Harmonic waves
A harmonic wave is a special kind of wave, described by a sinusoidal function, a sine or a cosine. Harmonic waves are the most fundamental kind of wave. They are the building blocks from which every other wave can be made.
Think of a piano. It has a fixed set of notes, and with them it can play a great many songs. Or think of any vector in space. It can be written as a combination of three basis vectors, \(\hat{i}\), \(\hat{j}\) and \(\hat{k}\). In the same way, any wave can be written as a combination of harmonic waves. The detailed method is the Fourier series. By it, any periodic wave can be written as a sum of harmonic waves.
So harmonic waves are the most fundamental waves, and the building blocks of all the others. This chapter restricts its analysis to harmonic waves only. Let us build the equation of one such wave.
A harmonic wave is a travelling wave whose displacement is a sine function of position and time. Every particle of the medium then performs simple harmonic motion, and any wave can be written as a combination of harmonic waves.
In 1807 Joseph Fourier sent a long paper on the flow of heat to the Academy of Sciences in Paris. In it he claimed that any function, even one with corners, can be written as a sum of sines and cosines. Lagrange and Laplace, who judged the paper, doubted the claim, and it was published in full only in 1822. The method is now used everywhere: in music software, in the compression of photographs, and in the analysis of every wave in this book.
A snapshot at \(t = 0\)
Imagine taking a snapshot of a harmonic wave at \(t = 0\). Fig 5.1 shows the shape. We now find the function that describes it.
From the graph, the curve is a sinusoid with a maximum value of \(A\) and a minimum value of \(-A\). At \(x = \lambda/4\) the wave has its maximum value, \(A\). At \(x = 3\lambda/4\) it has its minimum value, \(-A\). A highest point is called a crest, and a lowest point a trough. The derivation below finds the argument of the sine from these facts.
\[ y(x, 0) = A\sin(bx) \]
Why. the curve is a sinusoid through the origin, rising at first, with maximum \(A\) and minimum \(-A\). The constant \(b\) is still unknown.
\[ b \cdot \dfrac{\lambda}{4} = \dfrac{\pi}{2} \]
Why. the maximum \(A\) is at \(x = \lambda/4\), and \(\sin\) first reaches \(1\) when its argument is \(\pi/2\).
\[ b = \dfrac{2\pi}{\lambda} \]
Why. solve step 2 for \(b\).
\[ y(x, 0) = A\sin\left(\dfrac{2\pi}{\lambda}x\right) \]
Read it. put \(b\) into step 1. Check: at \(x = 3\lambda/4\) the argument is \(3\pi/2\), and \(y = -A\), as the graph shows.
Here \(A\) is the amplitude, the maximum displacement from equilibrium. The factor \(2\pi/\lambda\) is the wave number. It is tied to the spatial period of the wave, which is called the wavelength \(\lambda\). When \(x\) grows by \(\lambda\), the argument grows by \(2\pi\), and the shape repeats.
True or false? If we take a photograph of a sinusoidal wave, freezing time, the distance between two consecutive crests gives us the time period of the wave.
Setting the snapshot in motion
Now let this wave travel in the positive \(x\) direction with speed \(v\). By Lesson 3, we obtain the travelling wave by replacing \(x\) with \(x - vt\) in the snapshot. So the equation of the wave is
\[ y(x, t) = A\sin\left(\frac{2\pi}{\lambda}(x - vt)\right) = A\sin\left(\frac{2\pi}{\lambda}x - \frac{2\pi}{\lambda}vt\right) \]
This is equation (3) of the chapter. It is a function of the form \(f(x - vt)\), so it is a valid wave function. It is finite everywhere too, since a sine never leaves the range \(-1\) to \(1\). The function holds all the information about the displacement of every particle of the medium, at every position \(x\) and every time \(t\).
One particle of the wave
To see how one particle moves, put a fixed value of \(x\) into equation (3). For simplicity, take the particle at \(x = 0\). Then only the time is left in the function.
\[ y(0, t) = A\sin\left(-\dfrac{2\pi}{\lambda}vt\right) \]
Why. put \(x = 0\) in equation (3). The displacement of this particle is a function of time alone.
\[ y(0, t) = A\sin(\omega t + \pi), \qquad \omega = \dfrac{2\pi v}{\lambda} \]
Why. \(\sin(-\theta) = \sin(\theta + \pi)\). This is the form \(x = A\sin(\omega t + \varphi)\) of the chapter Kinematics of SHM, with initial phase \(\pi\). So the particle performs simple harmonic motion with angular frequency \(2\pi v/\lambda\).
?
Your turn. the period of an SHM is \(2\pi\) divided by its angular frequency. Which line is it?
\[ \lambda = vT \]
Read it. multiply step 3 by \(v\).
So the motion of one particle in a harmonic wave is simple harmonic motion. Its angular frequency is \(2\pi v/\lambda\), and its time period is \(T = \lambda/v\). In other words, \(\lambda = vT\). The wave travels one wavelength in the time that a particle of the medium takes to complete one period.
In Fig 5.2 the wavelength is \(4\,\text{m}\) and the period is \(2\,\text{s}\), so \(v = \lambda/T = 2\,\text{m/s}\). Watch \(P\) start downward, as \(y(0, t) = A\sin(-\omega t)\) says. At \(t = 2\,\text{s}\) the readout shows that the wave has moved \(4.00\,\text{m}\), exactly one wavelength. At that same instant \(P\) has completed exactly one oscillation and is back at \(y = 0\).
