The Timpson Lyceum · University Physics I (2025)

Periodic Motion and Simple Harmonic Motion - Lecture 14 (University Physics I)

Overview

This lecture opens the study of oscillations by examining simple harmonic motion (SHM) — the back-and-forth motion produced whenever a restoring force is directly proportional to displacement. We first see that tracking the $x$-coordinate of an object in uniform circular motion traces out a sinusoidal curve, which motivates a sinusoidal guess for the solution. Applying Hooke's law and Newton's second law to a mass on a spring yields a differential equation whose solution is $x(t)=A\cos(\omega t+\phi)$, with angular frequency $\omega=\sqrt{k/m}$. We then analyze the energy of an oscillator, learn how to extract amplitude and phase angle from initial conditions, and extend SHM to the simple pendulum (via the small-angle approximation), the physical pendulum, and finally damped oscillations. The central, recurring theme is that radically different physical systems all reduce to the same equation of motion.

Key Concepts

Connection to uniform circular motion

• An object moving around a circle of radius $A$ at constant angular speed $\omega$ has $\theta=\omega t+\phi$. Its $x$-coordinate is $x=A\cos(\omega t+\phi)$.
• Plotting $x$ versus time gives a cosine wave; the circle's radius becomes the wave's amplitude $A$, and $\omega$ controls how fast the cycle repeats.
• Sine and cosine differ only by a phase shift ($\cos(\theta-\tfrac{\pi}{2})=\sin\theta$), so either can describe the motion. Such curves are called sinusoidal.

Restoring force and the definition of SHM

• A mass on a spring obeys Hooke's law: $F=-kx$, where $k$ is the spring constant and $x$ is displacement from equilibrium.
• The minus sign makes the force a restoring force — always directed back toward equilibrium, and proportional to displacement (pull it twice as far, the spring pulls back twice as hard).
• Simple harmonic motion is exactly the motion that results when the net force is linear in displacement, $F=-kx$.

Equation of motion and its solution

• Newton's second law gives $m\dfrac{d^2x}{dt^2}=-kx$, i.e. $\dfrac{d^2x}{dt^2}+\dfrac{k}{m}x=0$.
• Guessing $x=A\cos(\omega t+\phi)$ and differentiating twice gives $\dfrac{d^2x}{dt^2}=-\omega^2 x$. Substituting forces $\omega^2=\dfrac{k}{m}$.
• Solution: $x(t)=A\cos(\omega t+\phi)$ with $\omega=\sqrt{\dfrac{k}{m}}$.
• $\omega$ is the angular frequency (radians per second of oscillation). A stiffer spring (larger $k$) oscillates faster; a larger mass oscillates slower. Notably, $\omega$ does not depend on amplitude — how far you pull it back does not change the period.

Frequency, period, angular frequency

• $f=\dfrac{1}{T}$ and $T=\dfrac{1}{f}$. Frequency $f$ (in hertz, $\mathrm{s}^{-1}$) is cycles per second; period $T$ is seconds per cycle.
• Angular frequency relates to frequency by $\omega=2\pi f$ (radians per second, since one cycle is $2\pi$ radians).
• For a mass-spring system: $f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ and $T=2\pi\sqrt{\dfrac{m}{k}}$.

Amplitude and phase angle from initial conditions

• The constants $A$ (max displacement) and $\phi$ (phase angle, sets the starting point) are fixed by the initial position $x_0$ and initial velocity $v_0$.
• Since $v(t)=-A\omega\sin(\omega t+\phi)$, evaluating at $t=0$ gives $\dfrac{v_0}{\omega x_0}=-\tan\phi$, so $\phi=\tan^{-1}\!\left(-\dfrac{v_0}{\omega x_0}\right)$.
• Combining $x_0$ and $v_0$ gives $A=\sqrt{x_0^2+\dfrac{v_0^2}{\omega^2}}$.

