Oscillations

Periodic motion is any motion that repeats itself identically at regular intervals of time $T$. Harmonic motion is periodic motion where the restoring mechanism is directly proportional to displacement.

1. Definition and Kinematics of SHM

A particle executes Simple Harmonic Motion (SHM) when the restoring force acting on it is directly proportional to its displacement from equilibrium and directed toward that equilibrium position: $$F = -k x$$ where $k$ is the force constant (stiffness) in N/m. By Newton's second law ($F = m \ddot{x}$): $$m \frac{d^2 x}{dt^2} + k x = 0 \implies \frac{d^2 x}{dt^2} + \omega_0^2 x = 0$$ where $\omega_0 = \sqrt{k/m}$ is the natural undamped angular frequency (rad/s). The general harmonic solution is: $$x(t) = A \cos(\omega_0 t + \phi)$$ where:

• $A$: Amplitude (maximum displacement from equilibrium).
• $\omega_0 = 2\pi f = \frac{2\pi}{T}$: Angular frequency.
• $\phi$: Initial phase constant (determined by initial conditions $x(0)$ and $v(0)$).
• $T = 2\pi \sqrt{\frac{m}{k}}$: Period of oscillation (independent of amplitude — isochronism).

2. Velocity and Acceleration

Differentiating displacement with respect to time: $$v(t) = \dot{x}(t) = -\omega_0 A \sin(\omega_0 t + \phi) = \omega_0 A \cos\left(\omega_0 t + \phi + \frac{\pi}{2}\right)$$ $$a(t) = \ddot{x}(t) = -\omega_0^2 A \cos(\omega_0 t + \phi) = -\omega_0^2 x(t) = \omega_0^2 A \cos(\omega_0 t + \phi + \pi)$$ Key phase relationships:

• Velocity leads displacement by $\frac{\pi}{2}$ radians (90°). Maximum speed $v_{max} = \omega_0 A$ occurs at equilibrium ($x = 0$).
• Acceleration leads displacement by $\pi$ radians (180°). Maximum acceleration $a_{max} = \omega_0^2 A$ occurs at maximum displacement ($x = \pm A$).

Eliminating time between $x$ and $v$ yields the elliptical phase space trajectory: $$v(x) = \pm \omega_0 \sqrt{A^2 - x^2} \implies \frac{x^2}{A^2} + \frac{v^2}{\omega_0^2 A^2} = 1$$

Simple harmonic motion is a conservative mechanical system where potential energy and kinetic energy continuously transform into one another while total mechanical energy remains invariant.

1. Kinetic and Potential Energy

• Kinetic Energy ($T$): $$T(t) = \frac{1}{2}m v^2 = \frac{1}{2}m \omega_0^2 A^2 \sin^2(\omega_0 t + \phi) = \frac{1}{2}k (A^2 - x^2)$$
• Potential Energy ($V$): Work done against the restoring force $F = -kx$: $$V(x) = -\int_0^x (-k x') dx' = \frac{1}{2}k x^2 = \frac{1}{2}k A^2 \cos^2(\omega_0 t + \phi)$$

2. Conservation of Total Energy

Summing kinetic and potential energies: $$E = T + V = \frac{1}{2}k A^2 \left[ \sin^2(\omega_0 t + \phi) + \cos^2(\omega_0 t + \phi) \right] = \frac{1}{2}k A^2 = \frac{1}{2}m \omega_0^2 A^2 = \text{constant}$$ The total mechanical energy is proportional to the square of the amplitude and independent of time and position. At the turning points ($x = \pm A$), velocity vanishes, and energy is purely potential ($E = V_{max} = \frac{1}{2}kA^2$). At the equilibrium position ($x = 0$), potential energy vanishes, and energy is purely kinetic ($E = T_{max} = \frac{1}{2}m v_{max}^2$).

3. Time-Averaged Energies and the Virial Theorem

Averaging over a complete oscillation period $T = 2\pi / \omega_0$: $$\langle \sin^2(\omega_0 t + \phi) \rangle = \frac{1}{T}\int_0^T \sin^2(\omega_0 t + \phi) dt = \frac{1}{2}$$ $$\langle \cos^2(\omega_0 t + \phi) \rangle = \frac{1}{2}$$ Therefore: $$\langle T \rangle = \frac{1}{4} k A^2 = \frac{1}{2}E, \quad \langle V \rangle = \frac{1}{4} k A^2 = \frac{1}{2}E$$ $$\langle T \rangle = \langle V \rangle = \frac{1}{2}E$$ This exact equipartition of average kinetic and potential energy is a direct consequence of the Virial Theorem for harmonic potentials ($V \propto x^2$).

