Module 3: Waves and Thermodynamics

Master every kind of wave: wave properties and the equation v = fλ, wave behaviour and interference, sound and the Doppler effect, the ray model of light with Snell's law, and heat and thermodynamics.

Pillar 1 of 5

Wave Properties

A wave is a travelling disturbance that carries energy without carrying matter with it. Learn the two wave types, the quantities that describe every wave, and the single equation, $v = f\lambda$ , that links them.

What is a Wave?

A wave transfers energy from one place to another without transferring matter. The particles of the medium simply oscillate about their rest positions as the disturbance passes through.

Mechanical waves

Need a medium to travel through: sound, water waves, and waves on a string.

Electromagnetic waves

Need no medium and can travel through a vacuum: light, radio waves, X-rays.

Worked Example

Classify each wave as mechanical or electromagnetic, and state whether it needs a medium: (a) sound in air, (b) light from the Sun, (c) a wave on a rope, (d) a radio signal.

(a) Sound — mechanical, needs a medium.
(b) Light — electromagnetic, no medium needed.
(c) Wave on a rope — mechanical, needs a medium.
(d) Radio signal — electromagnetic, no medium needed.

Practice Question

Q: A cork floats on a pond as ripples pass beneath it. It bobs up and down but does not travel across the pond. What does this show about a wave?

Show Answer

The wave transfers energy across the pond, but it does not carry the water (or the cork) with it. The medium only oscillates about its rest position as the disturbance passes.

Transverse and Longitudinal Waves

Waves are classified by the direction the particles move relative to the wave. In a transversewave the particles move at right angles to the wave (light, water surface); in a longitudinalwave they move back and forth along the wave (sound).

A transverse wave has crests and troughs; a longitudinal wave has compressions (particles bunched) and rarefactions (particles spread out).

Worked Example

Classify each wave as transverse or longitudinal: (a) sound, (b) light, (c) a wave on a string, (d) a slinky pushed and pulled along its length.

(a) Sound — longitudinal.
(b) Light — transverse.
(c) Wave on a string — transverse.
(d) Slinky pushed along its length — longitudinal.

Practice Question

Q: In a longitudinal wave, how do the particles move relative to the direction the wave travels?

Show Answer

They move parallel to the wave direction, oscillating back and forth along it. This bunches particles into compressions and spreads them into rarefactions.

Wave Characteristics

Four quantities describe any wave: the amplitude (maximum displacement from rest), the wavelength $\lambda$ (distance between successive crests), the period $T$ (time for one full cycle), and the frequency $f$ (cycles per second, in hertz).

Amplitude is measured from the rest line to a crest; wavelength is the distance from one crest to the next.

Worked Example

A transverse wave has crests that are $0.80\text{ m}$ apart, and its highest points reach $\;3.0\text{ cm}$ above the rest line. State its wavelength and amplitude.

Solution:

Wavelength = crest-to-crest distance: $\lambda = 0.80\text{ m}$

Amplitude = rest line to a crest: $A = 3.0\text{ cm} = 0.030\text{ m}$

Practice Question

Q: If the amplitude of a wave doubles but its wavelength stays the same, what has changed about the wave?

Show Answer

It carries more energy (energy grows with amplitude), but its wavelength, frequency and speed are all unchanged.

The Wave Equation

A wave travels one wavelength in one period, so its speed is wavelength times frequency. This single equation applies to every wave.

Wave Equation

v = f\lambda

$v$ = speed (m/s), $f$ = frequency (Hz), $\lambda$ = wavelength (m)

Worked Example

A sound wave has a frequency of $500\text{ Hz}$ and a wavelength of $\;0.68\text{ m}$ . Find its speed.

Solution:

$v = f\lambda = (500)(0.68) = 340\text{ m/s}$

This is the everyday speed of sound in air.

Practice Question

Q: A water wave travels at $1.5\text{ m/s}$ with a wavelength of $0.30\text{ m}$ . Find its frequency.

Show Answer

Rearrange $v = f\lambda$ for frequency:

$f = \frac{v}{\lambda} = \frac{1.5}{0.30} = 5.0\text{ Hz}$

Frequency and Period

Frequency and period are reciprocals: a wave that repeats many times each second has a short period.

Frequency and Period

f = \frac{1}{T}

Worked Example

A tuning fork vibrates at $400\text{ Hz}$ . Find its period.

Solution:

$T = \frac{1}{f} = \frac{1}{400} = 2.5\times10^{-3}\text{ s} = 2.5\text{ ms}$

Practice Question

Q: A wave completes one full cycle every $0.020\text{ s}$ . What is its frequency?

Show Answer

$f = \frac{1}{T} = \frac{1}{0.020} = 50\text{ Hz}$

Pillar 2 of 5

Wave Behaviour

All waves reflect, refract and diffract, and they can overlap. When two waves meet they add together by superposition, producing interference and, when trapped, standing waves.

