# Standing Waves

> AP Physics 2 · AP Physics 2
> Source: https://www.owlsprep.com/study/ap-physics-2-u6-standing-waves/

This module covers standing wave formation via superposition, boundary condition rules, harmonic series for strings and air columns, resonance, and AP exam problem-solving workflows.

**Prerequisites:** [Wave superposition principle](https://www.owlsprep.com/study/ap-physics-2-u6-wave-superposition/); [Traveling wave properties and speed calculations](https://www.owlsprep.com/study/ap-physics-2-u6-traveling-waves/)

## Learning objectives

- Distinguish between traveling and standing wave behavior
- Calculate node and antinode positions for strings and air columns
- Derive harmonic frequency relationships for fixed boundary systems
- Solve AP exam-style standing wave resonance problems

## Formation of Standing Waves via Superposition

A standing wave pattern emerges when two identical traveling waves of equal frequency and amplitude move in exactly opposite directions through the same medium. The waves continuously interfere with each other, producing a fixed pattern of oscillation that does not appear to travel across space.

**Standing Wave** — A non-propagating wave pattern formed by the perfect constructive and destructive interference of two counter-propagating identical traveling waves, with no net energy transfer across the medium.

$$y_{total} = 2A \sin(kx) \cos(\omega t)$$

> **info**
>
> Every point on a standing wave oscillates around a fixed equilibrium position, but the amplitude of oscillation depends entirely on the point's x coordinate along the medium.

**Worked example:** Two traveling waves on a string are described by $y_1 = 0.02 \sin(5x - 20t)$ m and $y_2 = 0.02 \sin(5x + 20t)$ m. Find the amplitude of the resulting standing wave at $x = 0.314$ m.

1. Use the position-dependent amplitude formula for superposed counter-propagating waves:

   $$A(x) = 2A \sin(kx)$$
2. Substitute given values, noting $k=5$ and $A=0.02$ m:

   $$A(x) = 2(0.02) \sin(5 * 0.314) = 0.04 * \sin(\pi/2) = 0.04 \text{ m}$$

**Check your understanding**

Test your understanding of standing wave formation

1. At what position in the example above would the oscillation amplitude be 0 m?

   - x=0 m
   - x=0.1 m
   - x=0.314 m
   - x=1 m

   *Why:* At x=0, $\sin(0) = 0$, so the amplitude is zero, forming a node.

## Nodes, Antinodes and Boundary Conditions

Nodes are points of zero oscillation, while antinodes are points of maximum oscillation. The positions of nodes and antinodes are completely determined by the boundary conditions of the medium, which define what motion is allowed at each end of the string or air column.

**Node / Antinode** — Node: Zero amplitude point from total destructive interference. Antinode: Maximum amplitude point from total constructive interference.

*Notation:* N / A

> **mnemonic**
>
> NA-NA: Nodes have No motion, Antinodes have All motion.

| Boundary Type | Allowed Motion | Node/Antinode at Boundary |
| --- | --- | --- |
| Fixed string end | Zero displacement | Node |
| Free string end | Maximum displacement | Antinode |
| Closed air column end | Zero air motion | Node |
| Open air column end | Maximum air motion | Antinode |

**Worked example:** A string of length 1.2 m is fixed at both ends. What is the distance between adjacent nodes on the fundamental standing wave?

1. For the fundamental mode, exactly half a wavelength spans the full length of the string, with nodes only at the two ends:

   $$L = \frac{\lambda_1}{2}$$
2. Adjacent nodes are always separated by $\lambda/2$, which equals the full length of the string for the fundamental mode:

   $$d_{node} = 1.2 \text{ m}$$

## Harmonic Series for Strings and Air Columns

**Exam command terms**

AP exam questions use specific terminology to signal which harmonic rules apply:

- **Fixed-fixed string** — Both ends are nodes, all integer harmonics allowed

- **Open-open air column** — Both ends are antinodes, all integer harmonics allowed

- **Closed-open air column** — One end node, one end antinode, only odd harmonics allowed

$$f_n = \frac{nv}{2L}, \quad n = 1, 2, 3... \quad (\text{Fixed-fixed / Open-open})$$

$$f_n = \frac{nv}{4L}, \quad n = 1, 3, 5... \quad (\text{Closed-open})$$

**Worked example:** A 0.8 m long closed-open air column has a sound wave speed of 343 m/s. Calculate the 3rd harmonic frequency.

