Interference and Superposition: Standing Waves & Beats

Curriculum note (2026-27): This page keeps its long-standing AP Physics 1 URL for existing readers, but wave interference, standing waves and sound in tubes are not among the eight units in the current College Board AP Physics 1 course framework. The current AP Physics 2 course and exam description includes waves, sound and physical optics. Use this guide for general physics learning or AP Physics 2 preparation; use the current College Board framework to decide what to prioritize for your own exam.

Two ripples meet on water. A guitar string reflects a pulse from its fixed end. A sound wave travels down a pipe and returns. In each case, the medium can respond to more than one disturbance at once. The principle of superposition gives a simple rule for the combined disturbance, while the pattern that follows is called interference. When reflection and interference repeatedly reinforce particular patterns, a standing wave can appear. These are related ideas, but they are not interchangeable.

This guide builds the ideas from individual pulses to strings, air columns and beats. You will learn how to identify a node, choose the right boundary condition, calculate a resonant frequency and check whether an answer makes physical sense. The explanations use algebra and diagrams you can sketch on paper; no calculus is required. For a broader map of current exam topics, use Sly Academy’s AP Physics 2 study guide.

What superposition actually says

At a particular position and instant, add the displacements caused by the individual waves to obtain the net displacement. If one upward pulse would move a point on a string 3 cm above equilibrium and a second pulse would move it 2 cm upward, their overlap puts the point 5 cm above equilibrium. If the second pulse instead displaces it 2 cm downward, the net displacement is 1 cm upward. The signs matter: a downward displacement is negative when upward is chosen positive.

This is an algebraic addition of displacements, not a claim that the pulses collide like solid objects. In an approximately linear medium, the waves pass through one another and continue after the brief overlap. Their temporary combined shape can look dramatic without either wave permanently changing. Real strings and air can depart from this simple model at extreme amplitudes, but the linear approximation explains most classroom examples. OpenStax’s superposition treatment gives the same point-by-point rule.

To draw a result, place both pulses on the same horizontal position scale. At several matching positions, read each vertical displacement relative to equilibrium and add the signed values. Plot the sums and join them smoothly. Do not add wavelengths, frequencies or speeds when the question asks for the instantaneous shape. Those quantities describe individual waves; the sum rule applies to the displacement at each point.

Constructive and destructive interference

Constructive interference occurs where the disturbances reinforce. Two identical upward crests that align momentarily produce twice the displacement of one crest. Destructive interference occurs where displacements oppose. Equal and opposite displacements cancel at that place and time. Partial cancellation is also possible: a 4 cm crest and a 1 cm trough leave 3 cm upward. Interference is therefore a spectrum of possible combinations, not just two all-or-nothing categories.

For two coherent sinusoidal waves of equal amplitude, a phase difference of a whole number of cycles gives a maximum and a half-cycle difference gives a minimum. If paths through the same medium differ by a whole number of wavelengths, their crests can arrive together; a half-integer number of wavelengths can place a crest against a trough. This path-difference shortcut requires attention to source phase and reflection phase changes. It is not safe to apply blindly to every diagram. If the source conditions are unclear, draw the waves or compare their phase at the observation point.

Interference patterns are evidence of wave behavior. They appear with water waves, sound and light, although the thing oscillating differs. A string’s displacement is visible; sound in air involves changing pressure and particle motion; light is an electromagnetic wave. That distinction becomes essential when you label nodes in a tube. The word ‘node’ alone does not tell you whether a diagram means air displacement or pressure.

From traveling waves to standing waves

A pulse traveling along a string reaches a boundary and reflects. A continuing sinusoidal wave can therefore overlap its own reflection. If the incident and reflected waves have the same frequency and similar amplitude while traveling in opposite directions, the combination can form a standing-wave pattern. The pattern has positions that remain still, called nodes, and positions with the largest oscillation amplitude, called antinodes. Individual pieces of string still move up and down: ‘standing’ describes the pattern, not a frozen string.

The separation between neighboring nodes is half a wavelength. The distance from a node to its nearest antinode is one quarter of a wavelength. Those geometric facts are powerful because many exam-style problems give the length and show a number of loops rather than stating the wavelength. A loop on a fixed-end string is the region between adjacent nodes, so one loop occupies half a wavelength. Count loops carefully before using a formula.

A standing wave does not mean that energy from the source has disappeared. In an ideal equal-amplitude pair, energy travels in opposite directions and there is no net average energy transport along the standing pattern. A real driven string loses energy to its surroundings; the driver continually replaces it. The distinction prevents the common mistake of equating a zero-displacement node with zero energy everywhere or all the time.

