Mechanical Waves & Representations

Flick one end of a long stretched spring and a hump runs to the far end. Tie a ribbon to one coil and watch that instead: it moves aside by a few centimetres, comes back, and stays where it was. That is the defining feature of a mechanical wave. The disturbance, and the energy that goes with it, crosses the medium; the matter does not. It works because each piece of the medium is coupled to the next by a restoring force, tension in a string or pressure in a gas, so a displaced piece pulls its neighbour after it with a delay set by how stiff the coupling is and how much mass has to be moved. That delay, not the hand, fixes the wave speed. If the hump crosses a 9.0 m spring in 1.5 s then v = 6.0 m s⁻¹, and a harder flick makes a taller hump that still takes 1.5 s. Remove the medium and there is nothing to couple, which is what separates these waves from light.

Classify a wave by comparing two directions: the way the disturbance travels and the way the particles are displaced. On a string the particles move at right angles to the travel direction, so the wave is transverse. Push and pull the end of a slinky along its length and bunched-up coils run down it while each coil shuffles forwards and backwards: a longitudinal wave, the kind sound is. A solid resists both squeezing and shearing, so it carries both types, at different speeds. In crustal rock an earthquake's longitudinal P wave travels at about 6.0 km s⁻¹ and its transverse S wave at about 3.5 km s⁻¹, so 100 km away they arrive 16.7 s and 28.6 s after the rupture, and the 11.9 s gap tells a seismologist the distance. The body of a fluid cannot pull back sideways, so it carries only longitudinal waves. For those, plot the displacement s along the travel direction; the graph looks transverse, but the medium is not shaped like it.

y(x, t) has two inputs and a sheet of paper has one horizontal axis. Hold the time fixed and plot y against x: that is a snapshot graph, a photograph of the entire medium at one instant. Hold the position fixed and plot y against t: that is a history graph, the log of one particle's motion for the whole experiment. Read the axis label before anything else, because for a sinusoidal wave the two graphs look identical. From a snapshot you read the amplitude A, the maximum displacement from equilibrium (half the crest-to-trough height), and the wavelength λ, the distance after which the shape repeats. From a history you read the same A and the period T, the time for one particle to complete one cycle, with f = 1/T. Neither graph alone contains the wave speed. You need one of each, or two snapshots a known time apart, or two histories a known distance apart.

Stand at one point of a periodic wave as a crest passes. One period later that particle is at the top again, which means the next crest has arrived: the pattern has moved forward by exactly one crest spacing. Distance λ in time T gives v = λ/T = fλ. The equation is easy to misread as saying that raising f raises v. It does not, because v belongs to the medium. Shake a spring on which waves travel at 3.0 m s⁻¹ at 4.0 Hz and the crests are 3.0 ÷ 4.0 = 0.75 m apart; shake it at 8.0 Hz and they close up to 0.375 m while the wave still covers 3.0 m every second. The source chooses f, the medium chooses v, and λ = v/f is the consequence. When a wave crosses into a different medium the frequency is the quantity that survives, because each particle at the boundary drives the next at the rate it is itself driven.

Take a pulse moving towards +x at 4.0 m s⁻¹. The snapshot at t = 0 shows it occupying 2.0 m to 5.0 m with a 3.0 cm peak at x = 4.0 m: a steep 1.0 m front and a gentle 2.0 m tail. What does the particle at x = 9.0 m do? Use Δt = Δx/v feature by feature. The front foot has 4.0 m to go and arrives at 1.00 s; the peak has 5.0 m to go and arrives at 1.25 s; the tail foot has 7.0 m to go and leaves at 1.75 s. So the history is flat, rises in 0.25 s, then falls over 0.50 s. The steep side, on the right of the snapshot, is on the left of the history, because whatever is furthest along in x arrives earliest in t. Algebra says the same: y(9.0 m, t) = F(9.0 m − vt), and as t grows the argument of F shrinks, so time runs through the shape backwards. For a wave moving towards −x the argument is x + vt and there is no mirroring.

A particle's velocity is the rate of change of its own displacement, vᵧ = ∂y/∂t, where the curly ∂ only means differentiate with respect to t while holding x fixed. That is the slope of the history graph. In the pulse above, the particle at 9.0 m rises 3.0 cm in 0.25 s, so vᵧ = +0.12 m s⁻¹, then falls 3.0 cm in 0.50 s, so vᵧ = −0.060 m s⁻¹. The snapshot holds the same information. Differentiating F(x − vt) by the chain rule gives ∂y/∂t = −v ∂y/∂x: the front has snapshot slope −0.030 m ÷ 1.0 m = −0.030, and −(4.0 m s⁻¹)(−0.030) = +0.12 m s⁻¹, as before. Two lessons follow. Ahead of an advancing crest the medium is moving up and behind it the medium is moving down, since the crest is about to reach the first and has just left the second. And the particle speed depends on the amplitude and steepness of the wave, while the wave speed of 4.0 m s⁻¹ depends on neither.