Wave Graphs: Decide Whether the Horizontal Axis Shows Distance or Time

Read the horizontal axis before measuring a wave

Two wave graphs can have almost identical shapes while answering different questions. A displacement–distance graph shows many positions at one instant. A displacement–time graph follows one position through successive instants. One gives a wavelength; the other gives a period.

The examples use original idealised transverse waves with a clearly defined equilibrium line. Treat the horizontal and vertical scales independently. A centimetre printed on the page is not automatically a centimetre of wave displacement or distance: the axis labels determine the physical quantities.

1. Identify what is changing

In either graph, amplitude A is the maximum magnitude of displacement from equilibrium. If the crests are at +3 mm and the troughs at −3 mm, amplitude is 3 mm. The full crest-to-trough vertical difference is 6 mm, which is twice the amplitude for this symmetric wave.

Equivalent neighbouring points must be at the same stage of a cycle: crest to next crest, trough to next trough, or a zero crossing to the next zero crossing with the same slope direction. Adjacent zero crossings with opposite slope directions are only half a cycle apart.

2. An original pair of graphs

Graph A is a snapshot of displacement y against position x from 0 to 1.60 m. Graph B records displacement y at one fixed position from 0 to 0.50 s. They represent the same steady wave in the same medium. Their common amplitude is 3 mm. The chosen time record begins with a downward crossing of equilibrium; the snapshot begins with an upward crossing as position increases.

Native alternative for graph A: The horizontal axis is position x in metres and the vertical axis is displacement y in millimetres. The equilibrium line is y = 0. The smooth sinusoid passes through (0, 0), (0.20, +3), (0.40, 0), (0.60, −3), (0.80, 0), (1.00, +3), (1.20, 0), (1.40, −3), and (1.60, 0). There are two complete spatial cycles. The zeros at 0, 0.80 and 1.60 m have upward slope.

Native alternative for graph B: The horizontal axis is time t in seconds and the vertical axis is displacement y in millimetres at one fixed point. The smooth sinusoid passes through (0, 0), (0.0625, −3), (0.125, 0), (0.1875, +3), (0.250, 0), (0.3125, −3), (0.375, 0), (0.4375, +3), and (0.500, 0). There are two complete time cycles. The zeros at 0, 0.250 and 0.500 s have downward slope.

The coordinate lists supply every reading needed for the calculations, even if the graphics are unavailable. The curves between the listed points are smooth sinusoidal arcs, not straight-line segments. The different starting slope directions are compatible with a rightward-travelling wave sampled at x = 0, although travel direction is not needed to find the speed magnitude here.

Worked case 1: read wavelength and amplitude

Use graph A to find amplitude and wavelength. Explain why 0.40 m is not the wavelength.

Solution: The crest displacement is +3 mm and equilibrium is 0, so A = 3 mm = 0.003 m. Consecutive crests occur at x = 0.20 m and x = 1.00 m. Their separation is λ = 1.00 − 0.20 = 0.80 m. As a cross-check, the 1.60 m span contains two cycles, giving 1.60/2 = 0.80 m per cycle.

The separation from the crest at 0.20 m to the next trough at 0.60 m is 0.40 m. That is half a wavelength, because a crest and trough are opposite stages of the oscillation. The vertical 6 mm crest-to-trough difference is a different quantity again; it is neither wavelength nor amplitude.

Worked case 2: read period and calculate frequency

Use graph B to find period and frequency. Then explain why counting the visible peaks alone can be misleading.

Solution: Neighbouring troughs occur at 0.0625 s and 0.3125 s. Thus T = 0.3125 − 0.0625 = 0.250 s. Frequency is cycles per second: f = 1/T = 1/0.250 = 4.0 Hz. Equivalently, two complete cycles in 0.500 s give 2/0.500 = 4.0 Hz.

A visible peak is a point, whereas a completed cycle is an interval between equivalent points. A graph that begins and ends at peaks may show three peaks but only two crest-to-crest intervals. Use intervals and their stated endpoints, not a count of decorative humps.

The negative displacement early in this record means the point is below equilibrium. It does not mean negative amplitude or negative frequency. Both amplitude and frequency are positive magnitudes in these calculations.

Worked case 3: combine compatible graphs

Find the propagation speed of the wave represented by graphs A and B. Show two equivalent calculations.

Solution: The distance advanced by one full pattern in one period is one wavelength. Hence v = λ/T = 0.80/0.250 = 3.2 m/s. Using frequency gives v = fλ = 4.0 × 0.80 = 3.2 m/s.

This is the speed at which the disturbance propagates along the medium. It is not the vertical speed of the single point shown in graph B. At a crest that point can be momentarily at rest vertically while the pattern continues travelling. Confusing those two motions is a reason to keep the graph description beside the formula.

