Cards Explorer

A gravitational field is a region of space in which a mass experiences a gravitational force. It is an example of a field of force.

Gravitational field strength at a point is the gravitational force per unit mass exerted on a small test mass placed at that point. Vector, unit N kg⁻¹.

The gravitational force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the separation between their centres.

A point mass is an object with mass but negligible volume. Two bodies may be treated as point masses when their dimensions are negligible compared with their separation.

Gravitational potential at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point, without change in kinetic energy. Scalar, unit J kg⁻¹.

Gravitational potential energy at a point is the work done by an external force in bringing a mass from infinity to that point, without change in kinetic energy. Scalar, unit J.

A geostationary orbit is a circular orbit in which a satellite appears stationary when observed from a fixed point on the Earth's surface.

Escape velocity is the minimum velocity at which a mass must be projected from a planet's surface in order to reach infinity, that is, to completely escape the planet's gravitational field.

Combining Newton's law of gravitation with the definition g = F/m gives the field strength of a point mass M at distance r as g = GM/r², directed towards M.

Near the Earth's surface g is approximately constant because h is negligible compared with R, so (R + h)² ≈ R². Its value, 9.81 N kg⁻¹, is numerically equal to the acceleration of free fall, 9.81 m s⁻².

When two situations share the same M, divide the equations: g₂/g₁ = (r₁/r₂)². If instead the DENSITY is fixed, M ∝ r³ so g ∝ r.

Between two masses there is a point where the two fields are equal in magnitude and opposite in direction, so the resultant field strength is zero. It lies closer to the smaller mass.

Far from a planet the field lines are radial, pointing to the centre, non-uniform with spacing widening outwards. Near the surface they are parallel, equally spaced and uniform. Line density represents field strength, and equipotentials are perpendicular to field lines with no work done moving along one.

The test mass must be small because it produces its own gravitational field, which would distort the field being measured. A small mass keeps that disturbance negligible.

F = GMm/r² (vector, N) and g = GM/r² (vector, N kg⁻¹) fall off as 1/r² and are always attractive. U = −GMm/r (scalar, J) and φ = −GM/r (scalar, J kg⁻¹) fall off as 1/r and are always negative. Linked by g = −dφ/dr and F = −dU/dr.

Potential at infinity is defined as zero. The gravitational field is attractive, so as a mass is brought in from infinity gravity does positive work and the external force does negative work. Hence φ (and U) fall below zero, and φ increases towards zero as r increases.

Gravitational field strength is minus the potential gradient: g = −dφ/dr. Equivalently F = −dU/dr. The field points in the direction of decreasing potential, and the magnitude of g equals the gradient of the φ–r graph.

Work done by an external force in moving a mass m between two points is W = ΔU = m(φ_final − φ_initial). Since φ ∝ −1/r, doubling r halves the potential, making it less negative, so moving outwards gives positive work.

Force-type curves (F, g) fall as 1/r²; energy-type curves (U, φ) fall as 1/r. Plotted with sign, U and φ are negative and rise towards zero, and the 1/r curve approaches zero more slowly than the 1/r² curve. All four tend to zero as r → ∞.

Escape means arriving at infinity with zero KE and zero PE, so by conservation of energy ½mv² + (−GMm/r) = 0, giving v_esc = √(2GM/r) = √(2gr), which is √2 × the orbital speed at the same radius.

Along the line joining two bodies, φ reaches a maximum (least negative) value at the point where the resultant field is zero, because g = −dφ/dr and a zero gradient means g = 0. A projectile only needs enough energy to reach that point; beyond it the far body pulls it in.

For any body in a circular orbit, the gravitational force provides the centripetal force: GMm/r² = mv²/r = mrω². Gravity is that centripetal force; it is not an additional force.

Setting gravity equal to the centripetal force, GMm/r² = mv²/r, m cancels, and rearranging gives v = √(GM/r) = √(gr). So a larger orbit means a slower satellite, independent of the satellite's mass.

From GMm/r² = mrω² and ω = 2π/T, we get T² = (4π²/GM) r³, so T² ∝ r³ for all bodies orbiting the same central mass. A graph of T² against r³ is a straight line through the origin with gradient 4π²/GM.

A geostationary satellite must have (1) period exactly 24 hours, matching the Earth's rotation, (2) orbit from west to east, the same sense as the Earth's rotation, and (3) lie in the plane of the equator. These fix r = 4.23 × 10⁷ m from the Earth's centre.

For a circular orbit, KE = +GMm/2r, GPE = −GMm/r and TE = KE + GPE = −GMm/2r. So KE = −TE = −½ GPE, and the total energy is always negative, which is what bound means.

