The wave in differential form
The wave function \(y = f(x - vt)\) of equation (1) can also be written in a differential form. To find it, we look at the derivatives of \(y(x, t)\). The displacement depends on two variables, so there are two kinds of derivative. Each one changes one variable and holds the other fixed.
A derivative taken with the other variable held fixed is a partial derivative, written with a curly \(\partial\). Hold \(t\) fixed, as in a photograph of the string, and \(\partial y/\partial x\) is the slope of the string at \(x\). Hold \(x\) fixed, watching one particle, and \(\partial y/\partial t\) is the velocity of that particle. From the chapter Motion in One Dimension, a velocity is the rate of change of a position with time. Here the position of the particle is its displacement \(y\).
The first derivatives
Write \(u = x - vt\), so that \(y = f(u)\). Let \(f'\) be the derivative of \(f\) with respect to its argument. The chain rule differentiates \(f\) with respect to \(u\), then multiplies by the derivative of \(u\).
\[ \dfrac{\partial u}{\partial x} = 1, \qquad \dfrac{\partial u}{\partial t} = -v \]
Why. \(u = x - vt\). With \(t\) fixed, \(u\) grows exactly as \(x\) does. With \(x\) fixed, \(u\) falls by \(v\) for every unit of time.
\[ \dfrac{\partial y}{\partial x} = f'(x - vt) \]
Why. the chain rule gives \(f'(u) \times \partial u/\partial x = f'(u) \times 1\).
\[ \dfrac{\partial y}{\partial t} = -v\,f'(x - vt) \]
Why. the chain rule gives \(f'(u) \times \partial u/\partial t = f'(u) \times (-v)\).
\[ \dfrac{\partial y}{\partial t} = -v\,\dfrac{\partial y}{\partial x} \]
Read it. step 3 is \(-v\) times the result of step 2, at every \(x\) and every \(t\).
This is the relation between the two first derivatives. It connects three things: the wave velocity \(v_{\text{wave}}\), the particle velocity \(v_{\text{particle}} = \partial y/\partial t\), and the slope \(\partial y/\partial x\) of the string. This is equation (2) of the chapter.
\[ v_{\text{particle}} = -\,v_{\text{wave}}\,(\text{slope}) \]
A pulse travels to the right along a string. Consider a small piece of the string on the front edge of the pulse, between the peak and the flat string ahead. At this instant, which way does the piece move?
Press play on the figure above and follow the arrows. On the front edge of the pulse the slope is negative, and the arrows point up. On the back edge the slope is positive, and the arrows point down. At the peak, and on the flat string far away, the slope is zero, and so is the velocity. Watch the painted bead \(P\) at \(x = 6\,\text{m}\). At \(t = 1.4\,\text{s}\) the readout gives a slope of \(-0.61\) there, so \(v_P = -2 \times (-0.61) \approx +1.23\,\text{m/s}\), upward.
Think of the pulse as a shape sliding to the right under a row of beads that can move only up and down. A bead just ahead of the peak is about to be lifted, so it moves up. A bead just behind the peak is about to be let down, so it moves down. The steeper the shape at a bead, the faster the bead must move to stay on it.
A wave travels towards \(+x\) at \(20\,\text{m/s}\) along a string. At one point the slope of the string is \(+0.05\). What is the velocity of the particle of string at that point, in m/s? Take upward as positive.
The second derivatives
Differentiate once more. Each partial derivative of \(f'\) again brings the factor \(\partial u/\partial x = 1\) or \(\partial u/\partial t = -v\). Let \(f''\) be the second derivative of \(f\) with respect to its argument.
\[ \dfrac{\partial^2 y}{\partial x^2} = f''(x - vt) \]
Why. differentiate \(\partial y/\partial x = f'(x - vt)\) with respect to \(x\). The chain rule gives \(f''\) times \(\partial u/\partial x = 1\).
\[ \dfrac{\partial^2 y}{\partial t^2} = v^2 f''(x - vt) \]
Why. differentiate \(\partial y/\partial t = -v f'(x - vt)\) with respect to \(t\). The chain rule gives \(-v \times f'' \times (-v) = v^2 f''\).
\[ f''(x - vt) = \dfrac{1}{v^2}\,\dfrac{\partial^2 y}{\partial t^2} \]
Why. divide step 2 by \(v^2\).
\[ \dfrac{\partial^2 y}{\partial x^2} = \dfrac{1}{v^2}\,\dfrac{\partial^2 y}{\partial t^2} \]
Read it. steps 1 and 3 both equal \(f''(x - vt)\), so they equal each other.
Every wave \(y(x, t)\) that travels along the \(x\) axis with speed \(v\) and keeps its shape satisfies \(\dfrac{\partial^2 y}{\partial x^2} = \dfrac{1}{v^2}\dfrac{\partial^2 y}{\partial t^2}\). The same holds for a wave \(g(x + vt)\), because \(v^2\) does not depend on the direction.