Boundary conditions on a string

A string clamped at both ends cannot move at those ends, so both endpoints are displacement nodes. The longest wavelength that fits has one antinode in the middle: half a wavelength occupies the length L. Thus the fundamental wavelength is 2L and its frequency is f1 = v/(2L), where v is the wave speed along the string. The second allowed pattern has two loops; its wavelength is L and its frequency is 2f1. In general, n half-wavelengths fit in the length: L = nλ/2, λn = 2L/n and fn = nv/(2L), for positive integer n.

Here ‘first harmonic’ is the fundamental. The second harmonic has two loops and one interior node, not two interior nodes. An overtone is counted differently: the first overtone is the second harmonic. State which naming system you use when describing a pattern. Label the fixed ends, count loops, then calculate; that sequence is more reliable than memorizing a picture of each harmonic.

String speed depends on tension T and linear mass density μ through v = sqrt(T/μ). Linear density means mass per unit length, in kg/m. Increasing tension raises speed and the resonant frequencies when length and density remain fixed. Increasing linear density lowers them. Doubling tension does not double speed: it multiplies speed and every fixed-mode frequency by sqrt(2). Doubling length, in contrast, halves each resonant frequency if the other properties are unchanged. These proportional relationships help check a numerical result before reaching for a calculator.

Standing sound waves in air columns

A resonating tube contains air that moves back and forth as sound reflects from its ends. An open end allows the air to move relatively freely, making it approximately a displacement antinode; a closed rigid end restricts motion, making it approximately a displacement node. The corresponding pressure pattern is opposite: an open end is approximately a pressure node and a closed end a pressure antinode. For the standard introductory model, always identify which physical quantity the drawing shows before you place letters N and A. See OpenStax’s air-column explanation for the boundary model.

The simplest tube model treats the ends as exact boundaries. Real pipes have an ‘end correction’ because the oscillating air extends slightly outside an open end; mouthpieces and shape can also affect the sound. Introductory calculations normally neglect these details unless the question explicitly gives a correction. Do not apply a correction to one tube but not another in a comparison unless the problem asks for it.

A tube open at both ends

Both ends are displacement antinodes. The fundamental fits half a wavelength between them, so L = λ1/2, λ1 = 2L and f1 = v/(2L). The next mode fits one complete wavelength along the tube and has frequency 2f1. In the ideal open-open tube, all positive integer harmonics are allowed: fn = nv/(2L). The open ends share the same boundary type as each other, but do not use a string diagram without changing node labels; a clamped string has nodes at both ends.

For example, with sound speed 340 m/s and an ideal open-open tube 0.85 m long, the fundamental is 340/(2 x 0.85) = 200 Hz. The second harmonic is 400 Hz and the third is 600 Hz. A frequency of 300 Hz would not be an ideal harmonic of this unchanged tube. If a problem gives a higher resonant frequency and asks which mode it is, divide by the fundamental and look for a positive integer.

A tube closed at one end

The closed end is a displacement node and the open end an antinode. The shortest pattern runs from a node to the nearest antinode, one quarter wavelength. Consequently L = λ1/4, λ1 = 4L and f1 = v/(4L). The next permitted shape fits three quarter-wavelengths in the tube; its frequency is 3f1. The next is 5f1. An ideal closed-open tube supports odd harmonics: f = (2m – 1)v/(4L), where m = 1, 2, 3 and so on. Calling 3f1 the ‘second harmonic’ would be incorrect; it is the third harmonic and the first overtone.

If the same 0.85 m tube is closed at one end and v = 340 m/s, its fundamental becomes 100 Hz. Its next ideal resonances are 300 Hz and 500 Hz. The switch from two open ends to one closed end changes the lowest allowed wavelength and the harmonic sequence. It does not change the speed of sound merely because a cap was added, assuming the same air conditions.

The result ‘only odd harmonics’ belongs to the ideal single-closed-end boundary model. A real instrument can generate additional spectral components through how it is driven and the tube’s geometry. When solving a textbook problem, separate the ideal resonances of the air column from every frequency that might appear in a real recorded note. The distinction is useful for explaining why instrument timbre is richer than a single pure sine wave.

Displacement and pressure: a practical check

Imagine a rigid cap at the left of a tube. Air next to the cap cannot oscillate through the wall, so its displacement amplitude is minimal. As compression and rarefaction reflect, pressure variation is strong there. At an open right end, pressure stays close to the surrounding atmospheric pressure, so the pressure variation is small while air can move. If a diagram gives pressure nodes instead of displacement nodes, reverse the boundary labels without changing the physical tube. A student who memorizes ‘closed means node’ without specifying displacement may misread a pressure graph.

Resonance and the role of the driver

A boundary permits certain spatial patterns, but a sustained large-amplitude standing wave usually needs a source driving near one of the permitted frequencies. At resonance, energy supplied by the driver accumulates in the oscillation until damping balances the input. Away from resonance, the response is smaller. Thus an allowed frequency is a property of the system and boundary conditions; observed amplitude also depends on how the system is excited and how quickly it loses energy.