Moving to a larger orbit makes the satellite slower (v = √(GM/r)) with less kinetic energy, yet its total energy increases (becomes less negative), so work must be done on it: W = GMm/2 (1/r₁ − 1/r₂).

Drag does negative work, so total energy decreases (more negative). Since TE = −GMm/2r, a more negative TE means r decreases. But KE = +GMm/2r, so as r falls KE increases and the satellite speeds up.

True weightlessness is when the resultant gravitational force on a body is zero (at infinity, or at a neutral point). Apparent weightlessness is when a body exerts no contact force on its support, which happens when body and support have the same acceleration g, as in free fall or in orbit.

Gravitational and electric fields share the same inverse-square structure with mass replaced by charge and G replaced by 1/4πε₀. Both use field = force per unit (mass/charge), potential = work per unit (mass/charge) from infinity, and field = minus potential gradient. Key difference: gravity is only attractive so φ is always negative, while electric forces can attract or repel so V can be either sign.

ΔU = mgh applies only where g is approximately constant, i.e. h ≪ R. For large h use ΔU = (−GMm/r_f) − (−GMm/r_i). Using mgh over a large height overestimates the change, because g is smaller up there.

Since m cancels in GMm/r² = mrω², v, ω and T depend only on M and r. Two satellites with the same period must have the same radius, whatever their masses. Doubling a satellite's mass at fixed period leaves the radius unchanged.

v_esc = √(2GM/r) = √(2gr) contains no property of the escaping object. It is independent of the object's mass (m cancels) and independent of launch direction (energy is a scalar). It depends only on M and r of the planet.

The distance of the particle from its equilibrium position in a stated direction.

The maximum displacement of the particle from the equilibrium position.

Period T: time for one complete oscillation. Frequency f: number of complete oscillations per unit time.

The fraction of a cycle by which one oscillation leads or lags another, expressed as an angle (2π rad = one full cycle).

(1) Read T off either graph. (2) Pick the same feature on both curves — peak→peak, or zero-crossings going the same direction. (3) Measure the time offset Δt between them. (4) Convert fraction of cycle to angle. The curve whose feature comes earlier leads. From equations instead: φ = difference of the brackets — sin vs cos with the same argument is automatically π/2. Traps: up-crossing vs down-crossing silently adds π; take the nearest corresponding feature — if φ > π, quote 2π − φ and swap leads↔lags; φ = 0 in phase, φ = π antiphase.

Motion in which the acceleration is directly proportional to the displacement from a fixed point and is always directed towards that point (opposite in direction to the displacement). Both clauses required — proportionality AND direction.

The net force on the body that is always directed towards the equilibrium position, acting to restore it there. F–x graph: straight line through the origin, negative gradient. Describing it (4-mark N2016 pattern): pendulum — the component of the weight along the arc, towards the lowest point; floating object — displaced down, upthrust exceeds weight (net up); displaced up, weight exceeds upthrust (net down).

Same units (rad s⁻¹), different meanings. Angular velocity = rate of change of angular displacement of something actually rotating. Angular frequency ω = 2πf — rate the oscillation moves through phase; an SHM particle moves in a line, nothing rotates. Only for the circular-motion projection are they numerically the same thing.

The frequency at which a system oscillates freely when displaced and released, with no external driving force and negligible resistive forces.

The frequency of the external periodic force applied to the system. It belongs to the driver, not the system — a forced oscillation settles at the driving frequency, not at f₀. Amplitude of the response depends on how close f_d is to f₀.

The continuous loss of energy of an oscillating system to resistive forces, causing the amplitude to decrease progressively.

Occurs when the driving frequency equals the natural frequency of the system: there is maximum rate of energy transfer from driver to system and the amplitude is maximum.

Starts at equilibrium: x = x₀ sin ωt. Starts at an extreme: x = x₀ cos ωt. Differentiate for v, again for a. Calculator in radians, always.

The timing-free workhorse — no t needed, only position. Sign gives direction.

Each differentiation multiplies the amplitude by ω. Speed peaks at equilibrium; acceleration peaks at the extremes, directed towards equilibrium.

Spring: stiffness and mass only — g does not appear (same period on the Moon). Pendulum: length and g only — mass does not appear, valid for small angles (sin θ ≈ θ).

For true SHM the period is independent of amplitude — a larger swing takes the same time. Favourite true/false trap.

Trolley between two springs (both stay stretched): displace x → one pulls harder, other pulls less → F = 2kx, so k_eff = 2k, ω = √(2k/m). Spring with mass (not light): part of the spring also oscillates → effective m larger → f lower than 2π-formula predicts. Mass removed from a spring system (washing-machine concrete): M drops → f₀ = √(k/M)/2π rises — resonance now happens at a higher rotation speed.

