Intuition The big pictureA wave is a pattern that repeats in two ways at once : it repeats in space (you see the same shape every so many metres) and it repeats in time (at one spot, the medium bobs up and down on a regular beat). Every wave parameter is just a name for one of these repetitions, or for how big the wiggle is.
Repetition in space → wavelength λ \lambda λ
Repetition in time → period T T T frequency f f f
Size of the wiggle → amplitude A A A
How fast the pattern travels → wave speed v v v
Definition The five parameters
Amplitude A A A — the maximum displacement of a particle from its equilibrium position. Units: metres (or whatever the medium's displacement is measured in). It is a distance from the middle line , NOT peak-to-peak.Wavelength λ \lambda λ — the shortest distance (in metres) over which the wave shape repeats. Crest-to-crest, or trough-to-trough.Period T T T — the time (seconds) for one full oscillation of any single particle; equivalently the time for the wave to advance one wavelength.Frequency f f f — number of full oscillations per second. Units: hertz, 1 Hz = 1 s − 1 1\,\text{Hz}=1\,\text{s}^{-1} 1 Hz = 1 s − 1 Wave speed v v v — the speed (m/s) at which a point of fixed phase (e.g. a particular crest) moves through the medium.
f = 1 / T f=1/T f = 1/ T T T T 1 second you fit 1 / T 1/T 1/ T
Intuition WHY this must be trueSpeed = distance ÷ time. In one period T T T , a crest travels forward by exactly one wavelength λ \lambda λ (because after one full oscillation the pattern looks identical, just shifted by λ \lambda λ
Intuition A subtle but crucial pointFor a wave in a given medium, v v v f f f λ \lambda λ inversely linked: λ = v / f \lambda = v/f λ = v / f not change.
Worked example Example 1 — find the speedA wave on a string has f = 50 Hz f=50\,\text{Hz} f = 50 Hz λ = 0.40 m \lambda=0.40\,\text{m} λ = 0.40 m v v v
Step 1: Use v = f λ v=f\lambda v = f λ Why this step? We want speed and we already know frequency and wavelength — the wave relation links exactly these three.Step 2: v = 50 × 0.40 = 20 m/s v = 50\times0.40 = 20\,\text{m/s} v = 50 × 0.40 = 20 m/s Sanity check: 50 50 50 0.4 0.4 0.4 20 20 20
Worked example Example 2 — change the frequency, same mediumThe same string now carries f = 100 Hz f=100\,\text{Hz} f = 100 Hz
Step 1: The medium is unchanged, so v = 20 m/s v=20\,\text{m/s} v = 20 m/s Why this step? Speed depends on the medium, not on who is shaking the string.Step 2: λ = v / f = 20 / 100 = 0.20 m \lambda=v/f=20/100=0.20\,\text{m} λ = v / f = 20/100 = 0.20 m Forecast-then-verify: We doubled f f f λ \lambda λ halves (from 0.40 to 0.20). ✓ Confirmed.
Worked example Example 3 — from period and amplitudeA buoy bobs through a total vertical swing of 1.2 m 1.2\,\text{m} 1.2 m 4 s 4\,\text{s} 4 s 3 m/s 3\,\text{m/s} 3 m/s A A A f f f λ \lambda λ
Amplitude: total swing is peak-to-peak = 2 A =2A = 2 A A = 1.2 / 2 = 0.6 m A=1.2/2=0.6\,\text{m} A = 1.2/2 = 0.6 m Why? Amplitude is measured from the middle, half the full swing.Frequency: f = 1 / T = 1 / 4 = 0.25 Hz f=1/T=1/4=0.25\,\text{Hz} f = 1/ T = 1/4 = 0.25 Hz Wavelength: λ = v / f = 3 / 0.25 = 12 m \lambda=v/f=3/0.25=12\,\text{m} λ = v / f = 3/0.25 = 12 m λ = v T = 3 × 4 = 12 m \lambda=vT=3\times4=12\,\text{m} λ = v T = 3 × 4 = 12 m Why two routes agree? Because v / f = v T v/f=vT v / f = v T 1 / f = T 1/f=T 1/ f = T
Common mistake "Amplitude is the distance between crest and trough."Why it feels right: that vertical distance is the most visually obvious length on a wave diagram, and it is a real measurement (peak-to-peak).
