PHYS-230 · Lesson 1 of 12 · 25 min
Why waves and optics: oscillation, propagation and light
A guitar string, the air in a flute, a building swaying in the wind, the electrons in a radio antenna and the light reaching your eye from a distant galaxy all obey the same small set of mathematical ideas. Something is displaced from equilibrium, a restoring effect pulls it back, and the disturbance either oscillates in place or travels outward and carries energy with it. This course is about those ideas. It starts with a single oscillator, builds up to waves on strings and in air, moves to electromagnetic waves, and ends with the optics of lenses, interferometers, gratings and telescopes. Along the way the same equations keep reappearing, which is the main reason the subject is worth studying as a whole rather than as a list of separate topics.
PHYS-130 introduced simple harmonic motion, damping, resonance, waves on strings, standing waves, sound levels and the Doppler effect in one lesson, and PHYS-131 met electromagnetic waves, polarization and refraction near its end. Here each of those topics gets the depth it deserves: complete derivations, the use of complex exponentials, normal modes, Fourier analysis, dispersion, the full theory of interference and diffraction, and the matrix method for designing optical systems. Every lesson includes runnable Python that checks the formulas numerically, so you can see each result emerge from the equations rather than take it on trust.
Why everything oscillates
The reason oscillations are everywhere is a theorem about smooth functions. Suppose a system has one coordinate \(x\) and a potential energy \(U(x)\) with a minimum at \(x_0\). Near the minimum, Taylor's theorem gives
At a minimum \(U'(x_0) = 0\) and \(U''(x_0) > 0\). Dropping the constant and keeping the first surviving term, the potential is approximately \(\tfrac12 k (x - x_0)^2\) with an effective spring constant \(k = U''(x_0)\). The force \(F = -dU/dx\) is then \(-k(x - x_0)\), which is Hooke's law. So any system sitting near a smooth stable equilibrium behaves, for small enough displacements, like a mass on a spring. A pendulum, a molecule vibrating along its bond, a ball rolling in a bowl and a floating log bobbing in water are all harmonic oscillators to first approximation, each with its own \(k\) and its own inertia.
Robert Hooke published the law in 1678 as an anagram that decoded to the Latin for “as the extension, so the force”. He meant springs, but the Taylor argument shows why his law has such reach. It also shows its limits: the cubic and higher terms make real oscillators anharmonic, so that the period depends on amplitude, a pure tone acquires overtones, and a large swing of a pendulum takes longer than a small one. Much of physics consists of solving the harmonic problem exactly and then treating the anharmonic corrections as small perturbations.
For the pendulum of length \(\ell\), the potential energy is \(U(\theta) = mg\ell(1 - \cos\theta)\), and expanding \(\cos\theta = 1 - \theta^2/2 + \theta^4/24 - \cdots\) gives \(U \approx \tfrac12 mg\ell\,\theta^2\). The kinetic energy is \(\tfrac12 m\ell^2\dot\theta^2\), so the effective inertia is \(m\ell^2\) and the angular frequency of small swings is \(\omega_0 = \sqrt{mg\ell / m\ell^2} = \sqrt{g/\ell}\). The period \(2\pi\sqrt{\ell/g}\) does not depend on the mass or, at small amplitude, on the amplitude, which is what made the pendulum clock possible. Christiaan Huygens built the first accurate one in 1656 and analyzed its imperfections in his 1673 book on pendulum clocks.
From oscillation to wave
An oscillation stays in one place. A wave is what happens when many oscillators are coupled, so that each one drives its neighbors. Picture a row of pendulums joined by springs. Displace the first and it starts to swing; the spring to its neighbor stretches and starts the second swinging a moment later; the second drives the third, and so on. A disturbance travels down the line even though no pendulum goes anywhere. In the limit of many closely spaced oscillators, the row becomes a continuous medium, such as a string, a column of air or an elastic solid, and the equations of motion of the individual oscillators merge into a single partial differential equation, the wave equation
Lesson 5 derives it for a string, where \(v = \sqrt{T/\mu}\) depends on the tension \(T\) and the mass per unit length \(\mu\). Its solutions include traveling shapes \(f(x - vt)\) and \(g(x + vt)\), sinusoidal waves \(A\cos(kx - \omega t)\) with \(\omega = vk\), and standing waves. Lesson 4 builds the bridge explicitly: two coupled pendulums have two normal modes, and a chain of \(N\) coupled masses has \(N\) modes whose frequencies approach those of a continuous string as \(N\) grows.