Simple harmonic motion governs a multitude of oscillating mechanical and structural devices.

1. Simple Pendulum

A point mass $m$ suspended by an inextensible massless string of length $L$. Restoring torque about suspension point $O$: $$\tau = -m g L \sin\theta = I \alpha = (m L^2) \frac{d^2\theta}{dt^2}$$ $$\frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0$$ For small angular displacements ($\sin\theta \approx \theta$ in radians): $$\frac{d^2\theta}{dt^2} + \omega_0^2 \theta = 0 \implies \omega_0 = \sqrt{\frac{g}{L}}, \quad T = 2\pi \sqrt{\frac{L}{g}}$$

2. Compound (Physical) Pendulum

A rigid body of arbitrary shape and mass $M$ free to oscillate in a vertical plane about a horizontal knife-edge axis $O$. Let $d$ be the distance from the pivot $O$ to the center of mass $G$, and $I$ the moment of inertia about $O$. By the parallel axis theorem: $I = I_G + M d^2 = M (k_g^2 + d^2)$, where $k_g$ is the radius of gyration about $G$. The restoring torque is: $$\tau = -M g d \sin\theta \approx -M g d \theta$$ $$I \frac{d^2\theta}{dt^2} + M g d \theta = 0 \implies \frac{d^2\theta}{dt^2} + \left(\frac{M g d}{I}\right) \theta = 0$$ The period of oscillation is: $$T = 2\pi \sqrt{\frac{I}{M g d}} = 2\pi \sqrt{\frac{k_g^2 + d^2}{g d}} = 2\pi \sqrt{\frac{L_{eq}}{g}}$$ where $L_{eq} = \frac{k_g^2 + d^2}{d} = d + \frac{k_g^2}{d}$ is the length of the equivalent simple pendulum.

• Center of Oscillation ($O'$): A point lying along the line $OG$ at distance $L_{eq}$ from $O$. If the body is suspended from $O'$, its period of oscillation is identical to that about $O$ (Theorem of Reversibility, exploited in Kater's reversible pendulum to determine $g$ with parts-per-million accuracy).
• Minimum Period: Minimizing $L_{eq}(d)$ with respect to $d$: $$\frac{dL_{eq}}{dd} = 1 - \frac{k_g^2}{d^2} = 0 \implies d = k_g$$ The minimum period occurs when the suspension point is at a distance equal to the radius of gyration: $T_{min} = 2\pi \sqrt{2 k_g / g}$.

3. Torsional Pendulum

A disk or cylinder suspended by a thin elastic wire. Twisting by angle $\theta$ creates a restoring torque $\tau = -C \theta$, where $C = \frac{\pi \eta r^4}{2 L}$ is the torsional rigidity of the wire. $$I \frac{d^2\theta}{dt^2} + C \theta = 0 \implies T = 2\pi \sqrt{\frac{I}{C}}$$

Simple harmonic motion can be mathematically and visually understood as the one-dimensional orthogonal projection of uniform circular motion.

1. The Reference Circle and Phasor Representation

Consider a reference particle $P$ moving counterclockwise along a circle of radius $A$ (called the reference circle) with constant angular velocity $\omega_0$. At $t = 0$, the radius vector makes an angle $\phi$ with the positive x-axis. At subsequent time $t$, the angle is $\theta(t) = \omega_0 t + \phi$. Projecting point $P$ onto the horizontal x-axis yields point $Q$: $$x(t) = A \cos(\omega_0 t + \phi)$$ Projecting point $P$ onto the vertical y-axis yields point $Q'$: $$y(t) = A \sin(\omega_0 t + \phi) = A \cos\left(\omega_0 t + \phi - \frac{\pi}{2}\right)$$ Both projected points $Q$ and $Q'$ execute pure simple harmonic motion with amplitude $A$ and angular frequency $\omega_0$, separated by a 90° phase difference.

2. Kinematic Projections

• Velocity: The linear tangential speed of $P$ on the circle is $v_0 = \omega_0 A$. Its projection on the x-axis gives the SHM velocity: $$v_x = -v_0 \sin(\omega_0 t + \phi) = -\omega_0 A \sin(\omega_0 t + \phi)$$
• Acceleration: The centripetal acceleration of $P$ directed toward the center is $a_c = \omega_0^2 A$. Its projection on the x-axis gives the SHM acceleration: $$a_x = -a_c \cos(\omega_0 t + \phi) = -\omega_0^2 x(t)$$

This phasor technique enables geometric addition of multiple harmonic oscillations using planar vector addition.

When a particle is acted upon by two or more simultaneous harmonic restoring forces, its resultant trajectory is determined by the principle of superposition.