Reflection

When a wave meets a barrier it bounces back. The law of reflection states that the angle of incidence equals the angle of reflection, both measured from the normal (the line perpendicular to the surface). The wave's speed, frequency and wavelength are unchanged.

Worked Example

A wave strikes a barrier with an angle of incidence of $35^\circ$ (measured from the normal). What is the angle of reflection, and what happens to the wave's speed and wavelength?

Solution:

By the law of reflection, angle of reflection $= 35^\circ$ .

Speed, frequency and wavelength are unchanged (the wave stays in the same medium).

Practice Question

Q: A ray hits a mirror straight on, along the normal (angle of incidence $0^\circ$ ). Where does the reflected ray go?

Show Answer

The angle of reflection is also $0^\circ$ , so the ray reflects straight back on itself, retracing its path.

Refraction

Refraction is the bending of a wave as it changes speed on entering a new medium, for example water waves slowing in shallow water. The frequency stays the same, but the speed and wavelength both change ( $v = f\lambda$ ). A wave that slows down bends toward the normal.

Crossing into shallow water the wave slows, so its wavefronts crowd closer together and its direction bends toward the normal.

Worked Example

Water waves of frequency $5.0\text{ Hz}$ slow from $0.40\text{ m/s}$ in deep water to $0.30\text{ m/s}$ in shallow water. Find the wavelength in each region.

Solution (frequency is unchanged, use $\lambda = v/f$ ):

Deep: $\lambda = \frac{0.40}{5.0} = 0.080\text{ m}$

Shallow: $\lambda = \frac{0.30}{5.0} = 0.060\text{ m}$

The wave slows and its wavelength shrinks, but its frequency stays the same.

Practice Question

Q: When a wave refracts into a slower medium, which of speed, frequency and wavelength stay the same, and which change?

Show Answer

Frequency stays the same. The speed and wavelength both decrease (and the wave bends toward the normal).

Diffraction

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. It is most pronounced when the gap is about the same size as the wavelength, or smaller. This is why you can hear sound around a corner: its long wavelength diffracts strongly.

Straight wavefronts spread out into curves after passing through a narrow gap, diffracting most strongly when the gap is about one wavelength wide.

Worked Example

Sound ( $\lambda \approx 1\text{ m}$ ) and light ( $\lambda \approx 5\times10^{-7}\text{ m}$ ) both pass through a $1\text{ m}$ doorway. Which diffracts noticeably, and why?

Solution:

Sound diffracts strongly: its wavelength is about the size of the gap, so it spreads out and you hear it around corners.

Light barely diffracts: its wavelength is millions of times smaller than the gap, so it travels almost straight and casts sharp shadows.

Practice Question

Q: To make waves passing through a gap diffract more, should you make the gap wider or narrower?

Show Answer

Narrower. Diffraction is greatest when the gap is about the same size as the wavelength or smaller.

Superposition and Interference

When two waves overlap, the total displacement at each point is the sum of the two, a rule called the principle of superposition. Waves that meet in phase reinforce (constructive interference); waves that meet out of phase cancel (destructive interference).

In phase, two waves add to a larger resultant; exactly out of phase, they cancel to nothing.

Worked Example

Two pulses meet at a point. At one instant, one gives a displacement of $+4\text{ cm}$ and the other $+3\text{ cm}$ . Find the resultant displacement. What if the second were instead $\;-3\text{ cm}$ ?

Solution (add the displacements):

Same sign: $+4 + 3 = +7\text{ cm}$ (constructive)

Opposite: $+4 + (-3) = +1\text{ cm}$ (partial cancellation)

Practice Question

Q: Two identical waves meet exactly out of phase (each crest lands on a trough). What is the resultant?

Show Answer

They cancel completely, total destructive interference, giving zero displacement everywhere they overlap.

Standing Waves and Resonance

When a wave is trapped, for example on a string fixed at both ends, it reflects and superposes with itself to form a standing wave. Points that never move are nodes; points of maximum oscillation are antinodes. Standing waves form only at special resonantfrequencies where a whole number of half-wavelengths fits the length.

A standing wave: nodes (N) stay still while antinodes (A) oscillate between the two extreme shapes.

Worked Example

A standing wave on a $1.2\text{ m}$ string shows 3 antinodes, so 3 half-wavelengths fit along it. Find the wavelength.

Solution:

$3 \times \tfrac{\lambda}{2} = 1.2\text{ m}$

$\lambda = \frac{2(1.2)}{3} = 0.80\text{ m}$

Practice Question

Q: Two identical waves travelling in opposite directions overlap. At a node, what is happening to the medium, and why?

Show Answer

At a node the two waves always arrive exactly out of phase, so they interfere destructively at all times. The displacement there is permanently zero, and the medium does not move.

Pillar 3 of 5

Sound Waves

Sound is a longitudinal mechanical wave. See why it needs a medium, how pitch and loudness relate to frequency and amplitude, why moving sources change pitch (the Doppler effect), and how standing waves make music.