An object resting on a vertically oscillating plate stays in contact only while the plate's downward acceleration ≤ g (normal force N ≥ 0). Contact is lost at the highest point, where downward acceleration is largest, when ω²x₀ exceeds g. Condition to stay in contact: ω²x₀ ≤ g. N = 0 at the moment a = g.

v leads x by π/2 (max speed through equilibrium); a leads v by π/2, so a is in antiphase (π) with x — the graphical face of the minus sign. Given one graph, get the next by taking gradients, not from memory.

At any point: v = sign of the gradient; a = opposite sign of x (always towards centre). So: moving away from equilibrium → v and a opposite (slowing down); moving towards equilibrium → v and a same direction (speeding up). That's the whole "at which point are v and a opposite" MCQ.

From v = ±ω√(x₀² − x²). Cuts the x-axis at ±x₀ (v = 0 at extremes), the v-axis at ±ωx₀ (v_max at equilibrium); traversed clockwise (x right, v up). Label questions: release point sits on the x-axis at the release displacement; first return to equilibrium is the v-axis intercept. With damping the curve spirals inward — same centre, shrinking amplitude.

A peg on a turntable (radius r, angular speed ω) casts a shadow on a screen: θ = ωt, shadow displacement x = r sin ωt — SHM with amplitude r and the same ω. Shadow speed passing the centre = rω; shadow acceleration when instantaneously at rest (edges) = rω². This is why ω, rad s⁻¹ and phase-as-angle appear in a straight-line motion.

Tides, mass below a ceiling, floating tubes: the reading oscillates about a mean, not zero. Centre = (max + min)/2, amplitude = (max − min)/2, equation = mean + x₀ sin ωt. Trap: in a distance-from-ceiling graph the amplitude is NOT the max reading — subtract the centre first. Time between high and low = T/2.

Max speed (at centre) and the next max acceleration (at an extreme) are separated by T/4 — adjacent special points are always a quarter period apart: centre → extreme → centre → other extreme, each step T/4.

Total energy is constant (undamped) and proportional to amplitude² and to f² — double x₀ → four times E.

At x = x₀/2 → KE = ¾E_total. KE = PE at x = x₀/√2 ≈ 0.707x₀. One line each: KE/E = 1 − (x/x₀)².

KE and PE each oscillate at frequency 2f — the particle hits max speed twice per cycle. The energy graphs are NOT at f.

Track GPE, EPE, KE, Total (columns of the classic table). Lowest point: EPE max, GPE min (reference 0), KE 0. Equilibrium: KE max. Highest point: GPE max, EPE min (not zero if spring still stretched), KE 0. Total constant. Shortcut for everything else: measure x from equilibrium and use the combined PE = ½kx² — gravity is already absorbed into the equilibrium position, so all standard SHM formulas apply unchanged.

Pull the mass a distance A below equilibrium, release from rest → total oscillation energy is ½kA² (A measured from equilibrium, k the spring constant). Then v_max = ωA, ω = √(k/m). This sidesteps every GPE-vs-EPE headache the question tries to cause.

Mass lowered gently to equilibrium (extension e): GPE lost = mge, but EPE gained = ½ke² = ½mge — exactly half. The other half went into the hand (external force did negative work). Classic "explain why the two answers differ" 2-marker: the mass was not in free fall; an external force removed energy.

Amplitude decays x₀₁ → x₀₂ (read two peaks off the graph) → energy lost to resistive forces:

Light: still oscillates; amplitude decays exponentially (constant fraction lost per cycle); period almost unchanged. Critical: returns to equilibrium in the shortest time without oscillating — car suspension, meter needles. Heavy: returns slowly, no oscillation. Sketch tip: exponential envelope, but zero-crossings stay equally spaced.

Free: oscillates at its natural frequency f₀ after a single displacement. Forced: driven by an external periodic force, oscillates at the driving frequency with amplitude depending on how close f_d is to f₀.

More damping → peak lower and broader, at slightly below f₀; resonance amplitude is large but finite — never infinite.

(1) System has natural frequency f₀. (2) Periodic driver transfers energy to it. (3) As f_d → f₀ the rate of energy transfer increases, so amplitude grows. (4) At f_d = f₀, energy transfer is at the maximum rate and amplitude is maximum — resonance. (+ damping lowers and broadens the peak.)

Useful: microwave ovens (matched to water molecules) · radio tuning circuits · MRI · instrument sound boards. Destructive: bridges near f₀ (Tacoma Narrows 1940; soldiers break step) · buildings in earthquakes. Cures: dampers (Taipei 101's 660-tonne sphere) or shift f₀ by stiffening/adding mass.