Combining values from unrelated graphs would be unjustified. The problem explicitly tells us that these plots describe the same steady wave in the same medium. A wavelength from one situation and a frequency from another need not yield a meaningful speed.

Worked case 4: know what one graph cannot tell you

A separate distance snapshot has neighbouring crests at 0.15 m and 0.75 m. The question also supplies a propagation speed of 2.4 m/s. Find wavelength, frequency and period.

Solution: λ = 0.75 − 0.15 = 0.60 m. Frequency = v/λ = 2.4/0.60 = 4.0 Hz. Period = 1/f = 0.25 s. Without the independent speed or some timing information, the snapshot alone would determine wavelength but not frequency or period.

Likewise, a time record alone supplies period but not wavelength. You need speed or equivalent spatial information. A formula with two unknown quantities does not become solvable merely because one has recognised the wave shape.

3. A short graph-reading routine

1. Say what both axes measure and record their units.
2. Locate the equilibrium line before reading amplitude.
3. Choose two neighbouring points at the same stage of the cycle.
4. Subtract their horizontal coordinates.
5. Name the result from the horizontal unit: length gives wavelength; time gives period.
6. Convert units before combining data in v = fλ or v = λ/T.
7. Check that the data describe the same wave and medium.

Do not borrow area and gradient rules from a velocity–time graph. The ordinate here is displacement, not velocity. The area under either displayed wave graph is not automatically distance travelled. On the time graph a slope concerns the transverse motion of the observed point, rather than the horizontal propagation speed.

If a trace is shifted vertically on a diagram, identify its stated equilibrium value first. A crest at y = 7 mm with equilibrium at y = 2 mm has amplitude 5 mm. Reading the crest’s coordinate alone would incorrectly give 7 mm. Axis origin and equilibrium can differ.

Three tasks with complete solutions

Task 1: a distance plot with centimetres

A distance graph has equilibrium at 0, crests at +4 mm, troughs at −4 mm, and neighbouring crests at 10 cm and 70 cm. The wave frequency is 5.0 Hz. Find amplitude, wavelength and speed.

Solution: A = 4 mm. Wavelength = 70 − 10 = 60 cm = 0.60 m. Speed = 5.0 × 0.60 = 3.0 m/s. Using 60 directly would give 300 cm/s, which is equivalent only if its unit is retained and converted correctly.

Task 2: a time plot with milliseconds

At one point, neighbouring crests occur at 40 ms and 200 ms. Find period and frequency. Can wavelength be found from those times alone?

Solution: T = 200 − 40 = 160 ms = 0.160 s. Frequency = 1/0.160 = 6.25 Hz. Wavelength cannot be determined without speed or spatial information. The 160 ms separation must not be labelled as a length.

Task 3: repair a half-cycle reading

In a distance snapshot, a crest at 0.30 m is followed by a trough at 0.75 m. The period is 0.60 s. A learner uses 0.45 m as wavelength. Correct the wavelength and find speed.

Solution: Crest-to-next-trough separation is 0.75 − 0.30 = 0.45 m, which equals half a wavelength. Thus λ = 2 × 0.45 = 0.90 m. Speed = 0.90/0.60 = 1.5 m/s. The learner’s 0.75 m/s would be half the correct speed because the length interval was misidentified.

Frequently asked questions

Which graph gives wavelength directly?

A displacement–distance snapshot. Measure the horizontal distance between neighbouring equivalent points, such as consecutive crests, and retain the distance unit.

Which graph gives period directly?

A displacement–time record at one fixed position. Measure the time between neighbouring equivalent points and then use frequency = 1/period if needed.

Is crest-to-trough height the amplitude?

For the symmetric waves here, it is twice the amplitude. Amplitude is measured from equilibrium to a crest or trough, using the magnitude of that displacement.

Do any two zero crossings define one cycle?

No. Adjacent zero crossings with opposite slope directions are half a cycle apart. Use crossings with the same slope direction for a full cycle.

Can one distance snapshot give frequency?

Not without timing information or an independently known wave speed. The snapshot supplies wavelength, but it does not show how quickly the pattern moves.

Does the point on a time graph travel along the wave at v?

No. The graph records that point’s displacement over time. Propagation speed describes the moving disturbance, not the point’s transverse speed.

Can I multiply frequency from any graph by a wavelength?

Only when the values are compatible measurements of the same wave in the stated medium and situation. Unrelated graphs do not establish a valid speed calculation.

Continue revision

Browse the Physics topical revision collection; individual files may have separate access requirements. The OpenStax wave reference supports the relationships between period, wavelength and propagation speed. All graphs, numerical cases and questions here are original.