Sound as a Longitudinal Wave

Sound is a longitudinal mechanical wave: it travels as compressions and rarefactions of the particles in a medium. The strongest evidence that sound is a mechanical wave is that it cannot travel through a vacuum: with no particles to vibrate, there is no sound. Sound also reflects (echoes), refracts, diffracts and interferes, just like other waves.

Worked Example

In a bell-jar demonstration a ringing bell is heard clearly, but as the air is pumped out the sound fades to silence even though the bell is still visibly striking. Explain.

Solution:

Sound is a mechanical wave that needs particles to carry its compressions and rarefactions. Removing the air removes the medium, so no sound reaches you, while light (an electromagnetic wave, needing no medium) still shows the bell moving.

Practice Question

Q: Does sound travel faster in air or in steel? Why?

Show Answer

Steel. Its particles are much more tightly bound, so the disturbance passes from particle to particle far more quickly than in air.

Pitch, Loudness and the Speed of Sound

Pitch = frequency

A higher frequency is heard as a higher pitch.

Loudness = amplitude

A larger amplitude carries more energy and sounds louder.

Sound travels at about $340\text{ m/s}$ in air, and faster in denser media such as water and steel, because the particles are more tightly coupled.

Frequency sets the pitch (more cycles per second sounds higher); amplitude sets the loudness (a taller wave sounds louder).

Worked Example

A tuning fork of frequency $256\text{ Hz}$ sounds in air, where the speed of sound is $\;340\text{ m/s}$ . Find the wavelength of the note.

Solution:

$\lambda = \frac{v}{f} = \frac{340}{256} = 1.33\text{ m}$

Practice Question

Q: Two sounds have the same frequency, but one has twice the amplitude. How do they differ to your ear?

Show Answer

Same pitch (frequency unchanged), but the larger-amplitude sound is louder, because it carries more energy.

The Doppler Effect

When a source of sound moves, the wavefronts bunch up ahead of it and spread out behind. An observer the source approaches hears a shorter wavelength and so a higher pitch; as it passes and recedes, the pitch drops. This is the Doppler effect, the change in the siren of a passing ambulance.

A moving source bunches its wavefronts ahead (higher pitch) and stretches them behind (lower pitch).

Worked Example

An ambulance siren emits a steady $700\text{ Hz}$ . As it speeds toward a bystander they hear about $760\text{ Hz}$ . Roughly what will they hear once it has passed and is receding, more or less than $700\text{ Hz}$ ? Explain.

Solution:

Less than $700\text{ Hz}$ . Moving away stretches each wavelength (the wavefronts spread out behind the source), so the received frequency drops below the emitted $\;700\text{ Hz}$ .

Practice Question

Q: Does the Doppler effect change the frequency the source emits, or only the frequency an observer receives?

Show Answer

Only the received frequency. The source always emits the same frequency; its motion changes the wavelength that reaches each observer.

Standing Waves in Strings and Pipes

Musical instruments make notes with standing waves. On a string fixed at both ends, the lowest note (the fundamental) fits half a wavelength into the length $L$ , so $\;\lambda = 2L$ . Higher harmonics fit two, three or more half-waves, giving frequencies that are whole-number multiples of the fundamental.

A string fixed at both ends: the fundamental fits one half-wavelength; the second harmonic fits a full wavelength.

Worked Example

A guitar string $0.65\text{ m}$ long vibrates at its fundamental, so $\;\lambda = 1.30\text{ m}$ . If waves travel along it at $260\text{ m/s}$ , find the fundamental frequency.

Solution:

$f = \frac{v}{\lambda} = \frac{260}{1.30} = 200\text{ Hz}$

Practice Question

Q: A guitar string of length $0.65\text{ m}$ vibrates at its fundamental. What is the wavelength of the standing wave?

Show Answer

The fundamental fits half a wavelength into the length, so $\lambda = 2L = 2(0.65) = 1.3\text{ m}$ .

Pillar 4 of 5

Ray Model of Light

Treating light as straight-line rays explains reflection, refraction and total internal reflection, and the way brightness falls off with distance. Angles are always measured from the normal to the surface.

Reflection

The law of reflection: the angle of incidence equals the angle of reflection, both measured from the normal, and the incident ray, reflected ray and normal all lie in the same plane.

The angle of incidence θᵢ equals the angle of reflection θᵣ, both measured from the normal.

Worked Example

A ray strikes a plane mirror at $25^\circ$ to the mirror surface. Find the angle of incidence and the angle of reflection (both measured from the normal).

Solution:

Angle from the normal: $90^\circ - 25^\circ = 65^\circ$

By the law of reflection, angle of reflection $= 65^\circ$ .

Always measure these angles from the normal, not from the surface.

Practice Question

Q: Why does a ray hitting a mirror straight on (along the normal) reflect straight back the way it came?

Show Answer

The angle of incidence is $0^\circ$ , so the angle of reflection is also $\;0^\circ$ ; the ray simply retraces its path.

Refraction and Snell's Law

Light bends when it changes speed passing between media. The refractive index $\;n = c/v$ measures how much a medium slows light. Snell's law links the angles either side of the boundary. Going into a denser (higher $n$ ) medium, light slows and bends toward the normal.

