Chapter 14 — Oscillations

Whenever a system is displaced from equilibrium and the restoring force tries to bring it back, the system oscillates. The vibrations of a tuning fork, the swing of a pendulum, the rocking of atoms in a crystal, even the oscillation of charge in an LC circuit — all share the same underlying mathematics: simple harmonic motion (SHM).

This chapter develops SHM from scratch, links it to uniform circular motion, derives the period for springs and pendulums, builds the energy picture, and then opens the door to damped and forced oscillations — including the deeply important phenomenon of resonance.

Concept Map

• 14.1 Periodic and oscillatory motion — period, frequency, displacement.
• 14.2 SHM — definition, kinematic equations, phase.
• 14.3 SHM as a projection of uniform circular motion.
• 14.4 Force law $F = -kx$; energy in SHM.
• 14.5 Systems executing SHM — horizontal spring, vertical spring, parallel & series combinations.
• 14.6 Simple pendulum; compound pendulum (brief).
• 14.7 Damped SHM — under-, critical-, over-damped.
• 14.8 Forced oscillations and resonance.

14.1 Periodic and Oscillatory Motion

Definitions

• A motion that repeats itself in equal time intervals is periodic. The smallest such interval $T$ is the period. Examples: Earth's revolution, the hands of a clock, a planet around a star.
• A motion about a mean position in which the body keeps moving to and fro is oscillatory (or vibratory). Examples: a pendulum, a tuning fork, a guitar string.

Every oscillatory motion is periodic, but not every periodic motion is oscillatory.

Frequency, Angular Frequency

• Frequency $\nu = 1/T$, measured in hertz ($1\ \text{Hz} = 1\ \text{s}^{-1}$).
• Angular frequency $\omega = 2\pi\nu = 2\pi/T$, in rad/s.

Displacement

For one-dimensional oscillation, the displacement $x(t)$ from the mean position is a periodic function of time:

Any periodic function can be written as a sum of sines and cosines (Fourier theorem). The simplest such function is a single sinusoid — hence "simple" harmonic motion.

Pitfalls

• "Periodic" $\neq$ "oscillatory". Circular motion is periodic but not oscillatory.
• Frequency has units of Hz, not rpm in physics problems unless explicitly converted.

14.2 Simple Harmonic Motion

Definition

Motion in which the displacement varies sinusoidally with time:

where

• $A$ — amplitude (max displacement),
• $\omega$ — angular frequency,
• $\phi$ — phase constant (initial phase),
• $(\omega t + \phi)$ — phase at time $t$.

Velocity and Acceleration

Differentiate:

So:

This is the defining equation of SHM. Any system with acceleration proportional to negative displacement executes SHM.

Maximum Values

Velocity–Displacement Relation

Eliminate time from $x = A\sin\theta$, $v = A\omega\cos\theta$:

A plot of $v$ vs $x$ is an ellipse.

Phase

Two SHMs with the same $\omega$ but different $\phi$ are in phase if $\phi_1 = \phi_2$, out of phase if $\vert \phi_1-\phi_2\vert =\pi$, in quadrature if $\vert \phi_1-\phi_2\vert =\pi/2$.

Worked Example

A particle in SHM with $A=5\ \text{cm}$, $T=2\ \text{s}$. Find $v$ and $a$ when $x=3\ \text{cm}$.

$\omega = 2\pi/T = \pi\ \text{rad/s}$.

Pitfalls

• The sign in $a=-\omega^2 x$ is crucial — it says the acceleration is always toward the mean position.
• "Frequency" sometimes means $\nu$ and sometimes $\omega$. Read units. $\nu$ is Hz, $\omega$ is rad/s.
• Don't confuse $\phi$ (phase constant) with $(\omega t + \phi)$ (instantaneous phase).

14.3 SHM as a Projection of Uniform Circular Motion

Geometrical Picture

Consider a particle moving in a circle of radius $A$ at constant angular speed $\omega$. The position vector makes angle $\theta = \omega t + \phi$ with the $x$-axis. Its projection on the $y$-axis is

This is exactly the equation of SHM.

The projection of uniform circular motion onto a diameter is SHM. Conversely, every SHM can be embedded in a "reference circle" of radius $A$ with angular speed $\omega$.