1. Superposition of Two Collinear SHMs of Identical Frequency

Let two collinear oscillations along the x-axis be: $$x_1(t) = A_1 \cos(\omega t + \phi_1), \quad x_2(t) = A_2 \cos(\omega t + \phi_2)$$ The resultant displacement is $x(t) = x_1(t) + x_2(t) = A \cos(\omega t + \Phi)$, where: $$A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi_2 - \phi_1)}$$ $$\tan\Phi = \frac{A_1 \sin\phi_1 + A_2 \sin\phi_2}{A_1 \cos\phi_1 + A_2 \cos\phi_2}$$

• If in phase ($\Delta\phi = 2n\pi$): $A = A_1 + A_2$ (Constructive).
• If in antiphase ($\Delta\phi = (2n+1)\pi$): $A = |A_1 - A_2|$ (Destructive).

2. Superposition of Two Mutually Perpendicular SHMs: Lissajous Figures

Consider a particle subjected to two orthogonal oscillations: $$x(t) = A \cos(\omega_x t), \quad y(t) = B \cos(\omega_y t + \delta)$$ The resulting path $(x(t), y(t))$ in the 2D plane is called a Lissajous Figure (discovered by Jules Antoine Lissajous, 1857).

Case A: Equal Frequencies ($\omega_x = \omega_y = \omega$)

Expanding $y(t)$: $$\frac{y}{B} = \cos(\omega t)\cos\delta - \sin(\omega t)\sin\delta = \frac{x}{A}\cos\delta - \sqrt{1 - \frac{x^2}{A^2}}\sin\delta$$ Rearranging and squaring: $$\left(\frac{y}{B} - \frac{x}{A}\cos\delta\right)^2 = \left(1 - \frac{x^2}{A^2}\right)\sin^2\delta$$ $$\frac{x^2}{A^2} - \frac{2 x y}{A B}\cos\delta + \frac{y^2}{B^2} = \sin^2\delta$$ This is the general equation of an oblique ellipse bounded within the rectangle $[-A, A] \times [-B, B]$:

• $\delta = 0$: Straight line of positive slope $y = (B/A) x$.
• $\delta = \pi/2$: Symmetrical upright ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ (circle if $A = B$).
• $\delta = \pi$: Straight line of negative slope $y = -(B/A) x$.
• $\delta = 3\pi/2$: Upright ellipse traced clockwise.

Case B: Frequency Ratio 1:2 ($\omega_y = 2\omega_x$)

$$x = A \cos(\omega t), \quad y = B \cos(2\omega t + \delta)$$ Using $\cos(2\theta) = 2\cos^2\theta - 1$, when $\delta = 0$: $$y = B [2(x/A)^2 - 1]$$ This forms a parabola! For arbitrary phase differences $\delta$, the curve traces a figure-eight (lemniscate) or distorted loop. The frequency ratio is determined experimentally by counting tangencies: $$\frac{\omega_x}{\omega_y} = \frac{\text{Number of intersections with vertical line}}{\text{Number of intersections with horizontal line}}$$

Real physical oscillators experience dissipative resistive forces (viscous drag, friction) that continually remove mechanical energy.

1. The Damped Equation of Motion

Assuming viscous damping where the retarding force is proportional to velocity: $F_d = -b \dot{x}$, where $b$ is the damping coefficient (N·s/m). Newton's second law: $$m \ddot{x} = -k x - b \dot{x} \implies m \ddot{x} + b \dot{x} + k x = 0$$ $$\ddot{x} + 2\gamma \dot{x} + \omega_0^2 x = 0$$ where $\gamma = \frac{b}{2m}$ is the damping attenuation constant (s⁻¹) and $\omega_0 = \sqrt{k/m}$ is the natural frequency. Seeking solutions of the form $x(t) = e^{\lambda t}$ yields the auxiliary equation: $$\lambda^2 + 2\gamma \lambda + \omega_0^2 = 0 \implies \lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$$