Energy in SHM

• Total mechanical energy $E=\tfrac{1}{2}kx^2+\tfrac{1}{2}mv^2$ is conserved (no friction).
• Substituting the SHM solution and using $\cos^2+\sin^2=1$ gives the constant $E=\tfrac{1}{2}kA^2$ — fully determined by amplitude.
• Quick derivation: at maximum displacement $x=A$ the mass is momentarily at rest, so all energy is potential, $E=\tfrac{1}{2}kA^2$.
• Solving the energy equation for speed gives velocity as a function of position: $v=\sqrt{\dfrac{k}{m}\,(A^2-x^2)}$. Speed is maximum at $x=0$ and zero at $x=\pm A$. Acceleration $a=-\omega^2 x$ is maximum (in magnitude) at $x=\pm A$.

Simple pendulum

• A mass on a string of length $L$: the tangential restoring force is $-mg\sin\theta$, giving $\dfrac{d^2x}{dt^2}+g\sin\theta=0$ where the arc-length displacement is $x=L\theta$.
• Using the small-angle approximation $\sin\theta\approx\theta$ (valid below roughly $15^\circ$, justified by the Taylor series $\sin\theta=\theta-\tfrac{\theta^3}{3!}+\cdots$), this becomes SHM: $\dfrac{d^2x}{dt^2}+\dfrac{g}{L}x=0$.
• Therefore $\omega=\sqrt{\dfrac{g}{L}}$, and $T=2\pi\sqrt{\dfrac{L}{g}}$. The period depends on length and $g$ but not on the mass (the restoring force scales with mass just as inertia does).

Physical pendulum

• A rigid body of mass $m$ and moment of inertia $I$ pivoting at a point a distance $d$ from its center of mass. Using torque, $I\dfrac{d^2\theta}{dt^2}+mgd\sin\theta=0$.
• Small-angle approximation again gives SHM with $\omega=\sqrt{\dfrac{mgd}{I}}$ and $\theta(t)=\Theta\cos(\omega t+\phi)$, where $\Theta$ is the maximum angular displacement.

Damped oscillations

• Adding a velocity-proportional resistive force, $F=-kx-bv$, gives $\dfrac{d^2x}{dt^2}+\dfrac{b}{m}\dfrac{dx}{dt}+\dfrac{k}{m}x=0$.
• The solution is a decaying sinusoid $x=A\,e^{-(b/2m)t}\cos(\omega t+\phi)$ with $\omega=\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}$.
• Energy leaves the system at the rate $\dfrac{dE}{dt}=-bv^2$, proportional to the resistive coefficient $b$ and the square of the speed.

Worked Example

Problem. A glider of mass $m=0.5\ \mathrm{kg}$ is attached to a spring. A $6\ \mathrm{N}$ force stretches the spring $0.03\ \mathrm{m}$. The glider is given an initial displacement $x_0=0.015\ \mathrm{m}$ and an initial velocity $v_0=0.40\ \mathrm{m/s}$. Find the spring constant, angular frequency, amplitude, and phase angle.

Spring constant from Hooke's law, $k=F/x$:

Angular frequency, amplitude, and phase angle:

The motion is therefore $x(t)=0.025\cos(20t-0.93)\ \mathrm{m}$, with velocity $v(t)=-0.5\sin(20t-0.93)\ \mathrm{m/s}$ and acceleration $a(t)=-10\cos(20t-0.93)\ \mathrm{m/s^2}$.

Summary

• Simple harmonic motion arises whenever the net force is a linear restoring force, $F=-kx$; its solution is $x(t)=A\cos(\omega t+\phi)$.
• For a mass on a spring, $\omega=\sqrt{k/m}$, so the period $T=2\pi\sqrt{m/k}$ depends only on $k$ and $m$ — never on the amplitude.
• Frequency, period, and angular frequency are interchangeable descriptions linked by $f=1/T$ and $\omega=2\pi f$.
• Total energy is conserved at $E=\tfrac{1}{2}kA^2$; this yields $v=\sqrt{(k/m)(A^2-x^2)}$, with maximum speed at equilibrium and maximum acceleration at the turning points.
• A simple pendulum becomes SHM under the small-angle approximation, with $\omega=\sqrt{g/L}$ and a period independent of mass; a physical pendulum gives $\omega=\sqrt{mgd/I}$.
• Adding a velocity-proportional drag produces damped oscillations whose amplitude decays as $e^{-(b/2m)t}$ and whose energy drains at the rate $dE/dt=-bv^2$.