Snell's Law

n_1\sin\theta_1 = n_2\sin\theta_2

Worked Example

A ray in air ( $n_1 = 1.00$ ) strikes glass ( $n_2 = 1.50$ ) at $\;30^\circ$ to the normal. Find the angle of refraction inside the glass.

Entering the denser glass, the ray slows and bends toward the normal, so θ₂ < θ₁.

Solution:

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.00)\sin 30^\circ}{1.50} = 0.333$

$\theta_2 = \sin^{-1}(0.333) = 19.5^\circ$

Practice Question

Q: Light passes from water ( $n_1 = 1.33$ ) into air ( $n_2 = 1.00$ ) at $\;20^\circ$ to the normal. Find the angle of refraction in the air.

Show Answer

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.33)\sin 20^\circ}{1.00} = 0.455$

$\theta_2 = \sin^{-1}(0.455) = 27.1^\circ$

Moving into the less dense air, the ray bends away from the normal.

Total Internal Reflection

Going the other way, from glass into air, light bends away from the normal. Beyond a critical angle $\theta_c$ the ray cannot escape at all and is entirely reflected back inside, which is total internal reflection. This is how optical fibres trap light.

Critical Angle

\sin\theta_c = \frac{n_2}{n_1}

Beyond the critical angle, a ray inside the glass is totally reflected back rather than refracting out.

Worked Example

Find the critical angle for a water-air boundary ( $n_{\text{water}} = 1.33$ , $\;n_{\text{air}} = 1.00$ ).

Solution:

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.33} = 0.752$

$\theta_c = \sin^{-1}(0.752) = 48.8^\circ$

Practice Question

Q: Find the critical angle for light passing from glass ( $n_1 = 1.50$ ) into air ( $n_2 = 1.00$ ).

Show Answer

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

$\theta_c = \sin^{-1}(0.667) = 41.8^\circ$

The Inverse Square Law

Light from a point source spreads over an ever-larger sphere, so its intensity (power per unit area) falls off as the square of the distance. Double the distance and the light is only a quarter as bright.

Inverse Square Law

I \propto \frac{1}{d^2} \quad\Rightarrow\quad I_1 d_1^{\,2} = I_2 d_2^{\,2}

Light from a point source covers four times the area at twice the distance, so its intensity drops to one quarter.

Worked Example

A lamp gives an intensity of $100\text{ W/m}^2$ at $2.0\text{ m}$ . What is the intensity at $4.0\text{ m}$ ?

Solution:

$I_2 = I_1\left(\frac{d_1}{d_2}\right)^2 = 100\left(\frac{2.0}{4.0}\right)^2$

$I_2 = 100 \times \tfrac{1}{4} = 25\text{ W/m}^2$

Practice Question

Q: A light meter reads $400\text{ lux}$ at $1.0\text{ m}$ from a lamp. What will it read at $2.0\text{ m}$ ?

Show Answer

$I_2 = I_1\left(\frac{d_1}{d_2}\right)^2 = 400\left(\frac{1.0}{2.0}\right)^2$

$I_2 = 400 \times \tfrac{1}{4} = 100\text{ lux}$

Pillar 5 of 5

Thermodynamics

Heat is energy on the move. Connect temperature to the motion of particles, calculate the heat needed to warm a substance, and follow the three ways heat travels until everything reaches thermal equilibrium.

Temperature and Thermal Energy

Temperature measures the average kinetic energy of the particles in a substance, while thermal energy is the total kinetic energy of all its particles. That is why a bathtub of warm water holds far more thermal energy than a cup of boiling water, even though the cup is hotter.

The Kelvin scale starts at absolute zero ( $0\text{ K} = -273^\circ\text{C}$ ), the temperature at which particle motion is at a minimum. Convert with $T(\text{K}) = T(^\circ\text{C}) + 273$ .

Celsius to Kelvin

T(\text{K}) = T(^\circ\text{C}) + 273

Worked Example

Convert $25^\circ\text{C}$ to kelvin, and convert $300\text{ K}$ to degrees Celsius.

Solution:

$25 + 273 = 298\text{ K}$

$300 - 273 = 27^\circ\text{C}$

Practice Question

Q: A cup of boiling water is hotter than a warm bathtub, yet the bathtub holds more thermal energy. Explain.

Show Answer

Temperature is the average kinetic energy per particle (so the cup is hotter), but thermal energy is the total kinetic energy of all particles. The bath has vastly more particles, so its total thermal energy is greater.

Specific Heat Capacity

The specific heat capacity $c$ is the energy needed to raise $\;1\text{ kg}$ of a substance by $1^\circ\text{C}$ . Water has an unusually high value ( $4200\text{ J/kg}^\circ\text{C}$ ), which is why it heats up and cools down slowly.

