motion that repeats itself at regular time intervals.
  • The fixed interval after which the motion repeats is called the time period (T).
• Examples mentioned
  • Planet orbiting the Sun (annual period).
  • Hands of a clock (e.g.
  • Hour-hand $T = 12\text{ h}$,
  • Minute-hand $T = 1\text{ h}$).
  • Satellite revolving round Earth.
  • Earth’s own rotation (sidereal day $\approx 24\text{ h}$).
  • Car wheel spinning while driving.
• Special subclass – Oscillatory Motion: to-and-fro motion along the same path.
  • All oscillatory motions are periodic, but converse is not always true.

Oscillatory Motion

• Specifics: Object moves back & forth / up & down about an equilibrium point.
• Examples
  • Swinging pendulum.
  • Vibrating tuning fork.
  • Mass–spring system (vertical or horizontal).
  • Playground swing.
  • Vibrating string on musical instruments.

Simple Harmonic Motion (SHM)

• Rigorous definition
  • Motion in a straight line where acceleration $a$ is always directed toward a fixed point (centre of oscillation / equilibrium) and proportional to displacement $x$ from that point.
    $a = -\omega^{2}x$
  • Via Newton’s 2nd law $F = ma$, the restoring force is
    $F = -m\omega^{2}x = -kx \qquad (1)$
    where $k = m\omega^{2}$ is the spring (force) constant.
• Consequences
  • Restoring force vanishes at equilibrium $x = 0$.
  • Centre of oscillation = equilibrium position.

Differential Equation & General Solution

• Starting from $F = -kx$:
  $m\dfrac{d^{2}x}{dt^{2}} + kx = 0$
  $\dfrac{d^{2}x}{dt^{2}} + \omega^{2}x = 0 \qquad \big(\omega^{2}=k/m\big)$
• Solution for displacement
  $x(t) = A\sin(\omega t + \delta)$
• Velocity
  $v(t)=\dfrac{dx}{dt}=A\omega\cos(\omega t+\delta)$
• Acceleration
  $a(t)=\dfrac{d^{2}x}{dt^{2}}=-A\omega^{2}\sin(\omega t+\delta)=-\omega^{2}x(t)$

Alternative Derivation via Energy (work–energy)

• Starting with initial conditions $x<em>{0},\;v</em>{0}$, integrate $v dv = -\omega^{2}x dx$ to obtain $v^{2}=v<em>{0}^{2}+\omega^{2}\bigl(x</em>{0}^{2}-x^{2}\bigr)$
  • Introduce $A^{2}=\dfrac{v<em>{0}^{2}}{\omega^{2}}+x</em>{0}^{2}$ giving
    $v=\omega\sqrt{A^{2}-x^{2}}$ and, after separation, the usual sine solution.

Parameters

• Amplitude $A$: maximum displacement (takes values $\pm A$).
• Time period $T$: smallest time after which state repeats.
  • From requirement $\omega (t+T)+\delta = \omega t + \delta + 2\pi$
    $\boxed{T = \dfrac{2\pi}{\omega}} = 2\pi\sqrt{\dfrac{m}{k}}$
• Frequency $\nu$ & angular frequency $\omega$:
  $\nu = \dfrac{1}{T}=\dfrac{\omega}{2\pi}=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$
• Phase $\Phi = \omega t+\delta$ locates particle’s status.
  • $\Phi=0$ ⇒ particle crosses mean position with positive velocity $v = A\omega$.
  • $\Phi = \dfrac{\pi}{2}$ ⇒ particle at positive extreme $x=A,\;v=0$.

Worked Example (Problem 1)

• Given $x = 5\,\text{m}\,\sin\Bigl(\pi\,\text{s}^{-1}\,t + \dfrac{\pi}{3}\Bigr)$.
  • Amplitude $A = 5\,\text{m}$.
  • Angular frequency $\omega = \pi\,\text{s}^{-1}$.
  • Time period $T = \dfrac{2\pi}{\pi}=2\,\text{s}$.
  • Max speed $v_{\text{max}} = A\omega = 5\times \pi\; \text{m/s}$.
  • Velocity at $t=1\,\text{s}$:
    $v = A\omega\cos\Bigl(\pi\cdot1 + \dfrac{\pi}{3}\Bigr)=5\pi\cos\Bigl(\dfrac{4\pi}{3}\Bigr)=5\pi\left(-\dfrac{1}{2}\right)=-\dfrac{5\pi}{2}\,\text{m/s}$ (toward the centre).

