understand how wavelength may be determined from the positions of nodes or antinodes of a stationary wave

Stationary Waves

A stationary (or standing) wave is formed by the superposition of two travelling waves of the same frequency and amplitude moving in opposite directions. The resulting wave pattern does not appear to travel; instead it has fixed points of zero displacement (nodes) and points of maximum displacement (antinodes).

Key Features

• Node (N): Point where the displacement is always zero.
• Antinode (A): Point where the displacement reaches a maximum amplitude.
• The distance between two successive nodes (or two successive antinodes) is half a wavelength:

  \$d{NN}=d{AA}=\frac{\lambda}{2}.\$

• The distance between a node and the adjacent antinode is a quarter of a wavelength:

  \$d_{NA}=\frac{\lambda}{4}.\$

Why Determine Wavelength from Nodes/Antinodes?

In many A‑Level experiments (e.g., a vibrating string or an air column), it is easier to measure the positions of nodes and antinodes than to directly count wavelengths. By using the geometric relationships above, the wavelength \$\lambda\$ can be calculated from these measured distances.

Determining Wavelength from Measured Positions

General Procedure

1. Identify a series of consecutive nodes or antinodes along the medium.
2. Measure the distances between successive nodes (or antinodes). Denote this distance as \$d\$.
3. Use the relation \$d = \lambda/2\$ to obtain the wavelength:

   \$\lambda = 2d.\$

4. If only node–antinode distances are available, use \$d = \lambda/4\$:

   \$\lambda = 4d.\$

5. For increased accuracy, measure several intervals and take the average value of \$d\$ before calculating \$\lambda\$.

Worked Example – Vibrating String Fixed at Both Ends

Consider a string of length \$L = 1.20\ \text{m}\$ fixed at both ends. The first three nodes (including the ends) are observed at positions:

Distances between successive nodes:

• \$d_{12}=0.30\ \text{m}\$
• \$d_{23}=0.30\ \text{m}\$
• \$d_{34}=0.60\ \text{m}\$ (covers two half‑wavelengths)

Using \$d = \lambda/2\$, the wavelength from the first interval is

\$\lambda = 2 \times 0.30\ \text{m} = 0.60\ \text{m}.\$

The larger interval \$d_{34}=0.60\ \text{m}\$ corresponds to two half‑wavelengths, confirming the same result:

\$\lambda = \frac{2d_{34}}{2}=0.60\ \text{m}.\$

Worked Example – Open–Closed Air Column

An open–closed cylindrical tube of length \$L = 0.85\ \text{m}\$ resonates at its fundamental frequency. The pressure node is at the open end, and a pressure antinode is at the closed end. The distance from the open end to the first pressure node (which coincides with a displacement antinode) is measured as \$d = 0.2125\ \text{m}\$.

For an open–closed tube, the distance from the open end to the first node is \$\lambda/4\$, so

\$\lambda = 4d = 4 \times 0.2125\ \text{m} = 0.85\ \text{m},\$

which matches the physical length of the tube for the fundamental mode.

Summary Table

Practice Questions

1. A string fixed at both ends vibrates in its second harmonic. The distance between the first and third nodes is measured as \$0.45\ \text{m}\$. Determine the wavelength of the wave on the string.

2. In an open–open pipe of length \$1.20\ \text{m}\$, the distance between two consecutive antinodes is measured as \$0.30\ \text{m}\$. Find the wavelength of the standing wave.

3. A closed–closed tube resonates at a frequency where the distance from the left end to the first node is \$0.18\ \text{m}\$ and the distance from that node to the next antinode is \$0.18\ \text{m}\$. Calculate the wavelength.

Suggested Diagram