Heat Energy

Q = mc\Delta T

$Q$ = heat (J), $m$ = mass (kg), $c$ = specific heat, $\Delta T$ = temperature change

Worked Example

How much heat is needed to raise the temperature of $2.0\text{ kg}$ of water by $\;30^\circ\text{C}$ ? ( $c = 4200\text{ J/kg}^\circ\text{C}$ )

Solution:

$Q = mc\Delta T = (2.0)(4200)(30)$

$Q = 2.52\times10^5\text{ J} = 252\text{ kJ}$

Practice Question

Q: How much heat is released when $0.50\text{ kg}$ of water cools by $\;20^\circ\text{C}$ ? ( $c = 4200\text{ J/kg}^\circ\text{C}$ )

Show Answer

$Q = mc\Delta T = (0.50)(4200)(20) = 42{,}000\text{ J} = 42\text{ kJ}$

Conduction, Convection and Radiation

Heat moves from hot to cold in three ways:

Conduction

Energy passes particle to particle through direct contact, mainly in solids. Metals conduct well; air is a poor conductor (a good insulator).

Convection

Warm fluid expands, becomes less dense and rises while cooler fluid sinks, setting up a circulating current. It works only in liquids and gases.

Radiation

Energy travels as infrared electromagnetic waves and needs no medium, which is how the Sun's heat reaches Earth through empty space.

Worked Example

Name the main method of heat transfer in each case: (a) a metal spoon heating up in a pot, (b) the Sun warming your face, (c) hot air rising above a heater, (d) water circulating in a boiling saucepan.

(a) Metal spoon — conduction.
(b) Sun on your face — radiation.
(c) Hot air rising — convection.
(d) Circulating water — convection.

Practice Question

Q: Why does heat from the Sun reach Earth even though space is a vacuum?

Show Answer

By radiation: infrared electromagnetic waves need no medium, so they cross empty space, unlike conduction and convection, which both need particles.

Thermal Equilibrium

When two objects at different temperatures are in contact, heat flows from the hotter to the cooler one until they reach the same temperature. At that point they are in thermal equilibrium: heat still passes both ways, but there is no net flow.

Worked Example

A hot spoon is placed in $0.20\text{ kg}$ of water at $20^\circ\text{C}$ . The water absorbs $4200\text{ J}$ of heat before equilibrium. Find the water's temperature rise and its final temperature. ( $c = 4200\text{ J/kg}^\circ\text{C}$ )

Solution (rearrange $Q = mc\Delta T$ ):

$\Delta T = \frac{Q}{mc} = \frac{4200}{(0.20)(4200)} = 5^\circ\text{C}$

Final temperature: $20 + 5 = 25^\circ\text{C}$

Heat flowed from the hot spoon to the cooler water until both reached the same temperature.

Practice Question

Q: A hot metal spoon is placed in a cup of cold water. Describe the direction of heat flow and the final state.

Show Answer

Heat flows from the hot spoon to the cold water (hot to cold). The spoon cools and the water warms until both reach the same temperature, at which point they are in thermal equilibrium and net heat flow stops.

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Pillar 1 of 5

Wave Properties

What is a Wave?

A wave transfers energy from one place to another without transferring matter. The particles of the medium simply oscillate about their rest positions as the disturbance passes through.

Mechanical waves

Need a medium to travel through: sound, water waves, and waves on a string.

Electromagnetic waves

Need no medium and can travel through a vacuum: light, radio waves, X-rays.

Worked Example

Classify each wave as mechanical or electromagnetic, and state whether it needs a medium: (a) sound in air, (b) light from the Sun, (c) a wave on a rope, (d) a radio signal.

(a) Sound — mechanical, needs a medium.
(b) Light — electromagnetic, no medium needed.
(c) Wave on a rope — mechanical, needs a medium.
(d) Radio signal — electromagnetic, no medium needed.

Practice Question

Q: A cork floats on a pond as ripples pass beneath it. It bobs up and down but does not travel across the pond. What does this show about a wave?

Show Answer

The wave transfers energy across the pond, but it does not carry the water (or the cork) with it. The medium only oscillates about its rest position as the disturbance passes.

Transverse and Longitudinal Waves

A transverse wave has crests and troughs; a longitudinal wave has compressions (particles bunched) and rarefactions (particles spread out).

Worked Example

Classify each wave as transverse or longitudinal: (a) sound, (b) light, (c) a wave on a string, (d) a slinky pushed and pulled along its length.

(a) Sound — longitudinal.
(b) Light — transverse.
(c) Wave on a string — transverse.
(d) Slinky pushed along its length — longitudinal.

Practice Question

Q: In a longitudinal wave, how do the particles move relative to the direction the wave travels?

Show Answer

They move parallel to the wave direction, oscillating back and forth along it. This bunches particles into compressions and spreads them into rarefactions.

Wave Characteristics

Amplitude is measured from the rest line to a crest; wavelength is the distance from one crest to the next.

Worked Example

A transverse wave has crests that are $0.80\text{ m}$ apart, and its highest points reach $\;3.0\text{ cm}$ above the rest line. State its wavelength and amplitude.

