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Introduction to Waves · Lesson 3 of 5

The Travelling-Wave Function

14 min read2 ideas5 checks1 figure to playBy Asheshanand Acharya
What you will learn 0 / 2
The shape of a pulse is a function of x
Maths you need

The general form of a travelling wave

This lesson describes a wave on a string with mathematics. Lesson 1 chose the coordinates. The position of a piece of string along the string is \(x\). The displacement of that piece from the straight, undisturbed line is \(y\). The displacement depends on which piece we look at and on when we look. So \(y\) is a function of two variables, \(y(x, t)\). This function is called the wave function.

Imagine a pulse on a string, a bump with a peak height of \(2\,\text{mm}\) and its centre at \(x = 2\,\text{m}\) at time \(t = 0\). At that one instant, the displacement depends on \(x\) alone. So the shape of the pulse at \(t = 0\) is a function of \(x\), which we call \(f\).

\[ y(x, 0) = f(x) \]

A pulse moves without changing shapeA graph of the displacement y of a string, in millimetres, against the position x, in metres, from 0 to 26 metres. At t = 0 a single bump 2 millimetres high is centred at x = 2 metres. When played, the bump moves to the right at 10 metres per second without changing shape. A dashed copy of the t = 0 bump stays behind, and a growing arrow from 2 metres to the moving centre is marked vt. At t = 2 seconds the bump is centred at 22 metres and the arrow reads vt = 20 m. Two chips show the two instants.246810121416182022242612Ox (m)y (mm) t = 0: the starting shape
Fig 3.1The shape of the pulse at \(t = 0\) is a bump \(2\,\text{mm}\) high with its centre at \(x = 2\,\text{m}\); the function \(y(x, 0) = f(x)\) describes this shape. Press play: the pulse travels to the right at \(v = 10\,\text{m/s}\). The chip "t = 2 s" shows it two seconds later. The whole shape has shifted right by \(vt = 20\,\text{m}\), so its centre is now at \(x = 22\,\text{m}\).
A pulse moving right is f of x minus vt
Maths you need

The pulse two seconds later

Now suppose this pulse begins to travel to the right along the string, with a constant speed \(v = 10\,\text{m/s}\). After \(2\,\text{s}\), the whole pulse has moved a distance \(vt = 10 \times 2 = 20\,\text{m}\). The centre of the pulse, which was at \(x = 2\,\text{m}\), is now at \(x = 22\,\text{m}\). Fig 3.1 shows both instants. The dashed copy is the pulse at \(t = 0\), and the arrow from it is the shift \(vt\).

How do we describe the shape of the pulse at this later time? Every part of the shape that was at position \(x\) at \(t = 0\) is at \(x + vt\) after a time \(t\). Turn this round. The displacement found at \(x\) at time \(t\) is the displacement that was at \(x - vt\) at \(t = 0\). At \(t = 2\,\text{s}\), the displacement at position \(x\) equals the displacement at position \(x - 20\,\text{m}\) at \(t = 0\). In other words, \(y(x, 2) = f(x - 20)\).

A quick test confirms the sign. The peak of \(f\) is where its argument is \(2\). In \(f(x - 20)\) the argument is \(2\) when \(x - 20 = 2\), that is, at \(x = 22\,\text{m}\). The minus sign has moved the peak to the right, exactly as the figure shows.

The idea holds for any shape and any speed. Let the shape of a pulse at \(t = 0\) be any function, \(y(x, 0) = f(x)\). Let the pulse travel to the right with speed \(v\). After a time \(t\), the whole shape has shifted to the right by \(vt\). The derivation below gives the displacement at any position \(x\) and any time \(t\).

Deriving \(y(x, t) = f(x - vt)\) step 1 of 4
1

\[ y(x, 0) = f(x) \]

Why. the shape of the pulse at \(t = 0\) is given as a function of \(x\).

RememberA wave travelling along x
\[ y(x, t) = f(x - vt) \quad \text{(towards } +x\text{)}, \qquad y(x, t) = g(x + vt) \quad \text{(towards } -x\text{)} \]
Holds whenany pulse or wave that travels along a line at constant speed \(v\) without changing its shape.
Symbols\(y\) the displacement of the medium at position \(x\) and time \(t\); \(f\) and \(g\) the shapes at \(t = 0\); \(v\) the speed of the wave
Units\(x\) and \(y\) in m, \(t\) in s, \(v\) in m/s

This is equation (1) of the chapter, the general mathematical form of any wave pulse that travels in the positive \(x\) direction. To write it, simply replace \(x\) with \(x - vt\) in the function for the initial shape. In the same way, a wave that travels in the negative \(x\) direction is described by \(y(x, t) = g(x + vt)\). The plus sign moves the shape to the left, because the peak of \(g(x + vt)\) moves to smaller \(x\) as \(t\) grows.