Energy in SHM

• Potential energy (taking $U=0$ at $x=0$):
  $U(x)=\dfrac{1}{2}kx^{2}=\dfrac{1}{2}m\omega^{2}x^{2}$
• Kinetic energy:
  $K=\dfrac{1}{2}mv^{2}=\dfrac{1}{2}mA^{2}\omega^{2}\cos^{2}(\omega t+\delta)$
• Total mechanical energy (constant): $E=K+U=\dfrac{1}{2}mA^{2}\omega^{2}\Bigl[\sin^{2}(\omega t+\delta)+\cos^{2}(\omega t+\delta)\Bigr]=\boxed{\dfrac{1}{2}mA^{2}\omega^{2}}$
  • Energy continuously interchanges between $U$ & $K$.

Linear Harmonic Oscillator – Simple Pendulum (small-angle)

• Tangential equation: $m\dfrac{d^{2}x}{dt^{2}} = -mg\sin\theta$.
• For small amplitude $\sin\theta \approx \theta = \dfrac{x}{\ell}$ giving $\dfrac{d^{2}x}{dt^{2}} = -\dfrac{g}{\ell}x$.
  • Thus $\omega = \sqrt{g/\ell}$ and $T = 2\pi\sqrt{\dfrac{\ell}{g}}$.
• Same linear SHM form applies.

Damped Oscillations

• Real systems lose energy to friction / resistance ⇒ amplitude decays.
• Equation (mass $m$, damping coefficient $\mu$, spring constant $k$):
  $m\dfrac{d^{2}x}{dt^{2}} + \mu\dfrac{dx}{dt} + kx = 0$
• Assume trial solution $x = Ce^{\alpha t}$ leading to characteristic eqn
  $\alpha^{2} + 2\delta \alpha + \omega<em>{0}^{2}=0$ with $\delta = \dfrac{\mu}{2m}, \quad \omega</em>{0}=\sqrt{\dfrac{k}{m}}$.

Three Regimes

1. Overdamped (\delta > \omega_{0})
   • Two negative real roots $\alpha<em>{1,2}= -\delta \pm \sqrt{\delta^{2}-\omega</em>{0}^{2}}$.
   • Displacement: $x(t)=e^{-\delta t}\bigl(C<em>{1}e^{\lambda t}+C</em>{2}e^{-\lambda t}\bigr)$ with $\lambda = \sqrt{\delta^{2}-\omega_{0}^{2}}$.
   • No oscillation; system returns exponentially to equilibrium.
2. Critically damped $(\delta = \omega_{0})$
   • Double root $\alpha = -\delta$.
   • Solution: $x(t)=e^{-\delta t}(C<em>{1}+C</em>{2}t)$.
   • Fastest non-oscillatory return to equilibrium.
3. Underdamped (\delta < \omega_{0})
   • Complex roots $\alpha = -\delta \pm i\bar\omega$ where $\bar\omega = \sqrt{\omega_{0}^{2}-\delta^{2}}$.
   • Solution (decaying sinusoid):
     $x(t)=Ae^{-\delta t}\cos(\bar\omega t + \phi)$.
   • Amplitude drops exponentially.

• Qualitative plots (Page 17) compare displacement vs time for the three cases.

Forced (Driven) Oscillations

• External periodic force $f_{0}\sin\omega t$ supplies energy, compensating damping.
• Equation:
  $\dfrac{d^{2}x}{dt^{2}}+2\delta\dfrac{dx}{dt}+\omega<em>{0}^{2}x = \dfrac{f</em>{0}}{m}\sin\omega t$
• Motion has two phases
  1. Transient phase – natural frequency $\omega_{0}$; dies out due to damping.
  2. Persistent (steady-state) phase – oscillates at driving frequency $\omega$ with constant amplitude.

Steady-State Solution

• Assume $x = a\sin(\omega t - \Phi)$.
• Substitute to obtain algebraic relations
  $a(\omega<em>{0}^{2}-\omega^{2}) = f</em>{0}\cos\Phi$
  $2\delta a\omega = f_{0}\sin\Phi$
• Amplitude
  $\boxed{a = \dfrac{f<em>{0}}{\sqrt{(\omega</em>{0}^{2}-\omega^{2})^{2} + (2\delta\omega)^{2}}}}$
• Phase lag
  $\tan\Phi = \dfrac{2\delta\omega}{\omega_{0}^{2}-\omega^{2}}$.
• Special features
  • Resonance: amplitude maximal when $\omega \approx \sqrt{\omega<em>{0}^{2}-2\delta^{2}}$ (≈ $\omega</em>{0}$ for weak damping).
  • Phase passes from $0$ (low $\omega$) through $90^{\circ}$ at resonance to $180^{\circ}$ (high $\omega$).
• Full solution = transient + steady; textbook wrote
  $x(t)=A_{h}e^{-\delta t}\sin(\bar\omega t+\varphi)+a\sin(\omega t-\Phi)$.