Solution:

Wavelength = crest-to-crest distance: $\lambda = 0.80\text{ m}$

Amplitude = rest line to a crest: $A = 3.0\text{ cm} = 0.030\text{ m}$

Practice Question

Q: If the amplitude of a wave doubles but its wavelength stays the same, what has changed about the wave?

Show Answer

It carries more energy (energy grows with amplitude), but its wavelength, frequency and speed are all unchanged.

The Wave Equation

A wave travels one wavelength in one period, so its speed is wavelength times frequency. This single equation applies to every wave.

Wave Equation

v = f\lambda

$v$ = speed (m/s), $f$ = frequency (Hz), $\lambda$ = wavelength (m)

Worked Example

A sound wave has a frequency of $500\text{ Hz}$ and a wavelength of $\;0.68\text{ m}$ . Find its speed.

Solution:

$v = f\lambda = (500)(0.68) = 340\text{ m/s}$

This is the everyday speed of sound in air.

Practice Question

Q: A water wave travels at $1.5\text{ m/s}$ with a wavelength of $0.30\text{ m}$ . Find its frequency.

Show Answer

Rearrange $v = f\lambda$ for frequency:

$f = \frac{v}{\lambda} = \frac{1.5}{0.30} = 5.0\text{ Hz}$

Frequency and Period

Frequency and period are reciprocals: a wave that repeats many times each second has a short period.

Frequency and Period

f = \frac{1}{T}

Worked Example

A tuning fork vibrates at $400\text{ Hz}$ . Find its period.

Solution:

$T = \frac{1}{f} = \frac{1}{400} = 2.5\times10^{-3}\text{ s} = 2.5\text{ ms}$

Practice Question

Q: A wave completes one full cycle every $0.020\text{ s}$ . What is its frequency?

Show Answer

$f = \frac{1}{T} = \frac{1}{0.020} = 50\text{ Hz}$

Pillar 2 of 5

Wave Behaviour

All waves reflect, refract and diffract, and they can overlap. When two waves meet they add together by superposition, producing interference and, when trapped, standing waves.

Reflection

Worked Example

A wave strikes a barrier with an angle of incidence of $35^\circ$ (measured from the normal). What is the angle of reflection, and what happens to the wave's speed and wavelength?

Solution:

By the law of reflection, angle of reflection $= 35^\circ$ .

Speed, frequency and wavelength are unchanged (the wave stays in the same medium).

Practice Question

Q: A ray hits a mirror straight on, along the normal (angle of incidence $0^\circ$ ). Where does the reflected ray go?

Show Answer

The angle of reflection is also $0^\circ$ , so the ray reflects straight back on itself, retracing its path.

Refraction

Crossing into shallow water the wave slows, so its wavefronts crowd closer together and its direction bends toward the normal.

Worked Example

Water waves of frequency $5.0\text{ Hz}$ slow from $0.40\text{ m/s}$ in deep water to $0.30\text{ m/s}$ in shallow water. Find the wavelength in each region.

Solution (frequency is unchanged, use $\lambda = v/f$ ):

Deep: $\lambda = \frac{0.40}{5.0} = 0.080\text{ m}$

Shallow: $\lambda = \frac{0.30}{5.0} = 0.060\text{ m}$

The wave slows and its wavelength shrinks, but its frequency stays the same.

Practice Question

Q: When a wave refracts into a slower medium, which of speed, frequency and wavelength stay the same, and which change?

Show Answer

Frequency stays the same. The speed and wavelength both decrease (and the wave bends toward the normal).

Diffraction

Straight wavefronts spread out into curves after passing through a narrow gap, diffracting most strongly when the gap is about one wavelength wide.

Worked Example

Sound ( $\lambda \approx 1\text{ m}$ ) and light ( $\lambda \approx 5\times10^{-7}\text{ m}$ ) both pass through a $1\text{ m}$ doorway. Which diffracts noticeably, and why?

Solution:

Sound diffracts strongly: its wavelength is about the size of the gap, so it spreads out and you hear it around corners.

Light barely diffracts: its wavelength is millions of times smaller than the gap, so it travels almost straight and casts sharp shadows.

Practice Question

Q: To make waves passing through a gap diffract more, should you make the gap wider or narrower?

Show Answer

Narrower. Diffraction is greatest when the gap is about the same size as the wavelength or smaller.

Superposition and Interference

In phase, two waves add to a larger resultant; exactly out of phase, they cancel to nothing.

Worked Example

Solution (add the displacements):

Same sign: $+4 + 3 = +7\text{ cm}$ (constructive)

Opposite: $+4 + (-3) = +1\text{ cm}$ (partial cancellation)

Practice Question

Q: Two identical waves meet exactly out of phase (each crest lands on a trough). What is the resultant?

Show Answer

They cancel completely, total destructive interference, giving zero displacement everywhere they overlap.

Standing Waves and Resonance

A standing wave: nodes (N) stay still while antinodes (A) oscillate between the two extreme shapes.

