# C.4 Standing waves and resonance

> IB Physics HL · Theme C: Wave behaviour
> Source: https://www.owlsprep.com/study/ib-physics-hl-u3-c-4-standing-waves-and/

This sub-topic explains how two identical opposite-travelling waves superpose to form stationary standing waves, covers harmonic frequencies for strings and pipes, and explores the phenomenon of resonance. You will learn to solve common exam problems involving boundary conditions and resonant frequencies.

**Prerequisites:** [Progressive wave properties](https://www.owlsprep.com/study/ib-physics-hl-u3-c-1-traveling-waves/); [Superposition principle](https://www.owlsprep.com/study/ib-physics-hl-u3-c-3-superposition-interference/)

## Learning objectives

- Distinguish between standing and progressive waves
- Describe standing wave formation for different boundary conditions
- Calculate harmonic frequencies for strings and pipes
- Explain resonance and its practical applications

## Formation of Standing Waves

**Standing (stationary) wave** — A wave formed by superposition of two identical progressive waves traveling in opposite directions, resulting in fixed positions of zero and maximum amplitude, with no net energy transfer.

*Example:* An incident wave reflected off a fixed boundary interfering with the incoming wave

Key features of standing waves: nodes are fixed points of zero amplitude, and antinodes are fixed points of maximum amplitude. The distance between two adjacent nodes is $\frac{\lambda}{2}$, and the distance between a node and its adjacent antinode is $\frac{\lambda}{4}$.

**Worked example:** Adjacent antinodes of a standing wave are 12.5 cm apart. What is the wavelength of the original progressive waves?

1. Adjacent antinodes, like adjacent nodes, are separated by half a wavelength:
2. $$\frac{\lambda}{2} = 12.5 \text{ cm}$$
3. Solve for wavelength:
4. $$\lambda = 2 \times 12.5 = 25 \text{ cm} = 0.25 \text{ m}$$

> **Exam tip:** Never confuse standing waves with progressive waves: standing waves do not propagate energy or waveform.

## Standing Waves on Fixed Strings

**First harmonic (fundamental frequency)** — The lowest possible resonant frequency of a system, corresponding to the longest possible wavelength that fits the boundary conditions.

*Notation:* f_1

A string fixed at both ends has nodes at both ends. This means the length of the string $L$ is always an integer multiple of half wavelengths, so: $L = n \frac{\lambda_n}{2}$ where $n = 1,2,3...$ is the harmonic number. Rearranging gives the frequency formula:

$$f_n = \frac{nv}{2L}$$

Where $v$ is the speed of the wave on the string.

**Worked example:** A 1.5 m string fixed at both ends has a wave speed of 300 m/s. Calculate the frequency of the 2nd harmonic.

1. Use the nth harmonic formula for fixed strings:
2. $$f_n = \frac{nv}{2L}$$
3. Substitute $n=2$, $v=300$ m/s, $L=1.5$ m:
4. $$f_2 = \frac{2 \times 300}{2 \times 1.5} = 200 \text{ Hz}$$

## Standing Waves in Open and Closed Pipes

Standing waves form in air columns (pipes) with different boundary conditions depending on whether the end is open or closed:

- **Closed end**: Air cannot move, so this is a **node**
- **Open end**: Air can move freely, so this is an **antinode**

For pipes:  
- Open-open (both ends open): All harmonics exist, formula is the same as fixed strings: $f_n = \frac{nv}{2L}, n=1,2,3...$  
- Closed-open (one end closed, one open): Only odd harmonics exist, formula is $f_n = \frac{nv}{4L}, n=1,3,5...$

**Worked example:** A 0.75 m pipe is closed at one end, open at the other. Speed of sound is 340 m/s. Calculate the fundamental frequency.

1. For closed-open fundamental, $n=1$, so $L = \frac{\lambda}{4}$:
2. $$\lambda = 4L = 4 \times 0.75 = 3.0 \text{ m}$$
3. Calculate frequency using $f = \frac{v}{\lambda}$:
4. $$f_1 = \frac{340}{3.0} \approx 113 \text{ Hz}$$

## Resonance

**Resonance** — A phenomenon where a system oscillates at maximum amplitude when an external driving force matches the system's natural resonant frequency.

Resonance occurs when the driving frequency matches one of the natural harmonic frequencies of a system (string, air column, etc). This is the working principle behind all acoustic musical instruments, where resonance amplifies the sound at specific harmonic frequencies.

> **tip**
>
> In IB exam questions about resonance in adjustable closed pipes, the first resonance always occurs at $\frac{\lambda}{4}$, and the second at $\frac{3\lambda}{4}$.

**Worked example:** A 256 Hz tuning fork produces the first resonance in a closed pipe when the pipe length is 32 cm. Calculate the speed of sound.

1. First resonance for closed pipe: $L = \frac{\lambda}{4}$
2. $$\lambda = 4 \times 0.32 = 1.28 \text{ m}$$
3. Use $v = f\lambda$:
4. $$v = 256 \times 1.28 = 327.68 \approx 330 \text{ m/s}$$

## Common pitfalls

- **Wrong:** Assuming closed-open pipes have even harmonics
  - Why it fails: The node-antinode boundary condition only allows odd multiples of the fundamental wavelength
  - Correct: Only use odd values of $n$ (1, 3, 5...) for closed-open pipes
- **Wrong:** Claiming standing waves transfer net energy
  - Why it fails: Confused standing wave properties with progressive waves
  - Correct: Remember standing waves do not transfer net energy, energy is stored in nodes and antinodes
- **Wrong:** Treating an open pipe end as a node
  - Why it fails: Mixed up boundary conditions for open vs closed ends
  - Correct: Always remember: open ends are antinodes, closed ends are nodes
- **Wrong:** Using $\lambda = L$ for the fundamental frequency of a fixed string
  - Why it fails: Forgot that both ends are nodes, so only half a wavelength fits
  - Correct: For fixed string fundamental, $\lambda = 2L$

## Cheatsheet

| System | Boundary | Allowed $n$ | Frequency formula |
| --- | --- | --- | --- |
| String fixed both ends | Node-node | 1, 2, 3... | $f_n = \frac{nv}{2L}$ |
| Pipe open both ends | Antinode-antinode | 1, 2, 3... | $f_n = \frac{nv}{2L}$ |
| Pipe closed one end | Node-antinode | 1, 3, 5... | $f_n = \frac{nv}{4L}$ |

## What's next

Standing waves and resonance underpin the behavior of nearly all acoustic musical instruments, and are foundational for understanding wave phenomena across all areas of physics, from mechanical sound waves to electromagnetic standing waves in circuits and quantum mechanical matter waves. Mastery of boundary conditions and harmonic frequency calculations is frequently tested in both Paper 1 and Paper 2 IB Physics HL exams, often combined with superposition or wave speed concepts. Understanding resonance also helps explain real-world phenomena like structural resonance in bridges and resonant sound production.

- [C.6 Interference](https://www.owlsprep.com/study/ib-physics-hl-u3-c-6-interference/)
- [C.5 Doppler effect (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u3-c-5-doppler-effect/)
- [C.7 Diffraction and resolution (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u3-c-7-diffraction-and-resolution/)

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