"X changes; what happens to the amplitude?" Always: (1) name what changed → (2) trace it to f_d, f₀ or damping → (3) state the amplitude effect. Worked set (N94 floating block at resonance): bigger incident waves → driver amplitude up → larger amplitude, still at resonance. Crest spacing (λ) increases at same wave speed → f_d = v/λ falls below f₀ → off resonance → amplitude drops. Block absorbs water → m up → f₀ = √(k/m)/2π falls → mismatch (and heavier damping) → amplitude drops. Rotating machinery (washing machine): the imbalanced rotation IS the periodic driver; max amplitude when rev s⁻¹ = f₀ = 1/T.

Loudspeaker cone (N09): if the cone's natural frequency sat inside the audio range, signals near it would resonate — that frequency reproduced with exaggerated amplitude, distorting the sound. Same logic for machine mounts, buildings, bridges: design f₀ away from the driving frequencies the system will meet.

The propagation of a disturbance which transfers energy (and momentum) from one point to another by means of oscillations/vibrations, without the physical transfer of matter — particles only oscillate about fixed equilibrium positions.

A wave in which the oscillations of the wave particles are perpendicular to the direction of propagation (direction of energy transfer). Say oscillations — the word "movement" is marked WRONG.

A wave in which the oscillations of the wave particles are parallel to the direction of propagation of the wave. Sound is THE example; has compressions and rarefactions.

The distance in a specified direction of a particle from its equilibrium position. Vector — can be negative. Unlike y₀, λ, f, T and v (all constants of the wave), displacement differs from particle to particle at an instant.

The magnitude of the maximum displacement of a particle from its equilibrium position. Never "height of the wave".

The shortest distance between two points on a wave that are in phase (at the same instant) — equivalently, between successive crests/troughs or compressions/rarefactions. "Shortest" (or "successive/adjacent") is a mark-bearing word.

T: time for one complete oscillation of a particle. f: number of complete oscillations per unit time. Set by the source — every particle oscillates at the source's frequency.

The speed at which the wave profile (energy) travels in the direction of propagation. A constant for a given wave in a given medium — NOT the particle velocity.

An angle giving a measure of the fraction of a cycle completed by an oscillating particle or wave. One full cycle = 2π rad = 360°; half cycle = π (NOT π/2). Independent of amplitude.

A measure, in angular form, of how much one particle/wave is out of step with another — the fraction of a cycle one is ahead of or behind the other. Quote in the range 0 ≤ Δφ < 2π. Only meaningful for the same frequency.

In phase: Δφ = 0 (or 2nπ) — same displacement AND same direction of motion; particles nλ apart. Antiphase: Δφ = π — always opposite; particles (n + ½)λ apart. Same amplitude NOT required.

Wavefront: a line/surface joining points in phase (e.g. all crests); consecutive wavefronts one λ apart. Ray: direction of propagation, always perpendicular to wavefronts.

Compression: region of a longitudinal wave where particles are closest together — pressure above normal. Rarefaction: furthest apart — pressure below normal. (Not "refraction".)

The rate of energy transfer (power) per unit area normal (perpendicular) to the direction of energy transfer of the wave. The perpendicular clause carries a mark.

A wave is plane polarised when its oscillations are confined to one direction only, in a plane normal to the direction of energy transfer. A phenomenon of transverse waves only.

Sound is longitudinal: oscillations are parallel to the direction of energy transfer, so there are no oscillations in the plane normal to the propagation direction to restrict. Restricting them would stop the wave itself. Three-line template — learn the chain.

Mutually perpendicular oscillating E and B fields, both perpendicular to the propagation direction — hence transverse, needs no medium, travels at c = 3.00 × 10⁸ m s⁻¹ in vacuum. Polarisation direction of light = direction of the E-field oscillation.

Displacement–distance: photo of ALL particles at one instant → read y₀ and λ; T is unreadable (use T = 1/f). Displacement–time: one particle tracked → read y₀ and T; λ unreadable. Same shape, different universe. Look at the axes before anything else — the school lecturer repeats this "like a broken recorder".

Syllabus demands the deduction, not just the formula: in one period T the waveform advances one wavelength λ, so v = distance/time = λ/T; with f = 1/T, v = fλ. Applies to the wave profile, never to particles.

Wave velocity: constant, = fλ, direction of propagation. Particle velocity: SHM variable — v = y₀ω cos(ωt), equal to the gradient of the y–t graph, zero at crest/trough, max (= ωy₀) at equilibrium. Raising intensity raises PARTICLE max speed, never wave speed — wave speed is a property of the medium.