The fix: amplitude is measured from the equilibrium line to a peak, so it is half the crest-to-trough distance: A = 1 2 ( peak-to-peak ) A=\tfrac12(\text{peak-to-peak}) A = 2 1 ( peak-to-peak )
Common mistake "If I increase the frequency, the wave speed increases."Why it feels right: higher frequency feels like "more energetic / faster," and shaking faster does send wiggles out faster... it seems.
The fix: speed is set by the medium (v = T / μ v=\sqrt{T/\mu} v = T / μ f f f v v v λ \lambda λ v v v
Common mistake "Wavelength and period are the same thing."Why it feels right: both describe "how often the wave repeats," and both shrink when frequency rises.
The fix: λ \lambda λ space (metres), T T T time (seconds). Different dimensions! They are linked only through the speed: λ = v T \lambda=vT λ = v T
Recall Quick self-test (cover the answers)
Define amplitude in one sentence.
Derive v = f λ v=f\lambda v = f λ
If v v v f f f λ \lambda λ
Units of f f f T T T λ \lambda λ
Recall Feynman: explain to a 12-year-oldImagine a long jump-rope. Amplitude is how high you flick the rope above the floor. Wavelength is the distance from one hump to the next hump along the rope. Period is how long you take to make one full up-and-down flick of your hand. Frequency is how many flicks you do each second. And speed is how fast a hump zooms down the rope toward your friend. Since each second you make f f f λ \lambda λ f × λ f\times\lambda f × λ
Mnemonic Remember the master equation"Very Fast Llamas" → V = F Λ V = F\,\Lambda V = F Λ f f f T T T f = 1 / T f=1/T f = 1/ T
Simple Harmonic Motion — each wave particle executes SHM; amplitude & period come straight from there.Transverse and Longitudinal Waves — these parameters apply to both wave types.Speed of Waves on a String — explains why v = T / μ v=\sqrt{T/\mu} v = T / μ The Wave Equation y(x,t) — combines all parameters: y = A sin ( k x − ω t ) y=A\sin(kx-\omega t) y = A sin ( k x − ω t ) k = 2 π / λ k=2\pi/\lambda k = 2 π / λ ω = 2 π f \omega=2\pi f ω = 2 π f Doppler Effect — what happens to f f f λ \lambda λ Sound Waves and Electromagnetic Spectrum — applications of v = f λ v=f\lambda v = f λ
Define amplitude of a wave The maximum displacement of a particle from its equilibrium position (NOT peak-to-peak; peak-to-peak = 2A).
Define wavelength The shortest distance over which the wave shape repeats (e.g. crest to crest), measured in metres.
Define period of a wave Time for one complete oscillation of a single particle, in seconds.
Define frequency Number of complete oscillations per second, measured in hertz (Hz = s⁻¹).
Relation between f and T f = 1/T (they are reciprocals).
State the fundamental wave equation v = fλ = λ/T.
Derive v = fλ In one period T a crest moves forward by one wavelength λ, so v = distance/time = λ/T = fλ.
If wave speed is fixed and frequency doubles, what happens to wavelength? It halves, since λ = v/f.
What determines the speed of a wave? The properties of the medium (e.g. tension and linear density for a string), not the frequency.
Units of f, T, λ, v f in Hz, T in s, λ in m, v in m/s.
A wave has f = 25 Hz, λ = 2 m. Speed? v = fλ = 25×2 = 50 m/s.
Wave repeats in space and time
Intuition Hinglish mein samjhoDekho, ek wave do tarah se repeat karti hai — space mein aur time mein . Agar tum wave ki ek photo le lo (snapshot), to crest-to-crest ka distance hota hai wavelength λ \lambda λ period T T T Frequency f f f f = 1 / T f=1/T f = 1/ T Amplitude A A A 2 A 2A 2 A A A A
Sabse important formula hai v = f λ v=f\lambda v = f λ T T T λ \lambda λ v = λ / T = f λ v=\lambda/T=f\lambda v = λ / T = f λ
Ek bada confusion: students sochte hain ki frequency badhao to speed badhegi. Galat! Speed medium decide karta hai (string ke liye tension aur density). Agar medium same hai to v v v wavelength chhoti ho jaati hai (λ = v / f \lambda=v/f λ = v / f
Yaad rakhne ka tarika: "Very Fast Llamas " = V = F Λ V=F\Lambda V = F Λ f f f T T T v = f λ v=f\lambda v = f λ f = 1 / T f=1/T f = 1/ T
