The features of a standing wave
The pattern of a standing wave does not move, so its features stay at fixed places on the string. This lesson finds each of them from the amplitude function of equation (11), \(A_x = 2A\sin(kx)\). There are three: the points that never move, the points that move the most, and the stretches of string between them.
Nodes
Nodes are points on the string that have zero amplitude. They remain permanently at rest. The condition for a node is that the amplitude function is zero, \(2A\sin(kx) = 0\). The derivation below turns this condition into positions.
\[ 2A\sin(kx) = 0 \]
Why. the condition for a node: the amplitude function is zero.
\[ kx = n\pi, \qquad n = 0, 1, 2, \dots \]
Why. \(A\) is not zero, so \(\sin kx = 0\). The sine is zero only when its angle is a whole multiple of \(\pi\).
\[ \dfrac{2\pi}{\lambda}\,x = n\pi \]
Why. put in the wave number, \(k = 2\pi/\lambda\).
\[ x_{\text{node}} = n\left(\dfrac{\lambda}{2}\right) \]
Read it. solve step 3 for \(x\).
Two neighbouring nodes have values of \(n\) that differ by one. So the distance between two adjacent nodes is \(\lambda/2\), half a wavelength.
A node is a point of a standing wave where the amplitude is zero, so the medium there stays at rest at all times. On a string with \(y = [2A\sin kx]\cos\omega t\), the nodes are at \(x = n(\lambda/2)\), half a wavelength apart.
A standing wave on a string is \(y = 0.04\sin(5\pi x)\cos(100\pi t)\), with \(x\) and \(y\) in metres and \(t\) in seconds. Find the distance between two adjacent nodes, in m.
Antinodes
Antinodes are points that oscillate with the largest possible amplitude, \(2A\). The condition for an antinode is that the amplitude function is as large as it can be, \(|2A\sin(kx)| = 2A\). The derivation below finds where this happens.
\[ |2A\sin(kx)| = 2A \]
Why. the condition for an antinode: the amplitude function has its largest size.
\[ |\sin(kx)| = 1 \]
Why. divide by \(2A\).
\[ kx = \left(n + \dfrac12\right)\pi, \qquad n = 0, 1, 2, \dots \]
Why. the sine is \(+1\) or \(-1\) only when its angle is an odd multiple of \(\pi/2\).
\[ x_{\text{antinode}} = \left(n + \dfrac12\right)\left(\dfrac{\lambda}{2}\right) \]
Read it. put \(k = 2\pi/\lambda\) and solve for \(x\).
Compare the two results. The antinode positions are the node positions moved on by half a node spacing. So every antinode lies midway between two nodes. The distance between a node and the nearest antinode is half of \(\lambda/2\), which is \(\lambda/4\).
In Fig 2.1 the wavelength is \(4\,\text{m}\). The nodes are at \(0\), \(2\), \(4\) and \(6\,\text{m}\), \(2\,\text{m}\) apart. The antinodes are at \(1\), \(3\) and \(5\,\text{m}\). The node at \(4\,\text{m}\) is \(1\,\text{m}\) from the antinode at \(5\,\text{m}\), a quarter of the wavelength.
An antinode is a point of a standing wave where the amplitude is largest. On a string with \(y = [2A\sin kx]\cos\omega t\), the antinodes swing with amplitude \(2A\) at \(x = (n + \tfrac12)(\lambda/2)\), midway between nodes.
In a standing wave with wavelength \(\lambda\), what is the distance between a node and the nearest antinode?
For the standing wave \(y = 0.04\sin(5\pi x)\cos(100\pi t)\), with \(x\) in metres, which positions are antinodes?
Loops and phase relationships
The segment of the string between two consecutive nodes is called a loop. Within one loop, \(\sin kx\) keeps one sign, since it passes through zero only at the nodes. So every particle of the loop has the same sign in its amplitude function, and all share the factor \(\cos\omega t\).
This gives the two phase rules of a standing wave.
- Within a loop. All particles in a single loop oscillate in phase. They pass through the rest line together and reach their maximum displacements at the same instant.
- Between loops. Particles in adjacent loops oscillate \(180°\) out of phase. When one loop moves up, its neighbours move down. Across a node \(\sin kx\) changes sign, and \(-\cos\omega t = \cos(\omega t + \pi)\).
Picture a skipping rope turned by two children, but with a third child at its middle holding it still. The rope forms two loops. Whenever the left loop bulges up, the right loop bulges down, like the two ends of a see-saw. The still point in the middle is the node, and the see-saw is the \(180°\) phase difference between neighbouring loops.
True or false? In a standing wave, all particles between two consecutive nodes reach their maximum displacement at the same instant.
Point P is in one loop of a standing wave, and point Q is in the next loop. What is the phase difference between their oscillations?
Energy in standing waves
In contrast to a travelling wave, a standing wave does not transport energy along its length. The chapter Propagating Waves on a String found how a wave carries energy. The part of the string on one side of a point pulls on the part on the other side and does work on it. That work needs the point to move.
The nodes are points of zero displacement at every instant. They act as barriers to energy. No work can be done across a point that does not move, so no energy crosses a node. The energy of a standing wave is therefore confined within the loops.
Inside each loop, energy is conserved. It changes continuously between two forms.
- Kinetic energy is largest when the string segment passes through its rest position, \(y = 0\), at its largest speed.
- Potential energy is largest when the string segment reaches its largest displacement and stops for an instant. Then the string is most stretched.
Press play in Fig 2.2. At \(t = 0\) the string has its extreme shape and no velocity arrows, and the \(U\) bar is full. A quarter of a period later the string lies flat, the arrows are longest and the \(K\) bar is full. In between, the readout shows the two shares trading, while the two bars always add to the dashed line. The arrows in the two loops point opposite ways, as the phase rules say.
The energy of one loop can be found. At the flat instant each particle has speed \(2A\omega|\sin kx|\), and a piece \(dx\) has mass \(\mu\,dx\). So the loop's kinetic energy is \(\int_0^{\lambda/2} \tfrac12\mu(2A\omega\sin kx)^2\,dx = \tfrac12\mu\omega^2A^2\lambda\). Each travelling wave carries \(\tfrac12\mu\omega^2A^2\) per unit length on average, from the chapter Propagating Waves on a String. Two waves over half a wavelength carry \(2 \times \tfrac12\mu\omega^2A^2 \times \lambda/2\), the same amount. The energy of both waves sits in the loops.
Which statements about the energy of a standing wave on a string are true?
A microwave oven fills its box with electromagnetic waves of frequency \(2.45\,\text{GHz}\). They reflect from the metal walls and form standing waves. The wavelength is \(\lambda = c/f = (3 \times 10^8)/(2.45 \times 10^9) \approx 12.2\,\text{cm}\). So hot spots, at the antinodes, lie about \(\lambda/2 \approx 6.1\,\text{cm}\) apart, with cold spots between them. This is why the plate turns: it carries each part of the food through the hot spots. Stop the plate, heat a slab of chocolate, and the melted spots come out about \(6\,\text{cm}\) apart.
- Taking the distance between adjacent nodes as \(\lambda\).Nodes are at \(x = n(\lambda/2)\). Adjacent nodes are \(\lambda/2\) apart, and one wavelength holds two loops.
- Saying that all particles of a standing wave move in phase.Particles in one loop move in phase. Particles in adjacent loops are \(180°\) out of phase.
- Thinking that a standing wave carries energy like a travelling wave.No work crosses a node, so the energy of each loop stays in that loop. It only changes form, kinetic to potential and back.