Use of the Reference Circle

The reference circle is the fastest way to compute the time taken to go from one position to another in SHM.

Example. Find the time for a particle to go from $x=0$ to $x=A/2$ (moving outward), period $T$.

The angle on the reference circle from $0$ to $A/2$ is $\theta=\pi/6$. So

Pitfalls

• Time from $x=A/2$ to $x=A$ is not the same as $0\to A/2$. Use the circle: it is $T/4 - T/12 = T/6$.
• The reference circle has the same $\omega$ as the SHM, not $2\omega$.

14.4 Force Law and Energy in SHM

Force Law

By Newton's second law, $F = ma = -m\omega^2 x$. Writing $k = m\omega^2$:

This is Hooke's law. Hence

Kinetic Energy

Potential Energy

For a Hooke's-law force, $U = \tfrac{1}{2}kx^2$ (taking $U(0)=0$).

Total Energy

Independent of $x$ and $t$ — energy is conserved.

Energy vs Time

If $x = A\sin\omega t$:

Both oscillate with frequency $2\omega$ (period $T/2$). The averages over one period are

Graphs (described)

• $x(t)$: sinusoid, period $T$, amplitude $A$.
• $v(t)$: sinusoid, period $T$, amplitude $A\omega$, leads $x$ by $\pi/2$.
• $a(t)$: sinusoid, period $T$, amplitude $A\omega^2$, leads $x$ by $\pi$.
• $K(x)$: inverted parabola peaking at $x=0$.
• $U(x)$: upright parabola minimum at $x=0$.
• $E(x)$: horizontal line (sum is constant).

Worked Example

A 0.5 kg block on a frictionless surface is attached to a spring with $k=200\ \text{N/m}$. It is displaced 4 cm and released. Find $\omega$, $T$, total energy, and the speed at $x=2$ cm.

Pitfalls

• $k$ here is the spring constant, not Boltzmann's constant.
• $K$ and $U$ both oscillate at $2\omega$, not $\omega$.
• Total energy $\propto A^2$ — doubling amplitude quadruples energy.

14.5 Systems Executing SHM — Springs

Horizontal Spring

A block of mass $m$ on a frictionless surface attached to a spring of stiffness $k$:

Vertical Spring

Hang the same block from a vertical spring. At equilibrium the spring stretches by $x_0 = mg/k$. Let $y$ be the displacement from this new equilibrium:

Gravity cancels — the period is identical to the horizontal case:

The only effect of gravity is to shift the equilibrium position.

Springs in Parallel

Two springs $k_1$, $k_2$ pulling the same mass:

Springs in Series

Two springs $k_1$, $k_2$ joined end-to-end:

Each spring carries the same force $F$. Extensions: $x_1=F/k_1$, $x_2=F/k_2$. Total $x = x_1+x_2 = F(1/k_1 + 1/k_2)$.

Cutting a Spring

If a spring of constant $k$ is cut into $n$ equal pieces, each piece has constant $nk$ (shorter spring = stiffer).

Worked Example

A spring of natural length 1 m and constant 100 N/m is cut into a 30 cm piece and a 70 cm piece. Find the constants.

Long piece (70 cm): $k_1 = 100 \cdot (1.0/0.7) \approx 143\ \text{N/m}$. Short piece (30 cm): $k_2 = 100 \cdot (1.0/0.3) \approx 333\ \text{N/m}$.

Pitfalls

• Vertical-spring period is the same as horizontal — gravity only shifts equilibrium.
• "Series" springs have a smaller $k_\text{eq}$, hence longer $T$. Easy to flip mentally.
• A spring cut in half doubles the stiffness, not halves it.

14.6 Simple Pendulum

Setup

A point mass $m$ on a light inextensible string of length $L$, displaced by a small angle $\theta$ from vertical.

Derivation of $T$

Tangential restoring force: $F_t = -mg\sin\theta$. Arc displacement $s = L\theta$. For small $\theta$, $\sin\theta\approx\theta$:

Compare with $\ddot s = -\omega^2 s$:

Validity (Small-Angle Approximation)

$\sin\theta\approx\theta$ holds to within 1% for $\theta\lesssim 14^\circ$. For larger amplitudes, $T$ increases — the pendulum becomes anharmonic and the period depends on amplitude (anisochronism).

