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What happens when two waves overlap, and how do standing waves form on a string?

Topic 14.6 Wave Interference and Standing Waves: apply superposition to interference and find the harmonics of standing waves.

A focused answer to AP Physics 2 Topic 14.6, covering the superposition principle, constructive and destructive interference, the formation of standing waves with nodes and antinodes, the harmonics of a string and a pipe, and resonance, with full worked examples.

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  1. What this topic is asking
  2. The superposition principle
  3. Constructive and destructive interference
  4. Standing waves and harmonics
  5. Try this

What this topic is asking

The College Board (Topic 14.6) wants you to apply the superposition principle to constructive and destructive interference, and to describe how standing waves form, with their nodes, antinodes and harmonics on a string or in a pipe.

The superposition principle

Superposition is the rule that makes interference possible: where waves meet, you simply add their displacements at each instant. Two waves can reinforce or cancel depending on how their crests and troughs line up, but neither wave is permanently altered, they emerge on the other side as if nothing had happened. This adding of displacements is the foundation of all interference, both for sound and (in the later topics) for light.

Constructive and destructive interference

Whether overlapping waves reinforce or cancel depends on their relative phase. In phase, the peaks coincide and add (louder sound, brighter light); out of phase, a peak meets a dip and they cancel (silence, darkness). For two sources, this is set by the path difference: a whole number of wavelengths arrives in phase (constructive), while an extra half wavelength arrives out of phase (destructive). This path-difference rule is exactly what governs the double-slit pattern of light in Topic 14.8.

Standing waves and harmonics

A standing wave forms when two identical waves travel in opposite directions and superpose, typically a wave and its reflection off a boundary. The result is a stationary pattern with nodes (points of zero motion, where the waves always cancel) and antinodes (points of maximum motion, where they always reinforce). On a string fixed at both ends, only wavelengths that put a node at each end fit, giving the harmonics: λn=2Ln\lambda_n = \dfrac{2L}{n} and fn=nf1f_n = n f_1, where the fundamental (n=1n = 1) has L=λ/2L = \lambda/2. Driving the string at one of these frequencies produces large-amplitude resonance. The strategic role of this topic is that interference and standing waves are the clearest demonstration of wave behavior, and they reappear directly for light: the superposition and path-difference ideas here become the bright and dark fringes of double-slit interference and diffraction (Topics 14.7 to 14.9), tying the wave nature of sound to the wave nature of light.

Try this

Q1. State what happens at a point where a crest of one wave meets a crest of an equal wave. [1 point]

  • Cue. Constructive interference: the displacements add to double the amplitude.

Q2. A string fixed at both ends has a fundamental of 8080 Hz. State the frequency of its second harmonic. [1 point]

  • Cue. 160160 Hz (f2=2f1f_2 = 2 f_1).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). A string of length 1.21.2 m is fixed at both ends. Waves travel on it at 6060 m/s. (a) Calculate the wavelength of the fundamental (first harmonic). (b) Calculate the fundamental frequency. (c) State the condition for two overlapping waves to interfere constructively, in terms of their displacements.
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A 7-point FRQ on standing waves.

(a) Fundamental wavelength (3 points): for a string fixed at both ends, the fundamental has a node at each end and one antinode in the middle, so the length is half a wavelength: L=λ/2L = \lambda/2, giving λ=2L=2(1.2)=2.4\lambda = 2L = 2(1.2) = 2.4 m.
(b) Fundamental frequency (2 points): f=vλ=602.4=25f = \dfrac{v}{\lambda} = \dfrac{60}{2.4} = 25 Hz.
(c) Constructive interference (2 points): two waves interfere constructively where their displacements are in phase (in the same direction), so they add to give a larger amplitude (for example crest meeting crest).

Markers reward λ=2L\lambda = 2L for the fundamental, the frequency, and the in-phase condition for constructive interference.

AP 2023 (style)1 marksSection I (multiple choice). Two waves meet so that a crest of one aligns with a trough of the other, with equal amplitudes. What is the result at that point? (A) the amplitude doubles (B) the displacements cancel (C) the frequency doubles (D) a standing wave always forms. Justify your reasoning.
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A 1-point MCQ on destructive interference. The answer is (B).

By superposition, the displacements add. A crest (positive) aligned with an equal trough (negative) sums to zero: destructive interference cancels the displacement at that point. The trap is (A): that is constructive interference, where crest meets crest.

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