explain and use the principle of superposition

Stationary Waves

What is a Stationary Wave?

Imagine a guitar string that is plucked and then left to vibrate. The string doesn’t move forward or backward as a whole; instead, it oscillates in place. The points that never move are called nodes, and the points that move the most are called antinodes. This pattern of fixed and moving points is a stationary wave (or standing wave).

🎸 The key to a stationary wave is the principle of superposition: two waves travelling in opposite directions add together to create a pattern that stays in one place.

Principle of Superposition

When two waves of the same amplitude and frequency travel in opposite directions, their displacements add algebraically at every point.

Let the two waves be:

$y_1(x,t)=A\sin(kx-\omega t)$

$y_2(x,t)=A\sin(kx+\omega t)$

Adding them:

$y(x,t)=y_1+y_2=2A\sin(kx)\cos(\omega t)$

Notice the separation: $\sin(kx)$ depends only on position (nodes & antinodes), while $\cos(\omega t)$ depends only on time (oscillation).

Nodes and Antinodes

  • Node: Point where $y(x,t)=0$ for all $t$ → $\sin(kx)=0$ → $kx = n\pi$ → $x = n\frac{\lambda}{2}$
  • Antinode: Point where displacement is maximum → $\sin(kx)=\pm1$ → $kx = (n+\tfrac12)\pi$ → $x = (n+\tfrac12)\frac{\lambda}{2}$

For a string fixed at both ends, the ends are nodes. The number of nodes (excluding the ends) determines the mode number $n$.

Modes of a Fixed String

Mode $n$ Wavelength $\lambda_n$ Frequency $f_n$
1 $2L$ $f_1 = \dfrac{v}{2L}$
2 $L$ $f_2 = \dfrac{v}{L}$
3 $\dfrac{2L}{3}$ $f_3 = \dfrac{3v}{2L}$

Here, $v$ is the wave speed on the string, and $L$ is the length of the string.

Exam Tip Box

Remember: For a string fixed at both ends, the fundamental frequency (mode 1) has a wavelength equal to twice the length of the string. Higher modes have wavelengths that are integer fractions of this value.

When you see a diagram of a string with nodes, count the number of segments between nodes to identify the mode number $n$.

🧪 Practice problem: A 1.0 m long guitar string vibrates at 110 Hz. What is the wave speed $v$?

Solution: $f_1 = \dfrac{v}{2L} \Rightarrow v = 2Lf_1 = 2 \times 1.0 \times 110 = 220\ \text{m/s}$.

Real‑World Analogy

Think of a rope tied at both ends and you swing it up and down. If you swing it at the right frequency, you’ll see a pattern of stationary waves—just like the string on a guitar. The places where the rope doesn’t move at all are the nodes, and the places that swing the most are the antinodes.

🪢 The same principle works for sound waves in a closed tube, light waves in a cavity, and even quantum wavefunctions!

Quick Review Checklist

  1. State the principle of superposition.
  2. Write the equations for two counter‑propagating waves.
  3. Show how they combine to give $y(x,t)=2A\sin(kx)\cos(\omega t)$.
  4. Identify nodes and antinodes for a fixed string.
  5. Use the formula $f_n = \dfrac{nv}{2L}$ to calculate frequencies.

?? Tip: Practice drawing node/antinode diagrams for modes 1–4.

Revision

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