Compound (Physical) Pendulum

A rigid body of mass $M$ pivoted at a point at distance $\ell$ from its centre of mass, with moment of inertia $I$ about the pivot:

Effects on Pendulum Period

• Altitude: $g$ decreases, $T$ increases.
• Temperature: $L$ expands, $T$ increases (basis of compensated pendulums).
• Inside an accelerating lift: replace $g$ by $g_\text{eff}=g\pm a$.
• Pendulum in a freely falling lift: $g_\text{eff}=0$, $T\to\infty$ (no oscillation).
• Charged pendulum in a vertical electric field: $g_\text{eff}=g\pm qE/m$.

Worked Example

A simple pendulum of length 1 m has $T = 2\pi\sqrt{1/9.8} \approx 2.007\ \text{s}$. Inside a lift accelerating upward at $2\ \text{m/s}^2$, $g_\text{eff}=11.8$, so

Pitfalls

• The small-angle approximation is not optional; for $\theta=60^\circ$, the true period is about 7% larger than $2\pi\sqrt{L/g}$.
• For a physical pendulum, $\ell$ is the distance from pivot to centre of mass, not the length of the object.
• Inside a lift, use $g_\text{eff}$, taking signs of accelerations carefully.

14.7 Damped Simple Harmonic Motion

Setup

Add a velocity-proportional damping force $-b\dot x$ (e.g., viscous drag):

Divide by $m$:

Solutions — Three Regimes

The character of the solution depends on whether $\gamma\lessgtr\omega_0$:

• Under-damped $(\gamma<\omega_0)$:

The amplitude decays exponentially; oscillations continue at a slightly lower frequency $\omega'$.

• Critically damped $(\gamma=\omega_0)$:

Returns to equilibrium fastest without oscillating — used in door closers, car shock absorbers.

• Over-damped $(\gamma>\omega_0)$:

Slow return to equilibrium, no oscillation.

Amplitude Decay (Under-Damped)

The time for amplitude to drop to $1/e$ of the initial value is the damping time $\tau = 1/\gamma = 2m/b$.

Energy Decay

Energy decays twice as fast as amplitude (in the exponent).

Quality Factor

Large $Q$ = lightly damped (a tuning fork has $Q\sim 10^3$).

Pitfalls

• Damping reduces frequency by a tiny amount ($\omega'<\omega_0$) — usually negligible.
• Critical damping is the fastest non-oscillatory return — not zero damping.
• Energy decay constant is $b/m$, amplitude decay constant is $b/2m$.

14.8 Forced Oscillations and Resonance

Setup

Apply an external sinusoidal force $F(t)=F_0\cos\omega_d t$ to a damped oscillator:

After transients die out, the steady-state response is

Amplitude Formula

Resonance

The amplitude $A(\omega_d)$ has a maximum at the resonant frequency

At resonance ($\omega_d\approx\omega_0$) the amplitude becomes

A high-$Q$ system has a sharp, tall resonance peak; a low-$Q$ system has a broad, low peak.

Phase Lag

The driven response lags the drive by

• Below resonance: $\delta\approx 0$ (in phase).
• At resonance: $\delta=\pi/2$ (quadrature).
• Above resonance: $\delta\to\pi$ (out of phase).

Examples of Resonance

• Tuning a radio — LC circuit resonates at the broadcast frequency.
• Push a swing in time with its natural period — amplitude grows.
• Tacoma Narrows bridge — wind-driven resonance collapse (1940).
• NMR / MRI — spins resonate at their Larmor frequency.

Pitfalls

• Resonant frequency $\omega_R$ is slightly less than $\omega_0$ in the presence of damping; with zero damping they coincide.
• In an undamped driven oscillator at $\omega_d=\omega_0$ the amplitude grows linearly without bound (sometimes mis-quoted as "exponential").
• High $Q$ means narrow bandwidth — good for tuners, bad for shock absorbers.

Solved Problems

Problem 1. A 200 g particle executes SHM with amplitude 5 cm and period 0.4 s. Find the maximum force on the particle.