2. The Three Damping Regimes

1. Underdamped Case ($\gamma < \omega_0$): The roots are complex conjugates $\lambda = -\gamma \pm i \omega_d$, where $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$ is the damped angular frequency. $$x(t) = A_0 e^{-\gamma t} \cos(\omega_d t + \phi)$$ The system oscillates with period $T_d = \frac{2\pi}{\omega_d} > T_0$ while its amplitude decays exponentially: $A(t) = A_0 e^{-\gamma t}$. Logarithmic Decrement ($\delta$): The natural logarithm of the ratio of two consecutive peak amplitudes separated by one period $T_d$: $$\delta = \ln\left( \frac{x(t)}{x(t + T_d)} \right) = \ln\left( \frac{A_0 e^{-\gamma t}}{A_0 e^{-\gamma (t + T_d)}} \right) = \gamma T_d = \frac{2\pi \gamma}{\omega_d}$$
2. Critically Damped Case ($\gamma = \omega_0$): Repeated real root $\lambda = -\gamma$. The general solution is: $$x(t) = (C_1 + C_2 t) e^{-\gamma t}$$ The system returns to equilibrium in the shortest possible time without oscillating (vital for car shock absorbers, galvonometers, and door closers).
3. Overdamped Case ($\gamma > \omega_0$): Two unequal negative real roots. The motion is non-oscillatory and dies out sluggishly: $$x(t) = C_1 e^{-(\gamma - \sqrt{\gamma^2-\omega_0^2}) t} + C_2 e^{-(\gamma + \sqrt{\gamma^2-\omega_0^2}) t}$$

3. The Quality Factor ($Q$)

The Quality Factor $Q$ quantifies the sharpness of an oscillator and its ability to store energy relative to rate of dissipation: $$Q = 2\pi \left( \frac{\text{Energy Stored in System}}{\text{Energy Dissipated per Cycle}} \right) = \frac{\omega_0}{2\gamma} = \frac{\omega_0 m}{b} = \frac{\pi}{\delta}$$ High-Q oscillators (e.g., quartz crystals with $Q \sim 10^5$, or optical cavities with $Q \sim 10^9$) ring for many thousands of cycles before dying out.

When a damped oscillator is driven by a periodic external force $F(t) = F_0 \cos(\omega t)$, it undergoes forced oscillations.

1. Differential Equation and Steady-State Solution

$$\ddot{x} + 2\gamma \dot{x} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega t)$$ The complete solution consists of a transient complementary function $x_h(t)$ (which decays as $e^{-\gamma t}$) plus a steady-state particular solution $x_p(t)$ oscillating at the driving frequency $\omega$: $$x(t) = x_{\text{transient}}(t) + x_{\text{steady}}(t)$$ After transient decay, the steady-state response is: $$x(t) = A(\omega) \cos(\omega t - \phi)$$ where amplitude $A(\omega)$ and phase lag $\phi(\omega)$ are: $$A(\omega) = \frac{F_0 / m}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4 \gamma^2 \omega^2}}$$ $$\tan\phi(\omega) = \frac{2\gamma \omega}{\omega_0^2 - \omega^2}, \quad 0 \le \phi \le \pi$$

2. Amplitude and Velocity Resonance

• Amplitude Resonance: Maximizing $A(\omega)$ by minimizing the denominator: $$\frac{d}{d\omega}\left[ (\omega_0^2 - \omega^2)^2 + 4\gamma^2 \omega^2 \right] = 2(\omega_0^2 - \omega^2)(-2\omega) + 8\gamma^2 \omega = 0$$ $$\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$$ Resonance occurs slightly below the natural frequency $\omega_0$. The peak amplitude at $\omega = \omega_r$ is: $$A_{max} = \frac{F_0 / m}{2\gamma \sqrt{\omega_0^2 - \gamma^2}} \approx \frac{F_0 / m}{2\gamma \omega_0} = \frac{Q F_0}{m \omega_0^2} = Q \cdot x_{\text{static}}$$ At resonance, amplitude is magnified by exactly the Quality Factor $Q$!
• Velocity (Power) Resonance: Differentiating $x(t)$ gives velocity amplitude: $$v_{max}(\omega) = \frac{\omega F_0 / m}{\sqrt{(\omega_0^2 - \omega^2)^2 + 4\gamma^2 \omega^2}} = \frac{F_0 / m}{\sqrt{(\frac{\omega_0^2 - \omega^2}{\omega})^2 + 4\gamma^2}}$$ The velocity resonance peak occurs exactly at $\omega = \omega_0$, where the velocity is in phase with the driving force ($\phi = \pi/2$).

3. Sharpness of Resonance and Bandwidth (FWHM)

The average power absorbed by the oscillator is: $$\langle P(\omega) \rangle = \frac{1}{2} b v_{max}^2(\omega) = \frac{F_0^2 \gamma \omega^2 / m}{(\omega_0^2 - \omega^2)^2 + 4\gamma^2 \omega^2}$$ The half-power frequencies $\omega_1, \omega_2$ occur when $\langle P \rangle = \frac{1}{2} P_{max}$: $$\Delta \omega = \omega_2 - \omega_1 = 2\gamma = \frac{\omega_0}{Q}$$ The sharpness of resonance is inversely proportional to bandwidth: a high $Q$ produces an extremely narrow, sharp resonance peak.