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

Stationary Waves

Imagine a guitar string that is plucked and then left to vibrate. The string doesn’t move as a whole; instead, it forms a pattern of fixed points called nodes (where the string never moves) and points that swing the most called antinodes. This pattern is a stationary wave – the shape stays in place while the energy moves along the string.

Key Concepts

• Node: point of zero amplitude, never moves.
• Antinode: point of maximum amplitude, swings the most.
• Wavelength (\$\lambda\$): distance between two consecutive nodes or antinodes.
• Standing wave equation: \$y(x,t)=A\sin(kx)\cos(\omega t)\$, where \$k=2\pi/\lambda\$.

Finding the Wavelength from Node Positions

If you can mark the positions of two nodes along a string of length \$L\$, the distance between them is half a wavelength. Therefore:

1. Measure the distance between two adjacent nodes, call it \$d\$.
2. Wavelength is \$ \lambda = 2d \$.

Example: If the distance between two nodes is 12 cm, then \$ \lambda = 2 \times 12\,\text{cm} = 24\,\text{cm} \$. 🎵

Finding the Wavelength from Antinode Positions

The distance between two adjacent antinodes is also half a wavelength.

1. Measure the distance between two consecutive antinodes, call it \$d'\$.
2. Wavelength is \$ \lambda = 2d' \$.

Example: Two antinodes 15 cm apart → \$ \lambda = 30\,\text{cm} \$. 🧪

Using the Length of the String

For a string fixed at both ends, the allowed standing waves satisfy:

\$\$

L = n \frac{\lambda}{2}\quad \text{where } n = 1,2,3,\dots

\$\$

Rearranging gives the wavelength:

\$\$

\lambda = \frac{2L}{n}

\$\$

So if a string of length 1 m has its 3rd harmonic (\$n=3\$) excited, the wavelength is \$ \lambda = \frac{2 \times 1\,\text{m}}{3} \approx 0.667\,\text{m} \$. 📐

Example Problem

On a 1.2 m long string fixed at both ends, the 4th harmonic is produced. The distance between the first and second nodes is measured to be 15 cm.

1. Using the node spacing: \$d = 15\,\text{cm} = 0.15\,\text{m}\$ → \$ \lambda = 2d = 0.30\,\text{m}\$.
2. Check with the harmonic formula: \$ \lambda = \frac{2L}{n} = \frac{2 \times 1.2}{4} = 0.60\,\text{m}\$.
3. Since the two results differ, the measured node spacing must be incorrect or the harmonic number misidentified.

Lesson: Always cross‑check with the string length and harmonic number. 🔍

Exam Tips Box

Tip 1: Remember that one wavelength = two node–node or two antinode–antinode distances.

Tip 2: For a string fixed at both ends, the formula \$L = n\lambda/2\$ is your go‑to.

Tip 3: If you’re given the number of nodes or antinodes, count them to find \$n\$.

Tip 4: Always check units – centimeters vs. meters can trip you up!

Tip 5: Use a diagram: draw the string, label nodes and antinodes, and mark measured distances. It makes the maths clearer.

Good luck, and keep vibrating with confidence! 🚀

Quick Reference Table