explain the formation of a stationary wave using a graphical method, and identify nodes and antinodes

Stationary Waves

Objective

Explain the formation of a stationary (standing) wave using a graphical method and identify the positions of nodes and antinodes.

1. Introduction

A stationary wave results from the superposition of two waves of the same frequency, amplitude and wavelength travelling in opposite directions. The resultant displacement at any point does not travel; instead it oscillates in place, producing fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes).

2. Graphical Method of Formation

The graphical method visualises the addition of two travelling waves at successive instants. The steps are:

1. Draw the first travelling wave, e.g. \$y_1 = A\sin(kx-\omega t)\$, at a chosen instant \$t=0\$.
2. Draw the second wave travelling in the opposite direction, \$y_2 = A\sin(kx+\omega t)\$, at the same instant.
3. At each position \$x\$, add the vertical displacements of the two waves to obtain the resultant \$y = y1 + y2\$.
4. Repeat the process for later times (e.g. \$t = \frac{T}{4}, \frac{T}{2}\$) to see how the resultant pattern evolves.

When the two waves are added, the resulting expression can be written using the trigonometric identity:

\$y = 2A\cos(\omega t)\sin(kx)\$

Here the spatial part \$\sin(kx)\$ determines the fixed pattern of nodes and antinodes, while the temporal factor \$\cos(\omega t)\$ makes the whole pattern oscillate in time.

3. Identification of Nodes and Antinodes

From the expression \$y = 2A\cos(\omega t)\sin(kx)\$:

• Nodes occur where the spatial factor \$\sin(kx)=0\$. Thus,

  \$\sin(kx)=0 \;\Rightarrow\; kx = n\pi \;\;(n = 0,1,2,\dots)\$

  \$\Rightarrow\; x_n = \frac{n\pi}{k} = n\frac{\lambda}{2}\$

  Nodes are spaced half a wavelength apart.

• Antinodes occur where \$|\sin(kx)|\$ is maximum, i.e. \$\sin(kx)=\pm1\$:

  \$kx = \left(n+\tfrac12\right)\pi\$

  \$\Rightarrow\; x_{a} = \left(n+\tfrac12\right)\frac{\lambda}{2}\$

  Antinodes are also spaced half a wavelength apart, but are offset by \$\lambda/4\$ from the nodes.

4. Example: String Fixed at Both Ends

Consider a string of length \$L\$ fixed at \$x=0\$ and \$x=L\$. The boundary condition requires nodes at both ends, so the allowed wavelengths satisfy:

\$L = n\frac{\lambda}{2}\;\;(n=1,2,3,\dots)\$

The corresponding frequencies are:

\$f_n = \frac{n v}{2L}\$

where \$v\$ is the wave speed on the string.

5. Summary Table

6. Key Points to Remember

• A stationary wave is the result of two identical waves travelling in opposite directions.
• The graphical method helps visualise how the superposition creates fixed patterns.
• Nodes occur at positions where the two waves are always out of phase by \$180^\circ\$.
• Antinodes occur where the two waves are always in phase.
• For a string fixed at both ends, only wavelengths that fit an integer number of half‑waves are allowed.