C4-1. Fundamentals of Standing Waves

1. Progressive vs Standing Waves

There are fundamental differences between travelling waves and those that are stationary:

Progressive (Travelling) Wave

  • All points have the same amplitude (in turn).
  • Transfers energy along the wave.
  • The wave speed is the speed at which the wave moves through the medium.
  • Does not have nodes or antinodes.
  • Points exactly a wavelength apart are in phase. The phase of points within one wavelength can be between $0$ to $360^\circ$.
motion

Standing (Stationary) Wave

  • Produced by two waves as they travel in opposite directions along the same line with the same frequency.
  • The waves must have the same wavelength, a similar amplitude, and the same speed. For standing waves to form, the half-cycles of the wave profile must be symmetrical (i.e., the same but inverted).
  • Each point has a different amplitude depending on the amount of superposition.
  • Does not transfer net energy; instead, energy is stored within the loops.
  • Each point on the wave oscillates at a different speed. The overall wave doesn't move.
  • Has fixed nodes and oscillating antinodes.

2. Nodes, Antinodes, and Phase on Standing Waves

  • Node: amplitude $= 0$. These are locations of zero amplitude resulting from destructive interference where the crest of one wave meets the trough of another, causing the two waves to cancel each other out. Nodes are fixed and do not move.
  • Antinode: maximum amplitude. These are locations of maximum amplitude resulting from constructive interference where the crest of one wave meets the crest of another (or trough meets trough).
  • Node-to-node spacing $= \dfrac{\lambda}{2}$.
  • Node-to-adjacent-antinode spacing $= \dfrac{\lambda}{4}$.
  • All points in one loop are in phase.
  • Points separated by an even number of nodes are in phase.
  • Points separated by an odd number of nodes are in antiphase.
  • The phase difference between any two points on a standing wave can only be in-phase ($0$ phase difference) or anti-phase ($\pi$ out of phase).
L equilibrium N N N N A A A λ/2 λ/4 λ

3. Derivation of Standing Waves

A standing wave is formed by the superposition of two progressive waves travelling in opposite directions. Let the two waves be:

$$y_1 = A \sin(kx - \omega t)$$ $$y_2 = A \sin(kx + \omega t)$$

By the principle of superposition, the resultant displacement $y$ is:

$$y = y_1 + y_2 = A \left[ \sin(kx - \omega t) + \sin(kx + \omega t) \right]$$

Using the trigonometric identity $\sin(\alpha) + \sin(\beta) = 2 \sin\left(\dfrac{\alpha+\beta}{2}\right) \cos\left(\dfrac{\alpha-\beta}{2}\right)$, we obtain:

$$y = \left[ 2A \sin(kx) \right] \cos(\omega t)$$
  • The term $\left[ 2A \sin(kx) \right]$ represents the amplitude envelope, which is dependent strictly on position $x$. It proves that amplitude varies structurally from $0$ (nodes) to $2A$ (antinodes).
  • The term $\cos(\omega t)$ confirms that every particle between nodes oscillates in Simple Harmonic Motion with exactly the same phase and temporal frequency $\omega$.

Example 1

Problem: A standing wave is set up on a string by the superposition of two progressive waves given by the equations:

$$y_1 = 4.0 \sin(5\pi x - 120\pi t)$$ $$y_2 = 4.0 \sin(5\pi x + 120\pi t)$$

where $x$ and $y$ are in millimeters ($\text{mm}$) and $t$ is in seconds ($\text{s}$). Calculate:
(a) The distance between adjacent nodes in $\text{mm}$.
(b) The maximum velocity of a particle located at an antinode.

Solution:

(a) Distance between adjacent nodes
From the wave equation format $y = A \sin(kx \pm \omega t)$, we identify the wave number $k = 5\pi \text{ mm}^{-1}$.
We know that $k = \dfrac{2\pi}{\lambda}$. Rearranging for wavelength:

$$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{5\pi} = 0.40 \text{ mm}$$

The distance between adjacent nodes is exactly $\dfrac{\lambda}{2}$:

$$\text{Node Spacing} = \dfrac{0.40}{2} = \mathbf{0.20 \text{ mm}}$$

(b) Maximum velocity at an antinode
From the superposition derivation, the maximum amplitude of the standing wave at an antinode is $A_{max} = 2A = 2(4.0) = 8.0 \text{ mm}$.
The angular frequency is $\omega = 120\pi \text{ rad s}^{-1}$.
Because the particles undergo simple harmonic motion, the maximum velocity of a particle is given by $v_{max} = \omega \times \text{Amplitude}$:

$$v_{max} = (120\pi) \times (8.0) = 960\pi \text{ mm s}^{-1} \approx \mathbf{3016 \text{ mm s}^{-1} \text{ (or } 3.0 \text{ m s}^{-1})}$$