Solution. $\omega = 2\pi/0.4 = 5\pi$ rad/s. $F_\text{max}=m\omega^2 A = (0.2)(5\pi)^2(0.05)\approx 2.47\ \text{N}.$

Problem 2. A particle starts SHM from $x=A/2$ moving toward $+A$. Period $T$. Write $x(t)$.

Solution. $\phi$ satisfies $A\sin\phi = A/2 \Rightarrow\phi=\pi/6$. Since velocity is positive, $\cos\phi>0$ — choose $\phi=\pi/6$ (not $5\pi/6$). So $x(t)=A\sin(2\pi t/T+\pi/6)$.

Problem 3. A spring of constant $k$ and a mass $m$ on a frictionless table. The block is given speed $v_0$ at the natural length. Find the amplitude.

Solution. Energy conservation: $\tfrac{1}{2}mv_0^2 = \tfrac{1}{2}kA^2\Rightarrow A = v_0\sqrt{m/k}.$

Problem 4. Two identical springs of constant $k$ are attached side by side to a block of mass $m$. Find $T$.

Solution. Parallel: $k_\text{eq}=2k$. $T = 2\pi\sqrt{m/2k}.$

Problem 5. A simple pendulum of length 1 m is taken to the surface of the Moon (g/6). New period?

Solution. $T'=2\pi\sqrt{6L/g}=T\sqrt 6\approx 2.45T\approx 4.92\ \text{s}.$

Problem 6. A vertical spring elongates by 4 cm under a load. Find $T$ of SHM when the load is displaced.

Solution. At equilibrium $mg=kx_0\Rightarrow k/m=g/x_0$. $T = 2\pi\sqrt{m/k}=2\pi\sqrt{x_0/g}=2\pi\sqrt{0.04/9.8}\approx 0.402\ \text{s}.$

Problem 7. The amplitude of a damped oscillator drops to half in 50 s. Find the damping constant $\gamma$.

Solution. $e^{-\gamma t}=1/2\Rightarrow \gamma t=\ln 2\Rightarrow \gamma=\ln 2/50\approx 0.0139\ \text{s}^{-1}.$

JEE / NEET Edge Cases

• Two SHMs along the same line. Same $\omega$, different $\phi$: resultant SHM of the same frequency. Amplitude $A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos(\phi_1-\phi_2)}.$
• Perpendicular SHMs (Lissajous figures). Same $\omega$ in phase: straight line. $\pi/2$ apart, equal amplitudes: circle. General: ellipse.
• Pendulum in a uniformly accelerated train (acceleration $a$ horizontal). $g_\text{eff}=\sqrt{g^2+a^2}$, equilibrium tilted by $\tan\theta=a/g$.
• U-tube with liquid columns. Period $T = 2\pi\sqrt{L/(2g)}$ where $L$ is total length of liquid (deep classic).
• Liquid in a test tube floating vertically. Period $T = 2\pi\sqrt{m/A\rho g}$.
• Ball in a tunnel through Earth. SHM with $T = 2\pi\sqrt{R/g}\approx 84$ minutes.
• Energy ratio. At $x=A/n$, $K/U = n^2-1$.
• Time period of SHM between two collisions can be derived using the reference circle.
• Resonance peak width. The full-width at half-maximum is $\Delta\omega\approx 2\gamma$. Hence $Q=\omega_0/\Delta\omega$.

Quick Recap

• SHM: $a = -\omega^2 x$, $x = A\sin(\omega t+\phi)$.
• $v=\omega\sqrt{A^2-x^2}$, $a_\text{max}=\omega^2 A$, $v_\text{max}=\omega A$.
• $T=2\pi\sqrt{m/k}$ for any spring system (horizontal or vertical).
• Springs: parallel $k_\text{eq}=k_1+k_2$; series $1/k_\text{eq}=1/k_1+1/k_2$.
• Pendulum: $T=2\pi\sqrt{L/g}$, small-angle only.
• Energy in SHM: $E=\tfrac{1}{2}kA^2=\tfrac{1}{2}m\omega^2 A^2$; $K, U$ oscillate at $2\omega$.
• Damped SHM: amplitude $\propto e^{-\gamma t}$, energy $\propto e^{-2\gamma t}$.
• Resonance: $\omega_d=\omega_0$, amplitude $\propto Q$.

Formula Sheet