Module C.4 IBDP HL only Track

Standing Waves and Resonance

Examine boundary condition dynamics, string harmonics, open/closed pipe air columns, and resonance nodes.

01. Observe 02. Think 03. Apply 04. Achieve Excellence

Standing Waves and Resonance

Introduction

Standing waves form when identical waves travelling in opposite directions superpose. The guide connects this to musical strings, air columns, resonance, and damping, making boundary conditions the key to predicting allowed wavelengths and frequencies.

Guide Focus

  • Describe standing waves using nodes, antinodes, amplitude, and phase.
  • Apply boundary conditions for strings and open/closed pipes.
  • Explain resonance, natural frequency, and damping qualitatively.

Key Concepts

1. Formation of standing waves

A standing wave forms by superposition of two identical waves travelling in opposite directions. Nodes have zero displacement, antinodes have maximum displacement, and neighbouring points may differ in phase depending on their positions.

2. Strings and pipes

Allowed modes depend on boundary conditions. Fixed ends are displacement nodes; free ends are displacement antinodes. In pipes, open ends are displacement antinodes and closed ends are displacement nodes.

3. Harmonics

The lowest allowed frequency is the first harmonic. Once wavelength is found from the standing-wave pattern and length, frequency follows from f = v / lambda.

4. Resonance and damping

Resonance occurs when a system is driven near its natural frequency, producing large amplitude. Light damping gives a high, sharp resonance peak; heavier damping lowers and broadens the response.

Common Mistakes

  • Calling the first harmonic an overtone in IB answers.
  • Treating open and closed pipe ends as the same boundary condition.
  • Assuming damping only changes amplitude and never affects resonant frequency.

Exam Tips

  • Sketch nodes and antinodes before writing equations.
  • Use displacement nodes/antinodes for pipes; pressure nodes are not required here.
  • For a closed-open pipe, only odd harmonics appear in the simplest model.

Practice Questions

Question 1 (Multiple Choice)

A string fixed at both ends vibrates in its first harmonic. How many displacement antinodes are present?

A. 1 B. 2 C. 0 D. 3

Correct Answer: A

Solution Architecture

The first harmonic of a fixed-fixed string has nodes at both ends and one antinode in the middle.

Question 2 (Structured Paper 2 Style)

A 0.80 m string is fixed at both ends. Wave speed on the string is 120 m s-1.

(a) Calculate the wavelength of the first harmonic. [1 mark]

(b) Calculate the first harmonic frequency. [2 marks]

Paper 2 Structured Problem

Markscheme Breakdown

Part (a) Solution:

For a fixed-fixed string, L = lambda / 2, so lambda = 2L = 1.60 m.

Part (b) Solution:

f = v / lambda = 120 / 1.60 = 75 Hz.