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.”
• 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.
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
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.
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.The common argument means the two terms have the same k and ω. Their sum has amplitude √(3² + 4²) = 5 cm.
- 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.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.
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.
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.
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.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.A full spatial repeat requires λ = 2π/80.0 = 0.07854 m = 7.854 cm.
- 3.The time repeat is T = 2π/3.0 = 2.094 s, and f = 3.0/(2π) = 0.4775 Hz.
- 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.Thus y = 0.005 sin(−36.0) = 0.0049589 m, approximately +4.96 mm. This is close to, but below, the positive amplitude.
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.
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.
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.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.The amplitude is 3.0 cm; ω = 36 rad s⁻¹ gives f = 36/(2π) = 5.730 Hz and T = 0.17453 s.
- 3.At the origin initially the phase is π/4. The wavelength is 2π/0.018 = 349.07 cm ≈ 3.49 m.
- 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.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.
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.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.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.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.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
A harmonic wave has k = 4π rad m⁻¹ and ω = 20π rad s⁻¹. Its speed is
v = ω/k = 5 m s⁻¹.
In y = a sin(kx + ωt), with k,ω positive, a constant-phase point moves
kx + ωt constant gives dx/dt = −ω/k.
Which separation ensures the same phase in a harmonic progressive wave?
A full wavelength changes phase by 2π. Equal instantaneous displacement alone is insufficient.
Which quantity is read from a y-versus-t graph at one fixed point?
The horizontal axis is time, so the repeat interval is a period.
Why is 1/(x + vt) problematic as a global small-displacement wave?
Translation of the argument does not remove the pole.
Practice Problems
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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
• 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.