Worked Example

A standing wave on a $1.2\text{ m}$ string shows 3 antinodes, so 3 half-wavelengths fit along it. Find the wavelength.

Solution:

$3 \times \tfrac{\lambda}{2} = 1.2\text{ m}$

$\lambda = \frac{2(1.2)}{3} = 0.80\text{ m}$

Practice Question

Q: Two identical waves travelling in opposite directions overlap. At a node, what is happening to the medium, and why?

Show Answer

At a node the two waves always arrive exactly out of phase, so they interfere destructively at all times. The displacement there is permanently zero, and the medium does not move.

Pillar 3 of 5

Sound Waves

Sound as a Longitudinal Wave

Worked Example

In a bell-jar demonstration a ringing bell is heard clearly, but as the air is pumped out the sound fades to silence even though the bell is still visibly striking. Explain.

Solution:

Practice Question

Q: Does sound travel faster in air or in steel? Why?

Show Answer

Steel. Its particles are much more tightly bound, so the disturbance passes from particle to particle far more quickly than in air.

Pitch, Loudness and the Speed of Sound

Pitch = frequency

A higher frequency is heard as a higher pitch.

Loudness = amplitude

A larger amplitude carries more energy and sounds louder.

Sound travels at about $340\text{ m/s}$ in air, and faster in denser media such as water and steel, because the particles are more tightly coupled.

Frequency sets the pitch (more cycles per second sounds higher); amplitude sets the loudness (a taller wave sounds louder).

Worked Example

A tuning fork of frequency $256\text{ Hz}$ sounds in air, where the speed of sound is $\;340\text{ m/s}$ . Find the wavelength of the note.

Solution:

$\lambda = \frac{v}{f} = \frac{340}{256} = 1.33\text{ m}$

Practice Question

Q: Two sounds have the same frequency, but one has twice the amplitude. How do they differ to your ear?

Show Answer

Same pitch (frequency unchanged), but the larger-amplitude sound is louder, because it carries more energy.

The Doppler Effect

A moving source bunches its wavefronts ahead (higher pitch) and stretches them behind (lower pitch).

Worked Example

Solution:

Less than $700\text{ Hz}$ . Moving away stretches each wavelength (the wavefronts spread out behind the source), so the received frequency drops below the emitted $\;700\text{ Hz}$ .

Practice Question

Q: Does the Doppler effect change the frequency the source emits, or only the frequency an observer receives?

Show Answer

Only the received frequency. The source always emits the same frequency; its motion changes the wavelength that reaches each observer.

Standing Waves in Strings and Pipes

A string fixed at both ends: the fundamental fits one half-wavelength; the second harmonic fits a full wavelength.

Worked Example

A guitar string $0.65\text{ m}$ long vibrates at its fundamental, so $\;\lambda = 1.30\text{ m}$ . If waves travel along it at $260\text{ m/s}$ , find the fundamental frequency.

Solution:

$f = \frac{v}{\lambda} = \frac{260}{1.30} = 200\text{ Hz}$

Practice Question

Q: A guitar string of length $0.65\text{ m}$ vibrates at its fundamental. What is the wavelength of the standing wave?

Show Answer

The fundamental fits half a wavelength into the length, so $\lambda = 2L = 2(0.65) = 1.3\text{ m}$ .

Pillar 4 of 5

Ray Model of Light

Reflection

The law of reflection: the angle of incidence equals the angle of reflection, both measured from the normal, and the incident ray, reflected ray and normal all lie in the same plane.

The angle of incidence θᵢ equals the angle of reflection θᵣ, both measured from the normal.

Worked Example

A ray strikes a plane mirror at $25^\circ$ to the mirror surface. Find the angle of incidence and the angle of reflection (both measured from the normal).

Solution:

Angle from the normal: $90^\circ - 25^\circ = 65^\circ$

By the law of reflection, angle of reflection $= 65^\circ$ .

Always measure these angles from the normal, not from the surface.

Practice Question

Q: Why does a ray hitting a mirror straight on (along the normal) reflect straight back the way it came?

Show Answer

The angle of incidence is $0^\circ$ , so the angle of reflection is also $\;0^\circ$ ; the ray simply retraces its path.

Refraction and Snell's Law

Snell's Law

n_1\sin\theta_1 = n_2\sin\theta_2

Worked Example

A ray in air ( $n_1 = 1.00$ ) strikes glass ( $n_2 = 1.50$ ) at $\;30^\circ$ to the normal. Find the angle of refraction inside the glass.

Entering the denser glass, the ray slows and bends toward the normal, so θ₂ < θ₁.

Solution:

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.00)\sin 30^\circ}{1.50} = 0.333$

$\theta_2 = \sin^{-1}(0.333) = 19.5^\circ$

Practice Question

Q: Light passes from water ( $n_1 = 1.33$ ) into air ( $n_2 = 1.00$ ) at $\;20^\circ$ to the normal. Find the angle of refraction in the air.