Two-Body (Coupled) Harmonic Motion

• System: two equal masses $m$ connected by three springs (end springs $k<em>{1}$, middle spring $k</em>{2}$).
• Equations of motion (about points A & B):
  $m\ddot x<em>{1} + (k</em>{1}+k<em>{2})x</em>{1} - k<em>{2}x</em>{2} = 0 \qquad (1)$
  $m\ddot x<em>{2} + (k</em>{1}+k<em>{2})x</em>{2} - k<em>{2}x</em>{1} = 0 \qquad (2)$
• Assume $x<em>{j}=A</em>{j}e^{i\omega t}$ → linear homogeneous system.
• Determinant condition gives two normal-mode frequencies.

Normal Modes

1. First (symmetric) mode – masses oscillate in phase $x<em>{1}=x</em>{2}$ (middle spring slack).
   $\omega<em>{1}= \sqrt{\dfrac{k</em>{1}}{m}}$
2. Second (antisymmetric) mode – masses opposite phase $x<em>{1}=-x</em>{2}$ (middle spring stretched/compressed).
   $\omega<em>{2}= \sqrt{\dfrac{k</em>{1}+2k_{2}}{m}}$

• Express general motion as superposition using $q<em>{1}=x</em>{1}-x<em>{2} \quad (\text{fast mode}),\quad q</em>{2}=x<em>{1}+x</em>{2}\;(\text{slow mode})$$x<em>{1} = \tfrac{1}{2}(q</em>{1}+q<em>{2}), \; x</em>{2}=\tfrac{1}{2}(q<em>{2}-q</em>{1})$
  • Where $q<em>{1}(t)=A</em>{1}\sin(\omega<em>{1}t+\phi</em>{1})$,
    $q<em>{2}(t)=A</em>{2}\sin(\omega<em>{2}t+\phi</em>{2})$.

Forced Harmonic Oscillator – Compact Summary (Page 28)

• Generic differential form with linear viscous damping $b$:
  $m\ddot x + b\dot x + kx = F_{0}\sin(\omega t)$
• Solution structure $x(t)=A<em>{h}e^{-\gamma t}\sin(\omega</em>{h}t+\phi) + A<em>{p}\cos(\omega t + \phi</em>{p})$
  • $A_{h}e^{-\gamma t}\,\dots$ → transient.
  • $A<em>{p}\cos(\omega t+\phi</em>{p})$ → steady state.

Motion of a Linear Harmonic Oscillator – Alternate Pendulum Derivation (Page 34)

• Start from polar form: torque equation $\tau = -mg\sin\theta \cdot L$ with angular acceleration $\alpha$.
• For small $\theta$:
  $L\alpha = -g\theta \quad \Rightarrow \quad \ddot\theta + \dfrac{g}{L}\theta = 0$.
• Identical to simple pendulum result above.

Consolidated Formula Sheet

• SHM: $x=A\sin(\omega t+\delta),\;v=A\omega\cos(\omega t+\delta),\;a=-\omega^{2}x$.
• $\omega = \sqrt{k/m}, \; T=2\pi/\omega, \; \nu=1/T$.
• Energy: $E=\tfrac{1}{2}mA^{2}\omega^{2}$.
• Damped: $m\ddot x + 2\delta m \dot x + kx = 0$ with $\delta=\mu/2m$.
• Forced: $m\ddot x + 2\delta m \dot x + kx = f_{0}\sin\omega t$.
• Steady amplitude: $a = f<em>{0}/\sqrt{(\omega</em>{0}^{2}-\omega^{2})^{2}+(2\delta\omega)^{2}}$.
• Phase lag: $\tan\Phi = 2\delta\omega/(\omega_{0}^{2}-\omega^{2})$.
• Coupled masses: $\omega<em>{1}=\sqrt{k</em>{1}/m},\;\omega<em>{2}=\sqrt{(k</em>{1}+2k_{2})/m}$.

Conceptual & Practical Notes

• Oscillations and waves underpin numerous physical systems: clocks, seismology, AC circuits.
• Damping models energy dissipation; critical damping used in door closers & automotive suspensions.
• Forced oscillations and resonance explain bridge collapses (Tacoma Narrows) and tuning of musical instruments.
• Coupled oscillators form basis of molecular vibrations and phonons in solids.
• Ethical engineering: ensure structures avoid dangerous resonances; design damping appropriately.