Every point performs SHM: speed max at equilibrium crossings, zero at crest/trough; acceleration a = −ω²y — max at extremes, towards equilibrium, zero at equilibrium; energy all-kinetic at centre, all-potential at extremes. All points share the same amplitude and energy (1-D, no losses). Chains straight back to Topic 10.

Default: upward positive (transverse), rightward positive (longitudinal). If the question declares downward/leftward positive, the same motion plots as minus the default graph. Read the convention first; a sinusoidal y–t graph does NOT mean the wave is transverse — graph shape never reveals wave type.

Convert distance to time through the wave's motion: the profile moves λ per T. For "shortest time until P has displacement X": slide the profile in the travel direction until the required feature reaches P; t = (distance moved/λ) × T.

On the displacement–distance graph of a sound wave, compressions and rarefactions all sit at equilibrium (zero-displacement) positions — never at the peaks. The lecturer makes the class chant this; it is the #1 weak-student error.

At a zero crossing, check the neighbours (with + = right): left neighbour displaced towards the point AND right neighbour displaced towards it → converging → compression. Both displaced away → rarefaction. This is how you tell WHICH zero crossing is which.

Δp–x graph: centred on zero, positive at C, negative at R — λ/4 (π/2) out of step with the displacement graph. Absolute p–x graph: same shape but centred on atmospheric pressure — it can never start from zero (zero pressure = vacuum). Displacement wave and pressure wave differ in phase by π/2.

λ = C-to-next-C (or R-to-R). C to adjacent R = λ/2, not λ. Pressure ideas exist only for longitudinal waves — never sketch Δp for a transverse wave.

Space version (snapshot): Δφ = (Δx/λ) × 2π. Time version (y–t graphs): Δφ = (Δt/T) × 2π. If λ missing, get it from λ = v/f first. If Δφ > 2π, reduce into 0–2π and say so (9π/2 → π/2).

The particle nearer the source leads (it started first). On y–t graphs: whoever reaches the crest earlier leads — measure crest-to-crest, never at intersection points (a named examiner complaint). Cyclic equivalence: B leads A by φ ⇔ A leads B by 2π − φ ⇔ A leads B by −φ. Two separate y–x graphs = two different waves: the one further advanced in the travel direction leads by (Δx/λ)2π. y–t graphs always compare particles, never whole waves.

When x and λ can't be read: relabel the x-axis as a phase axis (1λ ↔ 360°). A particle at displacement y sits at angle given by sin θ = y/y₀ — at half amplitude, θ = 30°. Work out each particle's angle, subtract. The "hard" phase question is pure trig, not a new formula.

Two points along a boundary hit by wavefronts at angle θ: effective separation is Δx = d sin θ (perpendicular distance between wavefronts through the points), THEN Δφ = (Δx/λ)2π. Draw the right triangle — the setup mark is the hard mark (N95 sea-wall).

Each particle is an SHM oscillator: E = ½mω²y₀² → energy ∝ f²y₀² → wave = sum of oscillating particles → I ∝ y₀² at constant f, I ∝ f²y₀² in general. Doubling frequency at fixed amplitude ALSO quadruples intensity. Quote the SHM chain when asked to justify.

I ∝ y₀² is not an equation — solve by ratio: I₂/I₁ = (y₀₂/y₀₁)². If P and A are both unknown, I = P/A is unusable; ratios still work. Amplitude ratio = √(intensity ratio) — square/root in the right direction.

3-D spherical (point source, "all directions"): I = P/4πr², I ∝ 1/r², A ∝ 1/r. Hemisphere ("all directions in front"): P/2πr². 2-D ripple (pond surface): energy over circumference → I ∝ 1/r, A ∝ 1/√r — DHS prints the warning: the 3-D law CANNOT be used. 1-D beam/plane (laser): I, A constant. Only I ∝ A² is universal.

Power collected = I at that point × receiver area (perpendicular). Detector size changes the power collected, never the intensity there. Energy received = I × A × t. Two routes to I at a receiver: P_source/4πr² or P_received/A_receiver — use whichever pair the question supplies. Convert cm² → m² religiously.

Two inverse-square trips. (1) I at target = P/4πd². (2) Power intercepted = that × S. (3) Target re-emits fraction k as a NEW point source. (4) Back at transmitter: divide by 4πd² again →

Entering a new medium: frequency unchanged (set by the source), speed changes, so λ = v/f scales with v. With amplitude info, I ∝ A² finishes the row: v and A both halved → (0.25I, f, 0.50λ). Writing "frequency halves" = instant lost mark.