A guitar player changes pitch by shortening a vibrating string with a finger, changing tension with a tuning peg, or choosing a string with a different linear density. In a wind instrument, finger holes change the effective vibrating air-column length; the instrument’s bore and mouthpiece complicate the ideal open/closed labels. The simple formulas remain a useful first model, not a complete acoustic description. For related fundamentals, Sly Academy’s physics encyclopedia can help revisit wave speed and oscillation vocabulary.

The wave relation v = fλ applies to traveling waves in the medium and to the wavelengths and frequencies of the standing pattern built from them. Use a consistent speed: wave speed on a string is set by string tension and density; sound speed in air depends on the medium and conditions. Do not carry the value 340 m/s from an air-column question into a string question. Likewise, a 200 Hz driver does not automatically impose the same wavelength in two different media, because wavelength adjusts to each medium’s wave speed.

Beats: interference that changes over time

Two sound waves with nearby but unequal frequencies can alternate between reinforcement and cancellation at a listener. The resulting loudness swells and fades are beats. For frequencies fA and fB, the beat frequency is |fA – fB|. If a 440 Hz tone and a 444 Hz tone sound together, you hear four loudness cycles per second in the ideal model. This does not mean the individual tones are 4 Hz sounds; their pitches remain near 440 Hz, while the amplitude envelope changes at 4 Hz.

Beats help tune instruments. If the beat rate slows as you adjust a string, its frequency is approaching the reference, although the beat rate alone does not tell you whether the string began too high or too low. Adjust in a known direction and observe whether the beats speed up or slow down. When the tones coincide, the slow beating disappears. Very different frequencies do not yield the simple, easily counted slow beats used in this method.

Do not confuse beats with a standing wave. Beats describe a temporal amplitude variation from two nearby frequencies at a location. A standing wave describes a spatial pattern of nodes and antinodes arising from oppositely directed waves of the same frequency. Both use superposition, but their characteristic patterns and common formulas answer different questions.

Worked problems that connect the ideas

Problem 1: add two pulses without losing the signs

At one instant, two pulses overlap at three marked points along a string. Pulse A has displacements +2, +4 and +1 cm; pulse B has -1, -3 and +2 cm at those same points. Add point by point to obtain +1, +1 and +3 cm. The second point is not -7 cm: the +4 and -3 contributions partly cancel. A sketch of the resultant should pass through the three net values, and its shape between them should follow the underlying smooth pulses rather than a series of abrupt corners.

After the pulses move past one another in a linear string, each continues with its original direction and approximate shape. The +1 cm value at the second point does not mean either original pulse has shrunk permanently. The calculation described a particular location and instant. In a real apparatus, friction and imperfect reflections may change the waves over time, but those effects are separate from the superposition rule.

Problem 2: four loops on a fixed string

A 1.20 m string fixed at both ends has four equally sized loops while driven at 120 Hz. Four loops mean n = 4, so L = nλ/2. Solving gives λ = 2L/n = 2(1.20)/4 = 0.60 m. The wave speed is v = fλ = 120 x 0.60 = 72 m/s. The fundamental for the same string is f1 = f4/4 = 30 Hz, which also follows from v/(2L) = 72/2.40. Two independent routes agree, a useful arithmetic check.

Suppose the four-loop pattern remains at 120 Hz but you want eight loops on the same length. The required wavelength would be 2(1.20)/8 = 0.30 m, half the original. At unchanged driving frequency the wave speed would need to halve to 36 m/s. If the same string’s linear density is unchanged, v proportional to sqrt(T) means the tension must become one quarter of its starting value. Simply saying ‘lower the tension’ captures the direction but misses the quantitative relationship. Also note that whether the apparatus can practically sustain the new pattern depends on its driver and damping.

Problem 3: find a tension from a standing pattern

A 0.75 m string has linear density 0.0060 kg/m and shows three loops at 90 Hz. First calculate wavelength: 2L/3 = 0.50 m. Then v = fλ = 45 m/s. Rearranging v = sqrt(T/μ) gives T = μv squared = 0.0060 x 45 squared = 12.15 N, or about 12 N at two significant figures. The unit works: kg/m times m squared/s squared gives kg m/s squared, a newton. Check that 90 Hz is the third harmonic, so the fundamental is 30 Hz and v/(2L) = 45/1.5 = 30 Hz.

Problem 4: choose the correct tube model

A 0.50 m air column has a rigidly closed left end and an open right end. With sound speed 340 m/s, the lowest frequency is v/(4L) = 340/2.0 = 170 Hz. The next ideal resonance is 3(170) = 510 Hz, not 340 Hz. If the tube were instead open at both ends with the same length, its lowest resonance would be v/(2L) = 340 Hz. Before calculating, sketch a node at the closed end and an antinode at the open end. That one drawing prevents the most common factor-of-two mistake.

