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Lesson 2 of 8

Waves · Lesson 2 of 8

Describing Progressive Waves

“Read and construct harmonic wave equations, connect spatial and time graphs, and calculate phase and propagation speed.”

Learning Objectives

• Interpret amplitude, phase, wavelength, angular wave number, period and angular frequency with their units. • Derive the wavelength–wave-number, period–frequency and wave-speed relations. • Determine the direction of propagation by following a point of constant phase. • Distinguish particle velocity from wave speed and check whether a proposed wave function is physically admissible.

One function describes two different observations

Label each undisturbed string element by its equilibrium coordinate x. Its transverse displacement at time t is y(x,t), measured from equilibrium, not the distance travelled. Two independent inputs are needed because neighbouring elements need not have the same displacement at the same instant. Holding t fixed gives a spatial snapshot y versus x; holding x fixed gives the displacement history of one element, y versus t. A sinusoid can describe both, but one horizontal axis is distance and the other is time.

A harmonic wave moving towards increasing xLaTeX
a is the positive displacement amplitude (m), k > 0 the angular wave number (rad m⁻¹), ω > 0 the angular frequency (rad s⁻¹), and φ the phase constant (rad).

The sine varies between −1 and +1, so y varies between −a and +a. A crest has y = +a and a trough has y = −a. The phase θ = kx − ωt + φ identifies the stage of the local oscillation. At x = 0 and t = 0 it equals φ. Changing the coordinate origin or the instant called zero changes φ without changing the physical wave. But once initial conditions and origins are specified, φ cannot simply be dropped without changing those conditions.

At a fixed time t₀, the phase is kx plus a constant, so a photograph shows a sine curve. At a fixed location x₀, the phase is a constant minus ωt, so that particular element performs SHM. For example, with φ = 0, the element at the origin has y(0,t) = −a sin(ωt). The minus sign only sets the initial direction of its oscillation; it does not make its amplitude negative.

Combining sine and cosine terms

An equivalent descriptionLaTeX
Here θ = kx − ωt, A = a cos φ and B = a sin φ. Choose φ with the signs of A and B; tan φ = B/A alone does not determine the quadrant.

Expand sin(θ + φ) as sin θ cos φ + cos θ sin φ. Matching coefficients gives A = a cos φ and B = a sin φ. Squaring and adding removes φ because sin²φ + cos²φ = 1. This is why a combination of same-frequency, same-wave-number sine and cosine terms is still one harmonic travelling wave. If both A and B are zero there is no oscillation and its phase is undefined.

Combining a wave without losing its phase

Problem
For y = 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t), take x in metres, t in seconds and y in centimetres. Identify amplitude, direction and speed.

  1. 1.The common argument means the two terms have the same k and ω. Their sum has amplitude √(3² + 4²) = 5 cm.
  2. 2.Both coefficients are positive, so cos φ = 3/5 and sin φ = 4/5. The phase constant is in the first quadrant: φ ≈ 0.9273 rad.
  3. 3.For a constant phase, 5x − 0.5t is constant, so x increases at 0.5/5 = 0.10 m s⁻¹. The wave moves towards increasing x.

Wavelength is a full spatial repeat

At the same instant, neighbouring crests are separated by the smallest positive distance that increases phase by 2π. Write k(x + λ) − kx = 2π. This gives kλ = 2π. Points separated by λ have the same displacement and the same stage of motion. Merely sharing an instantaneous displacement is insufficient: two points on opposite sides of one crest can have the same height and different phases. A zero crossing can repeat after λ/2 with the opposite spatial slope.

Wavelength and wave numberLaTeX
λ is in metres. k counts phase change per distance, not simply the number of wavelengths per metre.

Period is a full time repeat

Hold the element coordinate fixed and let one period T pass. Its phase changes by −ωT. A complete oscillation repeats when the magnitude of this phase change is 2π, so ωT = 2π. The frequency f, also written ν, counts complete oscillations per second and is 1/T. All elements in an ideal uniform harmonic progressive wave have the same amplitude and frequency. Their phases at one time differ by kΔx.

Period, frequency and angular frequencyLaTeX
T is in seconds, f in Hz = s⁻¹, and ω in rad s⁻¹. A frequency in hertz must not be substituted for ω without the factor 2π.
Wavelength belongs to space; period belongs to timeSnapshot at one timeHistory at one positionyx / mλyt / sTVertical axis: signed displacement in both plots.Same-looking curves do not give the same horizontal quantity.
Spatial and temporal graphs— A wavelength is a spatial repeat; a period is a time repeat. Axes are schematic.