1. Use the closed-open air column harmonic formula, with n=3 for the 3rd odd harmonic:

   $$f_3 = \frac{3v}{4L}$$
2. Substitute the given values to compute the result:

   $$f_3 = \frac{3 * 343}{4 * 0.8} \approx 322 \text{ Hz}$$

**Check your understanding**

Verify your harmonic rule knowledge

1. Which of the following systems cannot produce a 2nd harmonic?

   - 1m fixed-fixed string
   - 1m open-open pipe
   - 1m closed-open pipe
   - 1m stretched wire

   *Why:* Closed-open systems only support odd harmonics, so n=2 is impossible.

## Standing Wave Resonance

Resonance occurs when an external driving force applies energy to the medium at a frequency that exactly matches one of the system's natural harmonic frequencies. This produces a stable, large-amplitude standing wave pattern that persists as long as energy is supplied to offset damping losses.

> **warning**
>
> Common AP exam distractors claim any driving frequency will produce a standing wave. In reality, only frequencies that satisfy the system's boundary condition rules will form stable standing waves.

**Worked example:** A 2m long open-open pipe has a sound speed of 340 m/s. Find the lowest driving frequency that will produce a standing wave.

1. The lowest resonant frequency is the fundamental, n=1 for open-open systems:

   $$f_1 = \frac{v}{2L}$$
2. Substitute given values to find the result:

   $$f_1 = \frac{340}{2*2} = 85 \text{ Hz}$$

## Common pitfalls

- **Wrong:** Applying the 2L denominator formula to closed-open air columns
  - Why it fails: Closed-open systems have a quarter-wavelength fundamental, not a half-wavelength fundamental
  - Correct: Always use 4L as the denominator for closed-open pipes, and only allow odd n values
- **Wrong:** Counting nodes to find harmonic number and getting n off by 1
  - Why it fails: For a fixed-fixed string, the nth harmonic has n+1 total nodes, not n nodes
  - Correct: Count the number of half-wavelength segments across the medium to get n directly
- **Wrong:** Treating open air pipe ends as nodes instead of antinodes
  - Why it fails: Open air ends allow maximum longitudinal air motion, so they are antinodes by definition
  - Correct: Map all boundary conditions to node/antinode first before doing any frequency calculations
- **Wrong:** Claiming standing waves transfer net energy across the medium
  - Why it fails: Counter-propagating waves carry equal energy in opposite directions, leading to zero net energy flow
  - Correct: Only traveling waves have net energy transport through the medium
- **Wrong:** Calculating distance between adjacent antinodes as full $\lambda$
  - Why it fails: Adjacent nodes and adjacent antinodes are always separated by $\lambda/2$, not a full wavelength
  - Correct: Remember consecutive points of identical phase on a standing wave are always $\lambda/2$ apart

## Cheatsheet

| System Type | Boundary Conditions | Fundamental Wavelength | Harmonic Formula | Allowed n values |
| --- | --- | --- | --- | --- |
| Fixed-fixed string | Both ends nodes | 2L | $f_n = nv/(2L)$ | 1, 2, 3... |
| Open-open pipe | Both ends antinodes | 2L | $f_n = nv/(2L)$ | 1, 2, 3... |
| Closed-open pipe | One end node, one end antinode | 4L | $f_n = nv/(4L)$ | 1, 3, 5... |

## What's next

Mastering standing waves is critical for scoring on both multiple choice and free response questions in the AP Physics 2 waves and optics unit, as this topic frequently appears alongside Doppler effect and wave interference questions. The harmonic rules you learned here also extend to other wave systems, including standing sound waves in musical instruments and standing electromagnetic waves in laser cavities. Next, you will apply your understanding of superposition and standing waves to solve problems involving thin film interference, another high-frequency exam topic. You will also build on these concepts to analyze diffraction patterns formed when light passes through single and double slits, connecting physical wave behavior to measurable optical effects.

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