Light is a wave
Whether light is a wave was argued for two centuries. Isaac Newton's Opticks of 1704 favored a stream of particles, partly because light seemed to travel in straight lines and cast sharp shadows. Huygens, in his Treatise on Light written in 1678 and published in 1690, had proposed instead that light is a wave in which every point of a wavefront acts as a source of secondary wavelets, a construction that explains reflection and refraction. The decisive experiments came after 1800. Thomas Young showed in 1801 and 1803 that light passing through two narrow slits produces alternating bright and dark bands, which only waves can do, and he used the band spacing to measure wavelengths of a few tenths of a micrometer. Augustin-Jean Fresnel's 1818 theory of diffraction explained in detail how light bends around edges, and when Siméon Poisson objected that Fresnel's theory predicted a bright spot at the center of the shadow of a disc, François Arago looked and found it.
What was waving became clear in 1865, when James Clerk Maxwell showed that his equations for electric and magnetic fields have wave solutions traveling at \(1/\sqrt{\mu_0 \varepsilon_0}\), a speed that matched the measured speed of light. Heinrich Hertz generated and detected electromagnetic waves with spark-gap circuits in 1887, and the radio age followed. The twentieth century added a twist: light also comes in quanta, photons, whose energy is \(hf\). That does not undo the wave picture. Interference and diffraction are still described by waves, and the photon picture tells us how the energy of the wave is delivered to a detector. In this course we work with the classical wave description, which explains nearly everything about lenses, interferometers and gratings.
| Region | Wavelength | Frequency | Typical source or use |
|---|---|---|---|
| Radio | above 1 m | below 300 MHz | broadcast, MRI coils |
| Microwave | 1 mm to 1 m | 300 MHz to 300 GHz | radar, Wi-Fi, ovens |
| Infrared | 700 nm to 1 mm | 300 GHz to 430 THz | thermal imaging, fiber links at 1550 nm |
| Visible | 400 nm to 700 nm | 430 THz to 750 THz | the eye, lasers, displays |
| Ultraviolet | 10 nm to 400 nm | 750 THz to 30 PHz | sterilization, lithography |
| X-ray | 0.01 nm to 10 nm | 30 PHz to 30 EHz | medical imaging, crystallography |
| Gamma ray | below 0.01 nm | above 30 EHz | nuclear transitions |
The language of sinusoids
A sinusoidal oscillation \(x(t) = A\cos(\omega t + \phi)\) has three parameters. The amplitude \(A\) sets the size, the angular frequency \(\omega\) in radians per second sets how fast the phase advances, and the phase constant \(\phi\) sets where in the cycle the motion starts. The frequency \(f = \omega / 2\pi\) is measured in hertz and the period is \(T = 1/f\). A sinusoidal wave adds a spatial dependence, \(y(x, t) = A\cos(kx - \omega t + \phi)\), with wavenumber \(k = 2\pi/\lambda\). The phase \(kx - \omega t\) is constant on a crest, so the crest moves at the phase velocity \(v_p = \omega / k = f\lambda\).
Adding sinusoids of the same frequency is the core operation of interference, and it is easiest with complex numbers. Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\) lets us write \(A\cos(\omega t + \phi) = \operatorname{Re}(A e^{i\phi} e^{i\omega t})\). The complex number \(\tilde A = A e^{i\phi}\) is the phasor: it records amplitude and phase in one quantity, and the common factor \(e^{i\omega t}\) can be carried along and dropped. Adding two oscillations of the same frequency becomes adding two phasors, which is vector addition in the complex plane. Differentiation becomes multiplication by \(i\omega\), so differential equations with constant coefficients become algebra. Electrical engineers usually write \(j\) for \(\sqrt{-1}\) and physicists write \(i\); we use \(i\).