For a longitudinal wave, use s(x,t) for the signed particle displacement along the propagation direction. The same functional form s = a sin(kx − ωt + φ) applies. A graph of s versus x shows forward or backward displacement of air elements, not a literal up-and-down profile of a sound wave. Pressure and density waves accompany that displacement, but their peaks need not coincide with displacement peaks.

Extracting every quantity from a specified wave

Problem
For y(x,t) = 0.005 sin(80.0x − 3.0t), with all constants in SI units, find amplitude, wavelength, period, frequency and y at x = 0.300 m, t = 20.0 s.

  1. 1.Compare directly with a sin(kx − ωt): a = 0.005 m = 5 mm, k = 80.0 rad m⁻¹ and ω = 3.0 rad s⁻¹.
  2. 2.A full spatial repeat requires λ = 2π/80.0 = 0.07854 m = 7.854 cm.
  3. 3.The time repeat is T = 2π/3.0 = 2.094 s, and f = 3.0/(2π) = 0.4775 Hz.
  4. 4.At the requested event, θ = 80.0(0.300) − 3.0(20.0) = −36.0 rad. Use radians in the calculator. Adding 12π gives the equivalent phase 1.6991 rad.
  5. 5.Thus y = 0.005 sin(−36.0) = 0.0049589 m, approximately +4.96 mm. This is close to, but below, the positive amplitude.
Track a crest and track one material elementt = 0xt = T/4xt = T/2xRed: one element at x = 0. Orange: a crest moving right.A crest advances λ while the fixed element completes one period T.
Progressive snapshots— A crest moves right while the element at x = 0 stays at its equilibrium x coordinate.

Following phase gives the propagation speed

Follow a crest, or any selected fixed phase, rather than a material element. At two nearby events on this moving feature, kx − ωt + φ = k(x + Δx) − ω(t + Δt) + φ. Subtracting the common terms gives kΔx − ωΔt = 0. Hence Δx/Δt = ω/k. In differential notation dx/dt is the instantaneous velocity of this selected phase point; no new physical assumption is hidden in taking the limit of a small interval.

Speed of a harmonic progressive waveLaTeX
The pattern travels one wavelength while an element makes one oscillation. Here v is a positive speed; the signed phase velocity depends on the propagation direction.

For a sin(kx + ωt + φ), holding phase constant gives kΔx + ωΔt = 0 and dx/dt = −ω/k: it moves towards decreasing x. Multiplying the entire phase by −1, or converting a cosine to a sine, can change the appearance of an equation without changing its propagation direction. Determine the direction from the relative signs and a constant-phase calculation, not from the sign of the displacement.

Particle velocity is a different derivative. At a fixed x, the rate of change of y = a sin(kx − ωt + φ) is uᵧ = −aω cos(kx − ωt + φ). The factor −ω appears because the phase decreases at that rate. This velocity changes during each cycle and is transverse for a rope; its maximum magnitude is aω. Wave speed ω/k describes the advancing pattern along x. A particle can momentarily be at rest at a crest even while the crest itself continues moving.

A left-moving wave and three particle histories

Problem
A string has y = 3.0 sin(36t + 0.018x + π/4), where x,y are in cm and t in s. Find its properties and describe y–t graphs at x = 0, 2 and 4 cm.

  1. 1.Both x and t terms have the same sign, so the constant-phase point moves left: dx/dt = −36/0.018 = −2000 cm s⁻¹ = −20.0 m s⁻¹. This is a travelling wave.
  2. 2.The amplitude is 3.0 cm; ω = 36 rad s⁻¹ gives f = 36/(2π) = 5.730 Hz and T = 0.17453 s.
  3. 3.At the origin initially the phase is π/4. The wavelength is 2π/0.018 = 349.07 cm ≈ 3.49 m.
  4. 4.At x = 0, 2 and 4 cm, the three histories are 3 sin(36t + π/4), 3 sin(36t + π/4 + 0.036) and 3 sin(36t + π/4 + 0.072), all in cm. They have equal amplitude and period but successively shifted phases.
  5. 5.To sketch them, use t on the horizontal axis, y on the vertical, mark ±3 cm and one period 0.17453 s, then shift the curves using their initial phases. The phase advances correspond to time shifts of 0, 0.001 and 0.002 s.
Three nearby particles have shifted historiesy / cmx=0 cmx=2 cmx=4 cm0.000.050.100.150.20t / s+3−3All three curves: amplitude 3 cm, period 0.17453 s.The phase offsets 0, 0.036 and 0.072 rad are small but nonzero.
Particle histories— The small phase shifts are shown with solid, dashed and dotted curves; none changes amplitude or period.
Phase differences and length units

Problem
For y = 2.0 cos[2π(10t − 0.0080x + 0.35)], x,y in cm and t in s, find phase differences for separations 4 m, 0.5 m, λ/2 and 3λ/4.