This is the general mathematical form, as a wave equation, of any wave pulse that travels along the \(x\) axis. The wave equation is a second-order partial differential equation. It contains second derivatives with respect to two variables. It is one of the fundamental equations of wave physics. The chapter Propagating Waves on a String applies Newton's second law to a small piece of string and finds what sets the speed \(v\).
A wave \(y = f(x - 4t)\) travels along a string, with \(x\) in metres and \(t\) in seconds. At some point and instant, \(\partial^2 y/\partial t^2\) is \(k\) times \(\partial^2 y/\partial x^2\). Find \(k\).
In 1747 the French mathematician Jean le Rond d'Alembert studied a stretched, vibrating string. He wrote down the equation that its displacement obeys, the equation of this lesson. He also found its solution: any function of \(x - vt\) added to any function of \(x + vt\). It was one of the first partial differential equations ever solved. Euler and Daniel Bernoulli then argued with him for years about which shapes the functions could have.
Conditions for a valid wave function
Not every function of \(x\) and \(t\) describes a wave. A real string has a finite displacement everywhere, and a real wave keeps its shape as it travels. These facts give a short list of conditions.
For a function \(y(x, t)\) to represent a travelling wave:
A pulse on a string is described by \(y(x, t) = \dfrac{1}{1 + (x - 2t)^2}\). Here \(x\) is in metres and \(t\) in seconds. (a) Is this function a valid wave function? (b) What is the velocity of the wave? (c) In what direction does the wave travel?
(a) Yes. The function has the form \(f(x - vt)\), so it satisfies the wave equation. It is important to note that it also stays finite for all \(x\) and \(t\): its denominator is never less than \(1\).
(b) Compare it with the general form \(y(x, t) = f(x - vt)\). The coefficient of \(t\) gives \(v = 2\,\text{m/s}\).
(c) The obtained value of \(v\) is positive, with the form \(x - vt\). So the wave travels in the positive \(x\) direction.
The same pulse tests equation (2). At \(t = 0\) and \(x = 1\,\text{m}\), the slope of \(y = 1/(1 + x^2)\) is \(-2x/(1 + x^2)^2 = -2/4 = -0.5\). So the particle there moves with \(v_{\text{particle}} = -2 \times (-0.5) = +1\,\text{m/s}\). It is on the front edge of the pulse, and it is rising, as the last section predicted.
For the pulse \(y = \dfrac{1}{1 + (x - 2t)^2}\), where is the peak at \(t = 3\,\text{s}\), in metres?
Which of these functions are valid wave functions? (a) \(y(x, t) = e^{-(x - 2t)^2}\). (b) \(y(x, t) = (x - 2t)^2\). (c) \(y(x, t) = \sin(x - 2t)\).
(a) As \(x \to \pm\infty\), the exponential makes \(y \to 0\), so the function stays finite. It has the form \(f(x - vt)\) and satisfies the wave equation. So \(y = e^{-(x - 2t)^2}\) is a valid wave function.
(b) As \(x \to \pm\infty\), the function \((x - 2t)^2\) does not approach \(0\); it grows without limit. It does not stay finite. So \(y = (x - 2t)^2\) is not a valid wave function.
(c) As \(x \to \pm\infty\), the function \(\sin(x - 2t)\) keeps oscillating between \(-1\) and \(1\), so it stays finite. So \(y = \sin(x - 2t)\) is also a valid wave function.
Notice what makes (b) fail. The parabola does satisfy the wave equation, since \(\partial^2 y/\partial x^2 = 2\) and \(\partial^2 y/\partial t^2 = 8 = 2^2 \times 2\). It even has the form \(f(x - 2t)\). It fails only the last condition. A string cannot be displaced by an infinite amount, so a function that grows without limit cannot describe a real wave.
The function \(y = (x - 2t)^2\) is not a valid wave function. Which condition does it fail?
A classmate says that \(y = \sin(x - 2t)\) cannot be a wave, because it never goes to zero far away as a pulse does. Is the classmate right? Explain.
No. The conditions ask for a function that stays finite for all \(x\) and \(t\), not one that goes to zero. The sine keeps oscillating between \(-1\) and \(1\), so it is finite everywhere. It has the form \(f(x - vt)\) with \(v = 2\,\text{m/s}\) and satisfies the wave equation. It is a valid wave that fills the whole string instead of a single pulse.
- Dropping the minus sign in \(v_{\text{particle}} = -v \times \text{slope}\).For a wave moving towards \(+x\), a negative slope gives an upward velocity. Test it on the front edge of a crest.
- Confusing the particle velocity with the wave velocity.The wave velocity is along the string and the same everywhere. The particle velocity is across the string and changes from point to point.
- Accepting any function of \(x - vt\) as a wave.Check the last condition too: the function must stay finite for all \(x\) and \(t\).