Worked problem
Adding two oscillations with a phasor
Find the amplitude and phase of \(x(t) = 3\cos(\omega t) + 4\cos(\omega t + \pi/2)\).
Show solution
- The phasors are \(3\) and \(4e^{i\pi/2} = 4i\).
- Their sum is \(3 + 4i\), with modulus \(\sqrt{9 + 16} = 5\) and argument \(\arctan(4/3) \approx 0.927\) rad, or about \(53.1^\circ\).
- So \(x(t) = 5\cos(\omega t + 0.927)\). Note that the amplitudes did not add to 7: the two oscillations are a quarter cycle out of step, so they add like perpendicular vectors.
Answer: \(x(t) = 5\cos(\omega t + 0.927)\). Amplitudes add as vectors, not as numbers, and this is the whole of interference in miniature.
The same idea explains why two loudspeakers playing the same tone can be loud at one seat and nearly silent at another. At each seat the two sound waves arrive with a phase difference set by the difference in path length, and the total amplitude is the length of the phasor sum. When the paths differ by a whole number of wavelengths the phasors line up; when they differ by half a wavelength they point in opposite directions and largely cancel.
Energy and intensity
Waves carry energy without carrying matter. For a harmonic oscillator, the energy \(\tfrac12 kA^2 = \tfrac12 m\omega^2A^2\) is proportional to the square of the amplitude. For a wave, the power crossing a surface per unit area, the intensity \(I\), is likewise proportional to the square of the amplitude. This has a consequence that runs through all of interference: when two waves combine, amplitudes add but intensities do not. Two equal waves in phase give twice the amplitude and four times the intensity; in antiphase they give zero. Averaged over a pattern, energy is conserved, because the bright regions get exactly the energy taken from the dark ones.
Intensity spreads as a wave moves away from a small source. A point source radiating power \(P\) equally in all directions has \(I = P / 4\pi r^2\), the inverse-square law. For sound, intensities span an enormous range, from about \(10^{-12}\ \mathrm{W/m^2}\) at the threshold of hearing to about \(1\ \mathrm{W/m^2}\) at the threshold of pain, which is why we use the logarithmic decibel scale \(\beta = 10\log_{10}(I/I_0)\). For light, the eye works from starlight to sunlight across a similar range.
The plan of the course
Lessons 2 to 4 treat oscillators: simple harmonic motion in full, damping and driving with resonance and the quality factor, and coupled oscillators with normal modes. Lessons 5 to 7 treat mechanical waves: the wave equation, energy and impedance, superposition, standing waves and sound, and then Fourier analysis, which explains how any wave can be built from sinusoids, with dispersion and the distinction between phase and group velocity. Lesson 8 turns to electromagnetic waves and polarization. Lessons 9 and 10 cover interference and diffraction, the two effects that prove light is a wave and that set the limits of every optical instrument. Lesson 11 treats geometrical optics, the short-wavelength limit in which light travels along rays, with the matrix method for lens systems. The capstone, lesson 12, designs and analyses real instruments: the eye, the camera, the microscope and the telescope, and it looks at modern photonics, including lasers, optical fibers and the interferometers that detect gravitational waves.