Show Answer

$\sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{(1.33)\sin 20^\circ}{1.00} = 0.455$

$\theta_2 = \sin^{-1}(0.455) = 27.1^\circ$

Moving into the less dense air, the ray bends away from the normal.

Total Internal Reflection

Critical Angle

\sin\theta_c = \frac{n_2}{n_1}

Beyond the critical angle, a ray inside the glass is totally reflected back rather than refracting out.

Worked Example

Find the critical angle for a water-air boundary ( $n_{\text{water}} = 1.33$ , $\;n_{\text{air}} = 1.00$ ).

Solution:

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.33} = 0.752$

$\theta_c = \sin^{-1}(0.752) = 48.8^\circ$

Practice Question

Q: Find the critical angle for light passing from glass ( $n_1 = 1.50$ ) into air ( $n_2 = 1.00$ ).

Show Answer

$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667$

$\theta_c = \sin^{-1}(0.667) = 41.8^\circ$

The Inverse Square Law

Inverse Square Law

I \propto \frac{1}{d^2} \quad\Rightarrow\quad I_1 d_1^{\,2} = I_2 d_2^{\,2}

Light from a point source covers four times the area at twice the distance, so its intensity drops to one quarter.

Worked Example

A lamp gives an intensity of $100\text{ W/m}^2$ at $2.0\text{ m}$ . What is the intensity at $4.0\text{ m}$ ?

Solution:

$I_2 = I_1\left(\frac{d_1}{d_2}\right)^2 = 100\left(\frac{2.0}{4.0}\right)^2$

$I_2 = 100 \times \tfrac{1}{4} = 25\text{ W/m}^2$

Practice Question

Q: A light meter reads $400\text{ lux}$ at $1.0\text{ m}$ from a lamp. What will it read at $2.0\text{ m}$ ?

Show Answer

$I_2 = I_1\left(\frac{d_1}{d_2}\right)^2 = 400\left(\frac{1.0}{2.0}\right)^2$

$I_2 = 400 \times \tfrac{1}{4} = 100\text{ lux}$

Pillar 5 of 5

Thermodynamics

Temperature and Thermal Energy

The Kelvin scale starts at absolute zero ( $0\text{ K} = -273^\circ\text{C}$ ), the temperature at which particle motion is at a minimum. Convert with $T(\text{K}) = T(^\circ\text{C}) + 273$ .

Celsius to Kelvin

T(\text{K}) = T(^\circ\text{C}) + 273

Worked Example

Convert $25^\circ\text{C}$ to kelvin, and convert $300\text{ K}$ to degrees Celsius.

Solution:

$25 + 273 = 298\text{ K}$

$300 - 273 = 27^\circ\text{C}$

Practice Question

Q: A cup of boiling water is hotter than a warm bathtub, yet the bathtub holds more thermal energy. Explain.

Show Answer

Specific Heat Capacity

Heat Energy

Q = mc\Delta T

$Q$ = heat (J), $m$ = mass (kg), $c$ = specific heat, $\Delta T$ = temperature change

Worked Example

How much heat is needed to raise the temperature of $2.0\text{ kg}$ of water by $\;30^\circ\text{C}$ ? ( $c = 4200\text{ J/kg}^\circ\text{C}$ )

Solution:

$Q = mc\Delta T = (2.0)(4200)(30)$

$Q = 2.52\times10^5\text{ J} = 252\text{ kJ}$

Practice Question

Q: How much heat is released when $0.50\text{ kg}$ of water cools by $\;20^\circ\text{C}$ ? ( $c = 4200\text{ J/kg}^\circ\text{C}$ )

Show Answer

$Q = mc\Delta T = (0.50)(4200)(20) = 42{,}000\text{ J} = 42\text{ kJ}$

Conduction, Convection and Radiation

Heat moves from hot to cold in three ways:

Conduction

Energy passes particle to particle through direct contact, mainly in solids. Metals conduct well; air is a poor conductor (a good insulator).

Convection

Warm fluid expands, becomes less dense and rises while cooler fluid sinks, setting up a circulating current. It works only in liquids and gases.

Radiation

Energy travels as infrared electromagnetic waves and needs no medium, which is how the Sun's heat reaches Earth through empty space.

Worked Example

(a) Metal spoon — conduction.
(b) Sun on your face — radiation.
(c) Hot air rising — convection.
(d) Circulating water — convection.

Practice Question

Q: Why does heat from the Sun reach Earth even though space is a vacuum?

Show Answer

By radiation: infrared electromagnetic waves need no medium, so they cross empty space, unlike conduction and convection, which both need particles.

Thermal Equilibrium

Worked Example

Solution (rearrange $Q = mc\Delta T$ ):

$\Delta T = \frac{Q}{mc} = \frac{4200}{(0.20)(4200)} = 5^\circ\text{C}$

Final temperature: $20 + 5 = 25^\circ\text{C}$

Heat flowed from the hot spoon to the cooler water until both reached the same temperature.

Practice Question

Q: A hot metal spoon is placed in a cup of cold water. Describe the direction of heat flow and the final state.

Show Answer

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