The equation of a harmonic wave
Substitute the time period \(T\) of the particle's SHM into equation (3). Since \(v/\lambda = 1/T\), the term \(2\pi vt/\lambda\) becomes \(2\pi t/T\). This gives equation (4).
\[ y(x, t) = A\sin\left(\frac{2\pi}{\lambda}x - \frac{2\pi}{T}t\right) \]
This is the general equation of a harmonic wave that travels in the positive \(x\) direction. The factor \(2\pi/T\) is the angular frequency \(\omega\) of the wave. The factor \(2\pi/\lambda\) is the wave number \(k\). In terms of \(k\) and \(\omega\), equation (4) becomes equation (5).
\[ y(x, t) = A\sin(kx - \omega t) \]
The table below lists the terms used in the equation of a harmonic wave. The phase is written \(\Phi\), as in the chapter Kinematics of SHM, where \(\varphi\) is kept for an initial phase.
Read Fig 5.3 with care. The left graph is a photograph: it has a ruler along its axis, and its crests are \(\lambda = 4\,\text{m}\) apart. The right graph is a record of one particle: it has a clock along its axis, and its crests are \(T = 2\,\text{s}\) apart. The right graph is \(y(0, t) = -A\sin(2\pi t/T)\), so its first crest comes at \(3T/4 = 1.5\,\text{s}\). A wavelength is found in space, and a period in time.
A harmonic wave travels in the positive \(x\) direction. Its equation is \(y(x, t) = 0.05\sin(4\pi x - 10\pi t)\), with \(x\) and \(y\) in metres and \(t\) in seconds. Find (a) the wavelength, (b) the time period and (c) the frequency. Find (d) the speed of the wave, (e) the angular frequency and (f) the wave number. (g) Find the phase difference between two points \(10\,\text{cm}\) apart along \(+x\) at a fixed time. (h) Find the change in phase of a particle at a fixed position \(x\) as \(t\) changes from \(0\) to \(0.1\,\text{s}\).
Compare the equation with the general form \(y(x, t) = A\sin(kx - \omega t)\). Then \(A = 0.05\,\text{m}\), \(k = 4\pi\,\text{rad/m}\) and \(\omega = 10\pi\,\text{rad/s}\).
Use \(\lambda = 2\pi/k\), \(T = 2\pi/\omega\), \(f = 1/T\), \(v = \lambda f\), and for the phase changes \(\Delta\Phi = k\,\Delta x\) and \(\Delta\Phi = \omega\,\Delta t\).
(a) \(\lambda = 2\pi/4\pi = 0.5\,\text{m}\). (b) \(T = 2\pi/10\pi = 0.2\,\text{s}\). (c) \(f = 1/0.2 = 5\,\text{Hz}\). (d) \(v = 0.5 \times 5 = 2.5\,\text{m/s}\). (e) \(\omega = 10\pi\,\text{rad/s}\), read from the equation. (f) \(k = 4\pi\,\text{rad/m}\), read from the equation. (g) \(\Delta\Phi = 4\pi \times 0.1 = 0.4\pi\,\text{rad}\). (h) \(\Delta\Phi = 10\pi \times 0.1 = \pi\,\text{rad}\).
Check the speed another way: \(v = \omega/k = 10\pi/4\pi = 2.5\,\text{m/s}\), the same. The time in (h), \(0.1\,\text{s}\), is half a period, so the phase changes by half of \(2\pi\).
A wave is described by \(y(x, t) = 0.02\sin(5\pi x - 20\pi t)\), where \(x\) and \(y\) are in metres and \(t\) is in seconds. What is the wavelength?
For the same wave, \(y(x, t) = 0.02\sin(5\pi x - 20\pi t)\), find the speed of the wave, in m/s.
In the same wave, \(y = 0.02\sin(5\pi x - 20\pi t)\), what is the time period of the particle at \(x = 3\,\text{m}\)?
An FM station in India broadcasts on \(93.5\,\text{MHz}\). Its signal is an electromagnetic wave, not a mechanical one, but it is described by the same equation, \(y = A\sin(kx - \omega t)\), with \(v = c = 3 \times 10^8\,\text{m/s}\). Its wavelength is \(\lambda = c/f = 3 \times 10^8 / 93.5 \times 10^6 \approx 3.2\,\text{m}\). So \(k = 2\pi/\lambda \approx 1.96\,\text{rad/m}\) and \(\omega = 2\pi f \approx 5.9 \times 10^8\,\text{rad/s}\). A car aerial about \(0.8\,\text{m}\) long is a quarter of this wavelength, a length that suits the signal well.
- Using the coefficient of \(x\) as the wavelength.The coefficient of \(x\) is the wave number \(k\). The wavelength is \(2\pi/k\).
- Reading a period from a snapshot, or a wavelength from the record of one particle.A snapshot has position on its axis and shows \(\lambda\). The record of one particle has time on its axis and shows \(T\).
- Thinking that particles farther from the source oscillate with a longer period.Every particle of a harmonic wave has the same period \(T = 2\pi/\omega\). Only their phases differ.