  1. 1.The spatial phase coefficient has magnitude k = 2π(0.0080) = 0.016π rad cm⁻¹. Therefore λ = 125 cm = 1.25 m. Convert metres to centimetres before multiplying by this k.
  2. 2.The magnitude kΔx is 6.4π rad for 400 cm and 0.8π rad for 50 cm. The signed phase at a point farther along +x is decreased by these values.
  3. 3.For λ/2, the phase advance magnitude is π. For 3λ/4, it is 3π/2. A signed difference of −3π/2 is equivalent modulo 2π to +π/2; do not confuse a principal equivalent phase with the unreduced magnitude.
  4. 4.To compare whether motions are in phase, reduce modulo 2π. To report the phase accumulation over a specified separation, keep the unreduced value and explain the convention.

A translating mathematical shape still needs a physical domain

A shape F(x − vt) translates to the right without changing form: when t increases by Δt, increasing x by vΔt keeps its argument fixed. F(x + vt) translates left. The function need not be sinusoidal. For a physical small-displacement wave, also check that its argument and output have proper units, that it is defined on the required domain, and that displacement and strain stay within the model. A formal solution can fail as a global physical disturbance.

For example, (x − vt)² is a translating parabola after a coefficient with suitable units is supplied. It is finite at every finite coordinate, but grows without bound over an infinite medium, so it is not a globally bounded small-amplitude disturbance. log[(x + vt)/x₀] requires x₀ to be a positive reference length and x + vt > 0; it is singular as that positive argument tends to zero. 1/(x + vt) has a pole at x + vt = 0. None is a globally bounded, regular small-amplitude wave on the entire real line. Restricted-domain descriptions need explicit boundaries and limits. Similarly, 2√(x − vt) is real only for x − vt ≥ 0 and has a singular slope at the edge; it is not an acceptable smooth wave over the whole infinite medium.

Quick Quiz

Quiz

Quick check

A harmonic wave has k = 4π rad m⁻¹ and ω = 20π rad s⁻¹. Its speed is

Explanation

v = ω/k = 5 m s⁻¹.

Quick check

In y = a sin(kx + ωt), with k,ω positive, a constant-phase point moves

Explanation

kx + ωt constant gives dx/dt = −ω/k.

Quick check

Which separation ensures the same phase in a harmonic progressive wave?

Explanation

A full wavelength changes phase by 2π. Equal instantaneous displacement alone is insufficient.

Quick check

Which quantity is read from a y-versus-t graph at one fixed point?

Explanation

The horizontal axis is time, so the repeat interval is a period.

Quick check

Why is 1/(x + vt) problematic as a global small-displacement wave?

Explanation

Translation of the argument does not remove the pole.

Practice Problems

Practice Problems
  1. For y = 0.020 sin(4πx − 10πt + π/6), with SI units, determine amplitude, wavelength, period, frequency, direction, wave speed and initial displacement at the origin.
  2. Derive λ = 2π/k and T = 2π/ω from full phase repetition. Explain why equal displacement at two positions does not by itself define a wavelength.
  3. For y = 3.0 sin(36t + 0.018x + π/4), x,y in cm and t in s, sketch y–t histories at x = 0, 2 and 4 cm, with labelled amplitudes, periods and initial phases.
  4. For y = 2.0 cos[2π(10t − 0.0080x + 0.35)], x in cm, find the signed phase difference at equal time between the farther and nearer point for separations 4 m, 0.5 m, λ/2 and 3λ/4. Also give equivalent values modulo 2π.
  5. A right-moving rope wave is y = a sin(kx − ωt). Derive the velocity of a fixed string element and explain why its maximum magnitude aω is distinct from the wave speed ω/k.
  6. Test (x − vt)², log[(x + vt)/x₀] and 1/(x + vt) as global bounded small-amplitude waves. State the units or coefficients needed and identify domain, growth or singularity limitations.
  7. Combine 3 sin(5x − 0.5t) + 4 cos(5x − 0.5t) into one sine wave, choosing the correct phase quadrant. Explain why two different coefficients do not imply two propagation speeds.

Key Takeaways

Key Takeaways

• A spatial snapshot measures wavelength; a time history at fixed x measures period. • kλ = 2π, ωT = 2π and v = ω/k = λf connect phase repetition with propagation. • Follow constant phase to determine direction; do not confuse particle velocity with wave speed. • A translating argument is a mathematical test; domain, units and physical amplitude limits must also be checked.