Read the amplitude at two positions/times off the decaying envelope → energy lost to resistive forces = ½mω²(y₀₁² − y₀₂²) — the Oscillations formula carried over. Intensity ratio between the same two points: I₂/I₁ = (y₀₂/y₀₁)².

Unpolarised light = random mix of all oscillation planes. An ideal polariser transmits exactly ½ the intensity, regardless of orientation (average of cos²θ = ½), output polarised along its axis. This ½ comes before any Malus step — forgetting it is THE classic error (I₀/8, not I₀/4).

I = I₀cos²θ applies only to already-polarised light; θ is between the light's polarisation plane and the analyser axis — i.e. between consecutive axes, computed from geometry, never the printed angle to the vertical (axis 40° to horizontal + vertically polarised light → θ = 50°). Chains: multiply cos²θ per filter, amplitude (A ≡ y₀ throughout) picks up cosθ per filter.

Crossed pair alone: zero. Insert a middle polaroid at 45°: I = I₁cos²45° × cos²45° = I₁/4 ≠ 0 — the middle filter rotates the polarisation plane, reopening the door. Minimum transmission through a chain ⇔ some consecutive pair perpendicular; give BOTH angle answers in 0–360° (e.g. 120° and 300°).

Rotate a polaroid between beam and detector through at least 180°: intensity falling to zero at some orientation ⇒ plane polarised; constant intensity ⇒ unpolarised. Stock 2-marker, exact phrasing.

cos² shape: maxima at 0°, 180°, 360°, zeros at 90°, 270°, period 180°, never negative. Unpolarised source through a polariser THEN a rotating analyser: maxima at ½I₀. A single polaroid rotating in unpolarised light gives constant ½I₀ — no zeros (that constancy IS the unpolarised verdict). Applications: polaroid sunglasses (cut horizontally-polarised glare), 3-D cinema (orthogonal filters per eye).

Microphone into a CRO. Adjust the time-base until 2–3 cycles fill the screen. T = (divisions spanned by ONE cycle) × time-base setting, then f = 1/T. Count intervals honestly — traces usually show 2.5 cycles, and μs/ms time-base conversions are part of the question.

Speaker facing a reflector; microphone moved between them; signal swings max ↔ min. Adjacent maxima are λ/2 apart, so λ = 2d for neighbouring maxima. The mic senses pressure — its maxima sit at displacement nodes. Then v = fλ, compare with 340 m s⁻¹.

Visible: 400 nm (violet) – 700 nm (red). f = c/λ: 4.3 × 10¹⁴ Hz (red, 700 nm) to 7.5 × 10¹⁴ Hz (violet, 400 nm). Trap: minimum frequency pairs with the LONGEST wavelength.

When two or more waves of the same kind meet at a point, the resultant displacement at that point is the vector sum of the individual displacements that each wave would produce at that point on its own.

While two pulses overlap their displacements add, but afterwards each pulse carries on with its original shape, amplitude, speed and direction. At the instant two equal and opposite pulses fully cancel, the medium is momentarily flat but the particles are still moving, so the energy is entirely kinetic and nothing has been destroyed.

For two waves of the same frequency and equal amplitude A meeting with phase difference Δφ, the resultant amplitude is 2A cos(Δφ/2): it is 2A when Δφ = 0 and zero when Δφ = π. If the amplitudes differ, the resultant runs from A₁ + A₂ (in phase) down to |A₁ − A₂| (antiphase).

Interference is the superposition of two or more coherent waves to give a resultant wave whose amplitude is given by the principle of superposition, producing a fixed pattern of points of maximum and minimum intensity.

Two waves or sources are coherent if there is a constant phase difference between them. This requires them to have the same frequency (and hence the same wavelength in a given medium). Coherent does not mean in phase, and does not require equal amplitudes.

Phase difference is a measure of the fraction of a cycle by which one oscillation leads or lags another, expressed as an angle, where one complete cycle corresponds to 2π rad (360°).

Path difference is the difference in the distances travelled by two waves from their respective sources to the point where they meet.

For waves that leave their sources in phase, the phase difference on arrival is set entirely by the path difference: Δφ = (2π/λ) × path difference. One whole wavelength of extra path corresponds to 2π rad, and half a wavelength to π rad.

Constructive interference occurs where the two waves arrive in phase, that is Δφ = 2nπ. For sources emitting in phase this means a path difference of nλ (n = 0, 1, 2, …), and the resultant amplitude is the sum of the two amplitudes.

Destructive interference occurs where the two waves arrive in antiphase, that is Δφ = (2n + 1)π. For sources emitting in phase this means a path difference of (n + ½)λ (n = 0, 1, 2, …), and the resultant amplitude is the difference of the two amplitudes, which is zero if they are equal.

