Tie one end of a long rope to a pole and shake the other end steadily at the right rate. The waves you send reflect from the pole and come back. The rope stops looking like a travelling wave. It swings in loops, and a few points between the loops hardly move at all. Waves pass through those points in both directions all the time, yet the points stay still.
A product does not travel
A travelling wave such as \(A\sin(kx - \omega t)\) depends on \(x\) and \(t\) only through the one combination \(kx - \omega t\). If \(t\) grows, the same value of \(kx - \omega t\) is found at a larger \(x\). So the whole shape slides along the string. A product \(f(x)\,g(t)\) behaves differently. At every instant its shape is the same curve \(f(x)\), only multiplied by the number \(g(t)\). The shape swells and shrinks in place, and nothing slides.
\[ y = A\sin(kx - \omega t) + A\sin(kx + \omega t) \]
Why. the principle of superposition: the two displacements add at every point and every instant.
\[ \sin P + \sin Q = 2\sin\dfrac{P + Q}{2}\cos\dfrac{P - Q}{2} \]
Why. the sum-to-product identity for two sines.
\[ \dfrac{P + Q}{2} = kx, \qquad \dfrac{P - Q}{2} = -\omega t \]
Why. take \(P = kx - \omega t\) and \(Q = kx + \omega t\), and find half their sum and half their difference.
\[ y = [2A\sin kx]\cos\omega t \]
Read it. put step 3 into step 2 and multiply by \(A\). The cosine is even, so \(\cos(-\omega t) = \cos\omega t\).
A standing wave is the sum of two identical waves travelling in opposite directions. It has the form \(y = [2A\sin kx]\cos\omega t\): a fixed shape whose size oscillates in time. The pattern does not travel along the medium.
Each factor has a job. The factor \(2A\sin kx\) contains no time, so it is a fixed number for each point: it is the amplitude of the point at \(x\). The factor \(\cos\omega t\) contains no position, so it is the same for every point. Every point of the string therefore oscillates with the same \(\omega\), each within its own amplitude.
Which of these describes a standing wave? Here \(y\) is in cm, \(x\) in m and \(t\) in s.
Nodes and antinodes
The amplitude \(|2A\sin kx|\) is zero wherever \(\sin kx = 0\), that is where \(kx = 0, \pi, 2\pi, \ldots\). With \(k = 2\pi/\lambda\), these points are \(x = 0, \lambda/2, \lambda, \ldots\), and they are the nodes. The amplitude is largest, \(2A\), where \(|\sin kx| = 1\), at \(x = \lambda/4, 3\lambda/4, \ldots\). These points are the antinodes. Neighbouring nodes are \(\lambda/2\) apart, and each antinode lies midway between two nodes.
The sign of \(\sin kx\) also matters. Within one loop, between two neighbouring nodes, \(\sin kx\) has one sign, so every point of the loop moves up and down together. In the next loop the sign is reversed. Points in neighbouring loops therefore move in opposite phase: when one loop is up, the next is down.
A standing wave on a string is \(y = 0.5\sin(2\pi x)\cos(40\pi t)\), with \(x\) in m. How far apart are two neighbouring nodes, in m?
In a standing wave, point P is at the middle of one loop and point Q is at the middle of the next loop. How do their motions compare?
The same answer from two phasors
Stand at one point \(x\) and watch only time. There, the first wave is \(A\sin(kx - \omega t)\), which equals \(A\sin(\omega t + \pi - kx)\), because \(\sin\theta = \sin(\pi - \theta)\). The second wave is \(A\sin(\omega t + kx)\). So at this point the two waves are two sinusoids of the same \(\omega\), with phases \(\pi - kx\) and \(kx\). Their phase difference, \(2kx - \pi\), depends on where you stand.
Add the two phasors by the rule for one frequency. The amplitude is given by \(A_R^2 = 2A^2[1 + \cos(2kx - \pi)] = 2A^2(1 - \cos 2kx) = 4A^2\sin^2 kx\). So \(A_R = 2A|\sin kx|\), the same first factor as the product. At a node the two waves always arrive in opposite phase and cancel. At an antinode they always arrive in phase and add to \(2A\).
The product form needs two waves of exactly equal amplitude. A real reflection loses a little energy, so the returning wave is weaker. Then the points of least motion still swing with amplitude \(A_1 - A_2\), and part of the wave keeps travelling. The positions of the nodes also depend on the written form. Two cosines, \(A\cos(kx - \omega t) + A\cos(kx + \omega t)\), give \([2A\cos kx]\cos\omega t\), which has an antinode at \(x = 0\). Apply the identity to the waves as they are given.
This is the numerical check of Standing Waves on a String, lesson 1, done with the tool. Two waves, \(y_1 = 3\sin(\pi x/6 - 50\pi t)\) and \(y_2 = 3\sin(\pi x/6 + 50\pi t)\), travel on one string, with \(y\) in mm, \(x\) in cm and \(t\) in s. Find the amplitude of the particle at \(x = 1\,\text{cm}\), its frequency, and the positions of the nodes.
Equal amplitudes, \(A = 3\,\text{mm}\), and the same \(k = \pi/6\) per cm and \(\omega = 50\pi\,\text{rad/s}\), travelling opposite ways.
\(y = [2A\sin kx]\cos\omega t = [6\sin(\pi x/6)]\cos(50\pi t)\) mm.
At \(x = 1\,\text{cm}\), the amplitude is \(6\sin(\pi/6) = 6 \times 0.5 = 3\,\text{mm}\). Every point has \(f = \omega/2\pi = 50\pi/2\pi = 25\,\text{Hz}\). The wavelength is \(\lambda = 2\pi/k = 12\,\text{cm}\), so the nodes are at \(x = 0, 6, 12, \ldots\) cm.
The particle at \(1\,\text{cm}\) swings with amplitude \(3\,\text{mm}\), half of the largest, \(6\,\text{mm}\), at the antinodes \(x = 3, 9, \ldots\) cm. All particles oscillate at \(25\,\text{Hz}\).
A standing wave on a string is \(y = 0.02\sin(\pi x)\cos(100\pi t)\), with \(y\) and \(x\) in metres. What is the amplitude of the particle at \(x = 1/6\,\text{m}\), in cm?
At a node of a standing wave, how do the two travelling waves arrive?
A student says the standing wave \(y = [2A\sin kx]\cos\omega t\) travels at the speed \(\omega/k\), like each of its two waves. Explain what is wrong.
A travelling wave depends on \(x\) and \(t\) only through \(kx - \omega t\), so its shape slides along. The standing wave is a product of a function of \(x\) alone and a function of \(t\) alone. Its shape \(\sin kx\) never moves; it only swells, shrinks and turns over. The nodes stay at the same places. Each of the two component waves travels at \(\omega/k\), but their sum does not travel at all.
Standing Waves on a String builds the standing wave from two opposite waves and reads its amplitude and time parts from this product. It uses the same identity for a string with one free end. Standing Sound Waves uses it for the resonance of an air column, where pressure and displacement have their nodes at different places. Every harmonic of a sitar string or an organ pipe is one such product.
- Calling \(2A\sin kx\) the displacement of the point at \(x\).It is the amplitude. The displacement is the whole product, which changes with time.
- Putting the nodes a whole wavelength apart.\(\sin kx\) is zero twice per wavelength. Neighbouring nodes are \(\lambda/2\) apart.
- Using \([2A\sin kx]\) for waves written with cosines.Apply the identity to the given forms. Two cosines give \([2A\cos kx]\cos\omega t\), with an antinode at \(x = 0\).