If an experiment records a resonance near 500 Hz instead of the ideal 510 Hz, do not automatically conclude the wave relation is wrong. An effective-length correction at an open end, a different actual sound speed, or measurement uncertainty can shift a real result. Calculate the ideal prediction first and then discuss plausible experimental differences, rather than adjusting numbers until they fit.

Problem 5: identify a beat frequency

A tuning fork is labeled 256 Hz. A nearby musical note produces six beats per second with it. The note could be approximately 250 Hz or 262 Hz because both differ from 256 Hz by six. If tightening the string raises its pitch and the beat rate falls, the note began below the fork; the likely starting frequency was 250 Hz. If tightening makes the beats faster, the note began above the fork. A beat count by itself gives a difference, not the sign of that difference.

How to read and draw wave diagrams

First identify the horizontal axis. A snapshot graph of displacement against position describes the entire string at one instant; a displacement-against-time graph describes one position as time passes. A crest moving right in the snapshot does not tell you whether a marked particle is currently moving up or down unless direction of propagation and phase information are supplied. The distinction between a pattern moving through space and medium particles oscillating locally underlies many wave questions.

Next mark the equilibrium line, units, wavelength and any boundary. On a fixed string, an endpoint must remain on the equilibrium line at all times. On a tube diagram, state whether the vertical variation represents displacement or pressure. For interference, align the individual waves on a shared position axis before adding signed displacements. For a standing pattern, count the half-wavelength segments between nodes. For beats, look for an amplitude envelope over time rather than fixed spatial nodes.

Graphs can show a standing string at a moment when every point crosses equilibrium simultaneously. That flat snapshot does not imply the standing wave vanished. A short time later the antinodes will reach large displacement again while nodes remain fixed. One photograph is insufficient to identify all nodes: compare positions over a cycle or use the boundaries and mode shape. Similarly, a quiet instant in a beat pattern is a phase relationship at that instant, not evidence that either tone stopped.

A simple investigation and what to record

To explore a string, secure both ends, drive one end gently at adjustable frequency and increase frequency slowly. At each clear resonance, note the driving frequency, the number of loops, string length and estimated uncertainty. A plot of frequency against loop number should be approximately straight for fixed tension and linear density; its slope estimates the fundamental frequency. If tension changes during the run, that relationship will shift. Keep amplitude modest and follow apparatus safety instructions.

For an air column, a speaker or tuning fork can drive a tube while its effective length changes. Record lengths at several resonances for the same frequency, then compare the spacing between successive resonant lengths. In the ideal model, that spacing relates to half a wavelength even for a closed-open tube. Avoid assigning a resonant length to the physical tube edge without considering how a real open end behaves. State temperature or the sound speed used, and show how measurement uncertainty affects the inferred wavelength.

Experimental design questions may ask what to vary, what to hold constant and how to improve precision. Vary one factor deliberately, such as length, frequency or tension. Hold string density and measurement method constant where possible. Repeat resonance identification rather than reporting a single peak, and explain how you know you found a node or antinode. A claim that ‘the wave got bigger’ is less informative than a measured change in loop count or resonance frequency.

Common mistakes to avoid

• Adding amplitudes everywhere: maximum constructive interference is a special phase arrangement. At other places or times, add signed displacements point by point.
• Confusing a pulse with its medium: a traveling disturbance can move right while particles of a string move mainly up and down.
• Counting a loop as one wavelength: one fixed-string loop runs between adjacent nodes and occupies half a wavelength.
• Calling a closed tube’s first overtone its second harmonic: the first overtone of the ideal closed-open tube is the third harmonic.
• Mixing pressure and displacement nodes: their boundary positions are reversed in a sound column.
• Assuming every named wave topic is on the AP Physics 1 exam: the current AP Physics 1 framework lists eight units without mechanical-wave interference; consult the College Board’s latest framework before prioritizing revision.

Study plan and next step

Begin with a two-pulse sketch and three signed-displacement sums. Then draw the first three fixed-string modes and calculate their wavelengths from the loop count. Compare an open-open and a closed-open tube of the same length, labeling displacement and pressure boundaries separately. Finish with one beat-frequency problem and explain in words why two candidate pitches can give the same beat count. Explaining each boundary condition aloud is often more useful than memorizing a chart.

If your immediate goal is the current AP Physics 1 exam, return to the AP Physics 1 syllabus-aligned guide for its assessed mechanics, oscillations and fluids topics. If you are taking AP Physics 2 or a general introductory course, continue from this page into waves, sound and optics using the official course description and the linked OpenStax chapters. The URL is retained for continuity, but the exam-priority distinction matters more than the historical folder name.