To observe a steady two-source interference pattern: (1) the sources must be coherent, that is have a constant phase difference and therefore the same frequency; (2) the waves must be of the same type and must overlap in the region of observation; (3) the amplitudes should be approximately equal, for good contrast; (4) for electromagnetic waves, they must be unpolarised or polarised in the same plane. For light in particular the slit separation must be small enough that the fringes are wide enough to see.

A stationary (standing) wave is formed by the superposition of two progressive waves of the same type, amplitude, frequency and speed (hence the same wavelength) travelling along the same line in opposite directions.

Draw the two opposite-travelling waves at t = 0, T/8, T/4, 3T/8, T/2 and add them point by point. At t = 0 and t = T/2 the components coincide and the resultant has twice the amplitude; at t = T/4 and 3T/4 the components are exactly antiphase everywhere and the resultant is zero at every position. The positions of the zeros never move, and these fixed zeros are the nodes.

A node is a point on a stationary wave where the amplitude is zero; the two component waves always arrive there in antiphase. An antinode is a point where the amplitude is a maximum (equal to twice the amplitude of one component wave); the two component waves always arrive there in phase.

Adjacent nodes are λ/2 apart, and so are adjacent antinodes. A node and the next antinode are λ/4 apart. Here λ is the wavelength of the component progressive waves, so λ = 2 × (node-to-node distance).

All particles between two adjacent nodes oscillate in phase with each other, reaching their extremes and passing through equilibrium at the same instant, although their amplitudes differ. Particles in adjacent loops oscillate in antiphase, that is π rad out of phase.

Stationary wave: wave profile does not move, no net energy transported, amplitude varies with position from zero at nodes to maximum at antinodes, particles in a loop are in phase and adjacent loops antiphase, adjacent nodes λ/2 apart. Progressive wave: profile moves with the wave speed, energy is transported in the direction of propagation, every particle has the same amplitude, particles within one wavelength have continuously varying phase, adjacent in-phase particles are λ apart. Both have the same frequency as the component waves.

A string fixed at both ends must have a node at each end, so L = n(λ/2). This gives λₙ = 2L/n and fₙ = nv/2L = n f₁, with all harmonics n = 1, 2, 3, … present. The fundamental (first harmonic) is a single loop with λ = 2L.

A pipe open at both ends has a displacement antinode at each end, so L = n(λ/2), giving λₙ = 2L/n and fₙ = nv/2L = n f₁ with all harmonics present. The fundamental has one node at the centre and λ = 2L.

A pipe closed at one end has a displacement node at the closed end and a displacement antinode at the open end, so L = (2n − 1)λ/4. This gives λ = 4L/(2n − 1) and f = (2n − 1)v/4L, so only the odd harmonics exist (f₁, 3f₁, 5f₁ …). The fundamental is λ = 4L, exactly one octave below an open pipe of the same length.

The displacement antinode at an open end actually lies a short distance c beyond the physical end of the pipe, so the effective length is L + c for a pipe with one open end and L + 2c for a pipe open at both ends. The correction is eliminated by taking the difference between two successive resonance lengths, which is exactly λ/2.

A microwave transmitter faces a metal reflector. The reflected wave has the same type, amplitude, frequency and speed as the incident wave but travels in the opposite direction along the same line, so they superpose to form a stationary wave. Moving a small detector along the line between them gives alternating maxima and minima; the distance between adjacent minima is λ/2, and f = c/λ.

A string runs from a mechanical vibrator over a pulley to a hanging mass that sets the tension. A node exists at the pulley because the string is fixed there, and approximately at the vibrator because its amplitude of vibration is very small compared with that of an antinode. Observable stationary waves only appear at resonance, when L = n(λ/2) = n(v/2f), so you must adjust either the frequency of the vibrator or the length of the string.

A loudspeaker of fixed frequency is held over a tube whose effective air column length is varied by a plunger or by raising the water level. A loud sound is heard whenever resonance occurs, first at L₁ ≈ λ/4, next at L₂ ≈ 3λ/4, so L₂ − L₁ = λ/2 and the speed of sound is v = fλ = 2f(L₂ − L₁). A node forms at the closed end and an antinode at the open end.

Diffraction is the spreading of a wave into the region beyond an obstacle, or after it passes through an aperture, so that the wave does not travel only in straight lines. The wavelength, frequency and speed are unchanged; only the shape of the wavefronts changes.

In a ripple tank with a straight-wave generator, a wide gap (b ≫ λ) lets the wavefronts pass through still almost straight, with slight curving only at the edges and a clear shadow region either side. A narrow gap (b ≈ λ) produces almost semicircular wavefronts spreading through the whole region beyond the barrier. In both cases the wavelength is unchanged.