Predict firstLocked

A pulse \(2\,\text{mm}\) high has its peak at \(x = 12\,\text{m}\) at \(t = 0\). The pulse is described by \(y = f(x + 10t)\), with \(x\) in metres and \(t\) in seconds. Where is the peak at \(t = 1\,\text{s}\)?

The sign in x − vt and x + vtThe same graph of displacement in millimetres against position in metres. A 2 millimetre bump starts with its peak at x = 12 metres, and a dashed copy stays there. Two chips choose the wave function: y = f(x − 10t) or y = f(x + 10t). When played for one second, the first moves the bump 10 metres to the right, to 22 metres, and the second moves it 10 metres to the left, to 2 metres. An arrow from 12 metres to the moving peak grows with it, and a readout gives the time and the position of the peak.246810121416182022242612Ox (m)y (mm) t = 0.0 s, peak at x = 12 m

Choose each chip in the figure above and press play. With \(f(x - 10t)\) the peak goes from \(12\,\text{m}\) to \(22\,\text{m}\) in one second. With \(f(x + 10t)\) it goes from \(12\,\text{m}\) to \(2\,\text{m}\). In both, the height and the shape of the pulse do not change. Only the sign between \(x\) and \(vt\) decides the direction.

Common mistake

The minus sign in \(x - vt\) means motion towards \(+x\), and the plus sign in \(x + vt\) means motion towards \(-x\). Students often read it the other way round. Always test with the peak: find where the argument keeps its value at \(t = 0\), and see which way that point moves.

CheckpointUnanswered

A wave pulse is described by \(y(x, t) = \dfrac{A}{1 + (x + 3t)^2}\), with \(x\) in metres and \(t\) in seconds. In which direction does it travel, and with what speed?

ExampleThe shape two seconds later

A pulse travels in the positive \(x\) direction with speed \(v = 10\,\text{m/s}\). Its shape at \(t = 0\) is \(y(x, 0) = f(x) = \sin x\). Which function gives the shape of the pulse at \(t = 2\,\text{s}\)?

Try it on paper first. The solution is one tap away when you are ready.

When x has a coefficient

The coefficient of \(t\) gives the speed only when \(x\) stands alone in the argument. Consider \(y = 3/[1 + (2x + 6t)^2]\). Take the factor \(2\) out of the argument: \(2x + 6t = 2(x + 3t)\). Now the function has the form \(g(x + 3t)\). It travels towards \(-x\) at \(3\,\text{m/s}\), not at \(6\,\text{m/s}\). In general, a function of \(ax - bt\) travels towards \(+x\) at the speed \(b/a\).

CheckpointUnanswered

A pulse is described by \(y = \dfrac{3}{1 + (2x + 6t)^2}\), with \(x\) in metres and \(t\) in seconds. With what speed, and in which direction, does it travel?

Checkpoint · numericUnanswered

A pulse is described by \(y = f(x - 5t)\), with \(x\) in metres and \(t\) in seconds. At \(t = 0\) its peak is at \(x = 1\,\text{m}\). Where is the peak at \(t = 3\,\text{s}\), in metres?

m± 2 %
Checkpoint · in your own wordsUnanswered

A friend says, "In \(f(x - vt)\) the minus sign makes the pulse move backwards, towards \(-x\)." Explain what is wrong.

In the world
Measuring a pulse with a phone camera

Stretch a slinky along a smooth floor and send a pulse along it. Record it with a phone at \(30\) frames a second, next to a metre scale. In one frame the peak is at the \(0.8\,\text{m}\) mark. Six frames later, that is \(0.2\,\text{s}\) later, it is at the \(2.0\,\text{m}\) mark. The pulse moved \(1.2\,\text{m}\) in \(0.2\,\text{s}\), so its speed is \(6\,\text{m/s}\). If its shape at the first frame is \(f(x)\), the slinky is described by \(y = f(x - 6t)\), with \(t\) counted from that frame.

Exam traps
  1. Reading the minus sign in \(x - vt\) as motion towards \(-x\).
    Follow the peak. Where \(x - vt\) keeps its value, \(x\) grows with \(t\), so the wave moves towards \(+x\).
  2. Taking the coefficient of \(t\) as the speed when \(x\) has a coefficient too.
    Write \(ax \pm bt = a(x \pm bt/a)\). The speed is \(b/a\).
  3. Changing the shape when shifting it.
    The function \(f\) stays the same. Only its argument changes, from \(x\) to \(x - vt\).