Checking the small-angle approximation
The program below compares the exact pendulum potential with its harmonic approximation, computes the exact period of a pendulum at several amplitudes using the arithmetic-geometric mean, and adds phasors to confirm the worked example. The exact period of a pendulum released from rest at angle \(\theta_0\) is \(T = T_0 / \operatorname{agm}(1, \cos(\theta_0/2))\), where \(T_0 = 2\pi\sqrt{\ell/g}\) is the small-angle period; this compact formula follows from the elliptic integral that the energy equation produces.
import math, cmath
# 1. Harmonic approximation to the pendulum potential U = 1 - cos(theta)
for deg in (5, 10, 20, 45, 90):
th = math.radians(deg)
exact = 1 - math.cos(th)
approx = th * th / 2
print(f"theta = {deg:3d} deg: U = {exact:.6f}, theta^2/2 = {approx:.6f}, error = {100*(approx-exact)/exact:6.3f} %")
# 2. Exact period T/T0 = 1/agm(1, cos(theta0/2))
def agm(a, b):
while abs(a - b) > 1e-15:
a, b = (a + b) / 2, math.sqrt(a * b)
return a
L, g = 1.0, 9.81
T0 = 2 * math.pi * math.sqrt(L / g)
print(f"\nsmall-angle period T0 = {T0:.5f} s for a 1 m pendulum")
for deg in (5, 15, 30, 60, 90, 150):
ratio = 1 / agm(1.0, math.cos(math.radians(deg) / 2))
print(f"amplitude {deg:3d} deg: T = {T0*ratio:.5f} s, T/T0 = {ratio:.5f}")
# 3. Phasor sum 3 cos(wt) + 4 cos(wt + pi/2)
z = 3 + 4 * cmath.exp(1j * math.pi / 2)
print(f"\nphasor sum: amplitude {abs(z):.4f}, phase {cmath.phase(z):.4f} rad = {math.degrees(cmath.phase(z)):.2f} deg")theta = 5 deg: U = 0.003805, theta^2/2 = 0.003808, error = 0.063 % theta = 10 deg: U = 0.015192, theta^2/2 = 0.015231, error = 0.254 % theta = 20 deg: U = 0.060307, theta^2/2 = 0.060923, error = 1.022 % theta = 45 deg: U = 0.292893, theta^2/2 = 0.308425, error = 5.303 % theta = 90 deg: U = 1.000000, theta^2/2 = 1.233701, error = 23.370 % small-angle period T0 = 2.00607 s for a 1 m pendulum amplitude 5 deg: T = 2.00702 s, T/T0 = 1.00048 amplitude 15 deg: T = 2.01469 s, T/T0 = 1.00430 amplitude 30 deg: T = 2.04099 s, T/T0 = 1.01741 amplitude 60 deg: T = 2.15287 s, T/T0 = 1.07318 amplitude 90 deg: T = 2.36784 s, T/T0 = 1.18034 amplitude 150 deg: T = 3.53510 s, T/T0 = 1.76220 phasor sum: amplitude 5.0000, phase 0.9273 rad = 53.13 deg
The output shows that at \(10^\circ\) the harmonic potential is within a fraction of one part in a hundred of the true one, and the period at \(15^\circ\) is only about 0.4 of a part in a hundred longer than \(T_0\). At \(90^\circ\) the period is about 18 parts in a hundred longer, and as the amplitude approaches \(180^\circ\) the period grows without bound, because a pendulum balanced upside down takes forever to fall. The phasor sum reproduces the amplitude 5 and phase \(53.13^\circ\) of the worked example.
How to study this course
Three habits help. First, check units and limits: a formula for a damped oscillator should reduce to the undamped one when the damping is zero, and a diffraction formula should give geometric shadows when the wavelength is tiny. Second, draw phasor diagrams, which turn trigonometric identities into pictures. Third, run and modify the programs; the numerical answer is a check on the algebra and the algebra is a check on the code. Most of the practice problems at the end of each lesson can be checked this way.
Common mistakes
MistakeAdding amplitudes of two oscillations as plain numbers.
InsteadAmplitudes add as phasors. Two equal oscillations a quarter cycle apart combine to \(\sqrt2\) times one amplitude, not twice.
MistakeAdding intensities when waves interfere.
InsteadFor coherent waves add the amplitudes first and then square. Intensities add only for incoherent sources.
MistakeAssuming the pendulum period is independent of amplitude at all amplitudes.