For a single slit of width b, the first diffraction minimum lies at the angle given by sin θ = λ/b. The mth minimum is at sin θ = mλ/b. The result comes from splitting the wavefront in the slit into two halves so that every wavelet in the top half is cancelled by a partner b/2 below it with a path difference of λ/2.

The central maximum runs from the first minimum on one side to the first minimum on the other, so its angular width is 2λ/b and its width on a screen a distance D away is approximately 2λD/b. It is twice as wide as each secondary maximum and much brighter.

Two images are just resolved when the central maximum of the diffraction pattern of one coincides with the first minimum of the diffraction pattern of the other. For an aperture of width b this gives a minimum angular separation θ ≈ λ/b; sources separated by more than this are well resolved, and by less are unresolved.

Monochromatic light passes through a single slit, which acts as a point source so that the light reaching the double slit is coherent across both slits. Diffraction at each of the two slits provides two coherent sources, and the diffracted beams overlap and interfere, giving evenly spaced bright and dark fringes on a screen a distance D away.

For double-slit interference, λ = ax/D, where a is the slit separation, x is the fringe separation (the distance between the centres of adjacent bright fringes) and D is the slit-to-screen distance. It is valid only when D ≫ a so the rays are effectively parallel, and a ≫ λ so the angles are small and sin θ ≈ tan θ.

Water: two dippers on one bar in a ripple tank give anti-nodal lines of large amplitude and nodal lines of calm water. Sound: two loudspeakers on one signal generator, walk along a line in front and hear alternating loud and soft. Light: Young's double slit, giving bright and dark fringes on a screen. Microwaves: a transmitter feeding two slits cut in an aluminium plate, with a detector and meter reading alternating maxima and minima as it moves along a line.

Fringe separation depends only on λ, D and a. Increasing D or λ, or decreasing a, widens the fringes. Contrast depends on how equal the two amplitudes are: covering one slit partly makes the bright fringes less bright and the dark fringes brighter because destructive interference is now incomplete, so contrast falls with no change in spacing. Increasing the source intensity changes neither the spacing nor the contrast, only the overall brightness.

For a diffraction grating with slit separation d, the principal maxima occur at angles given by d sin θ = nλ, where n = 0, 1, 2, … is the order. If the grating is specified as N lines per metre, then d = 1/N. Every pair of adjacent slits contributes a path difference of d sin θ, so at these angles all the slits are in phase.

Because sin θ ≤ 1, the grating equation requires n ≤ d/λ. The highest observable order is the largest whole number not exceeding d/λ, so you round down, never up. The total number of maxima seen is 2n_max + 1, counting the central maximum once and each order on both sides.

Shine the light normally on a grating of known d, measure the angle θ of a chosen order n, and calculate λ = d sin θ / n. Measure the same order on both sides of the normal and take half the angle between them to remove any error in locating the normal. Higher orders give a larger θ and therefore a smaller percentage uncertainty, but are dimmer and so harder to locate.

The zero order is white, because all wavelengths have zero path difference there. Every higher order is a continuous spectrum, with violet closest to the centre and red furthest out, since d sin θ = nλ means longer wavelength deviates more. Higher orders can overlap, for example the red end of the second order can fall beyond the violet end of the third order.

For a given d, the positions of the maxima are the same for two slits and for a grating, but as the number of slits increases the maxima become much narrower and brighter, separated by broad dark regions. Just off the exact angle, the many beams get progressively further out of phase and cancel, so the intensity collapses almost immediately. This makes grating maxima far easier to locate precisely.

When a wave diffracts at a gap or an edge, its speed, frequency and wavelength are all unchanged. Only the shape of the wavefronts and the amplitude change. The same is true for superposition and interference: the component waves keep their own λ, f and v throughout.

The double-slit result x = λD/a relies on small angles (sin θ ≈ tan θ), which holds because a is much larger than λ. On a grating, d is comparable to λ, the angles are large, and the maxima are not evenly spaced, so you must use d sin θ = nλ and work in angles, never in fringe separations.

Superposition happens whenever any two waves of the same type meet, coherent or not. What coherence buys you is a pattern that stays fixed, so it can be observed. Two independent lamps do superpose, but their phase relationship changes randomly every 10⁻⁸ s or so, so the maxima and minima reshuffle far faster than the eye or detector can follow and average out to uniform illumination.

Between the 1st and 5th bright fringe there are 4 gaps, not 5, so x = (separation)/4. A grating showing orders up to n_max gives 2n_max + 1 maxima in total. The nth bright fringe corresponds to path difference nλ, but the nth dark fringe corresponds to path difference (n − ½)λ.