Aim for FivePhysics 2

Unit 14 · Topic 14.6

14.6 Wave Interference and Standing Waves

New to AP Physics 2 since 2024–25: beats, plus standing waves on strings and in pipes. When waves overlap, they pass through each other, and the displacement at each point is the sum of the individual displacements. That's superposition: it can be constructive or destructive. Two tones with slightly different frequencies produce beats at fbeat=∣f1−f2∣f_{\text{beat}} = \lvert f_1 - f_2 \rvert. Waves confined to a string or pipe can form standing waves with fixed nodes and antinodes and only certain allowed wavelengths.

Key terms

  • superposition
  • constructive and destructive interference
  • beat frequency
  • standing wave
  • node and antinode
  • harmonic

Superposition

Send a pulse from each end of a rope. When they meet, they don't bounce off each other. They pass right through and continue on, unchanged. While they overlap, the rope's displacement at each point is the sum of the two pulses' displacements there.

Constructive interference happens when the displacements are in the same direction, so they add to a bigger displacement. Destructive interference happens when they're in opposite directions, so they partly or completely cancel. Two equal and opposite pulses make the rope flat for an instant, but the energy is still there, and both pulses reappear a moment later.

To find the shape at any instant, sketch each pulse where it would be on its own, then add their heights point by point.

Beats

Play two tones with slightly different frequencies, such as tuning forks at 440 Hz and 444 Hz. The waves drift in and out of step: at some moments their crests line up (loud), and at others a crest meets a trough (quiet). You hear a single tone that swells and fades.

The number of swells per second is the beat frequency, fbeat=∣f1−f2∣f_{\text{beat}} = \lvert f_1 - f_2 \rvert. Here that's 4 beats per second. Musicians tune by adjusting one string until the beats slow down and vanish.

Standing waves

When waves reflect back and forth in a confined region, such as a guitar string or an organ pipe, the waves going each way overlap. At certain frequencies they form a standing wave: a pattern that oscillates in place instead of traveling.

Nodes are points that never move (amplitude always zero). Antinodes are points that swing with the largest amplitude. Neighboring nodes are half a wavelength apart, and there's an antinode halfway between them.

The ends set the allowed wavelengths. A fixed end of a string must be a node. A free end is an antinode. In a pipe, the air at a closed end can't move, so it's a node; the air at an open end moves freely, so it's an antinode.

Harmonics

The pattern with the longest wavelength that fits is called the fundamental, or first harmonic. The pattern with the next-longest wavelength is the second harmonic, then comes the third, and so on. Each harmonic's frequency follows from f=vλf = \frac{v}{\lambda}.

These patterns aren't on the equation sheet. Draw the pattern and count how many half or quarter wavelengths fit in the length L.

SystemEndsAllowed wavelengthsHarmonics present
String fixed at both endsNode, nodeλn=2Ln\lambda_n = \frac{2L}{n}All (n = 1, 2, 3, …)
Pipe open at both endsAntinode, antinodeλn=2Ln\lambda_n = \frac{2L}{n}All (n = 1, 2, 3, …)
Pipe closed at one end, or string with one free endNode, antinodeλn=4Ln\lambda_n = \frac{4L}{n}Odd only (n = 1, 3, 5, …)

Why a closed pipe skips even harmonics

With a node at one end and an antinode at the other, the shortest fit is a quarter wavelength, so the fundamental has λ = 4L. The next pattern that fits adds another half wavelength: three quarters in the pipe, λ = 4L/3, which is 3 times the fundamental frequency. There's no way to fit a pattern at twice the fundamental frequency, so even harmonics are missing.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Overlapping pulses

    A pulse 3.0 cm high travels right along a rope, and a pulse 2.0 cm deep (a trough) travels left. What is the rope's maximum displacement when the pulses fully overlap, and what happens afterward?

    Show the solution
    1. Step 1: Superposition: add displacements with signs. (+3.0 cm) + (−2.0 cm) = +1.0 cm.
    2. Step 2: This is destructive interference, since the displacements are in opposite directions.
    3. Step 3: Afterward, the pulses pass through each other and continue unchanged: the 3.0 cm crest keeps moving right and the 2.0 cm trough keeps moving left.

    Answer: +1.0 cm while overlapping; afterward both pulses continue unchanged.

  2. Example 2

    Tuning with beats

    A guitar string and a 440 Hz tuning fork sounded together produce 4 beats per second. When the string is tightened slightly, the beat frequency increases to 6 per second. What was the string's original frequency?

    Show the solution
    1. Step 1: From fbeat=∣f1−f2∣f_{\text{beat}} = \lvert f_1 - f_2 \rvert, the string was either 440 − 4 = 436 Hz or 440 + 4 = 444 Hz.
    2. Step 2: Tightening raises the wave speed on the string, which raises its frequency.
    3. Step 3: If it started at 436 Hz, raising it would move it closer to 440 Hz and the beats would slow down. The beats sped up, so the string moved farther from 440 Hz.
    4. Step 4: So it started at 444 Hz (and is now 446 Hz).

    Answer: 444 Hz

  3. Example 3Calculator allowed

    Closed pipe (missing harmonics trap)

    A pipe 0.50 m long is closed at one end and open at the other. The speed of sound is 343 m/s. Find its three lowest resonant frequencies.

    Show the solution
    1. Step 1: Fundamental: a node at the closed end and an antinode at the open end means a quarter wavelength fits, so λ = 4L = 2.0 m.
    2. Step 2: f1=vλ=3432.0=171.5f_1 = \dfrac{v}{\lambda} = \dfrac{343}{2.0} = 171.5 Hz, about 172 Hz.
    3. Step 3: Only odd harmonics are allowed, so the next two are 3f₁ = 514.5 Hz and 5f₁ = 857.5 Hz.
    4. Step 4: The trap is listing 343 Hz (2f₁). That would need a node or antinode at both ends, which this pipe can't have.

    Answer: About 172 Hz, 515 Hz and 858 Hz.

Common mistakes

  • Thinking colliding pulses bounce off each other. They pass through each other and continue unchanged.
  • Including even harmonics for a pipe closed at one end. Only odd harmonics fit a node at one end and an antinode at the other.
  • Saying neighboring nodes are one wavelength apart. They're half a wavelength apart.
  • Assuming energy is destroyed when waves cancel. The cancellation is momentary or local; the energy is still carried by the waves.

On the exam

  • Standing wave questions often give a picture of the pattern. Count the loops (each is half a wavelength) to relate L to λ, then use v = fλ.
  • Beat questions often have two possible answers. Use extra information, such as a change in tension or a second tuning fork, to decide between them, and explain your choice.

Connected topics

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Check yourself

4 questions on 14.6 Wave Interference and Standing Waves. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

On a long string, a pulse with a displacement of +3.0 cm travels right and a pulse with a displacement of −2.0 cm travels left. At the instant their peaks overlap exactly, what is the string's displacement at that point?

Question 2 of 4Calculator allowed

A guitar string and a 440 Hz tuning fork sounded together produce 3 beats per second. When the string is tightened slightly, the beat frequency rises to 5 beats per second. What was the string's original frequency?

Question 3 of 4Calculator allowed

A string 0.80 m long is fixed at both ends, and waves travel on it at 240 m/s. What is the frequency of its third harmonic?

Question 4 of 4Calculator allowed

A pipe 0.25 m long is closed at one end and open at the other. Sound travels at 343 m/s. What is the next resonant frequency above the fundamental?

0 of 4 answered