InsteadThat holds only in the small-angle limit; at \(90^\circ\) the period is about 18 parts in a hundred longer.
MistakeConfusing angular frequency \(\omega\) with frequency \(f\).
Instead\(\omega = 2\pi f\). A 50 Hz oscillation has \(\omega \approx 314\ \mathrm{rad/s}\).
MistakeThinking a wave carries the medium along with it.
InsteadThe medium oscillates in place; the disturbance and its energy travel.
Summary and key terms
- Near any smooth stable equilibrium the potential is approximately \(\tfrac12 U''(x_0)(x - x_0)^2\), so small oscillations are simple harmonic.
- Coupling many oscillators produces waves; in the continuum limit they obey \(\partial_t^2 y = v^2\partial_x^2 y\).
- Young's two-slit experiment and Fresnel's diffraction theory established that light is a wave; Maxwell identified it as electromagnetic.
- A sinusoid has amplitude, frequency and phase; phasors \(\tilde A = Ae^{i\phi}\) turn adding oscillations into adding complex numbers.
- Energy and intensity are proportional to amplitude squared, so interfering amplitudes add but intensities do not.
- harmonic approximation
- Replacing a potential near its minimum by the quadratic term of its Taylor series.
- angular frequency
- The rate at which the phase of an oscillation advances, omega equals 2 pi f.
- phasor
- A complex number recording the amplitude and phase of a sinusoid of fixed frequency.
- wavenumber
- The spatial rate of change of phase, k equals 2 pi over the wavelength.
- phase velocity
- The speed of a crest of a sinusoidal wave, omega over k.
- intensity
- Power per unit area carried by a wave, proportional to the amplitude squared.
Practice problems
Work each problem on paper first, then reveal the answer to check.
A bead of mass \(m\) slides on a wire shaped as \(U(x) = U_0(x^4 - 2x^2)\). Find the equilibria and the small-oscillation angular frequency about the stable ones.
Show answer
\(U'(x) = U_0(4x^3 - 4x) = 0\) at \(x = 0, \pm1\). \(U''(x) = U_0(12x^2 - 4)\): negative at 0 (unstable), \(8U_0\) at \(\pm1\). So \(\omega = \sqrt{8U_0/m}\).
Find the small-angle period of a 2.45 m pendulum.
Show answer
\(T = 2\pi\sqrt{2.45/9.81} \approx 3.14\ \mathrm{s}\).
Write \(2\cos(\omega t) - 2\sin(\omega t)\) as a single cosine.
Show answer
\(-\sin\omega t = \cos(\omega t + \pi/2)\), so the phasor is \(2 + 2i\): amplitude \(2\sqrt2\), phase \(\pi/4\). So \(2\sqrt2\cos(\omega t + \pi/4)\).
Green light has wavelength 532 nm. Find its frequency and wavenumber in vacuum.
Show answer
\(f = c/\lambda = 3.00\times10^8/532\times10^{-9} \approx 5.64\times10^{14}\ \mathrm{Hz}\); \(k = 2\pi/\lambda \approx 1.18\times10^7\ \mathrm{m^{-1}}\).
Two coherent waves of equal intensity \(I_0\) meet with phase difference \(\delta\). Find the total intensity.
Show answer
Amplitude \(|1 + e^{i\delta}| = 2|\cos(\delta/2)|\) times one amplitude, so \(I = 4I_0\cos^2(\delta/2)\), ranging from 0 to \(4I_0\) with average \(2I_0\).
A speaker radiating 0.5 W uniformly is 10 m away. Find the intensity and sound level.
Show answer
\(I = 0.5/(4\pi\cdot100) \approx 3.98\times10^{-4}\ \mathrm{W/m^2}\); \(\beta = 10\log_{10}(3.98\times10^{8}) \approx 86\ \mathrm{dB}\).
Lesson quiz · 5 questions
Score 60% or more to complete this lesson.
Create a free account to save your progress, take quizzes and earn a certificate.