Cards Explorer

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 8.

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.

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.