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Waves · TOPIC LESSON

Standing Waves and Resonance

Follow the travelling phase, then let boundaries select standing waves and moving observers shift sound.

JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections

BY THE END, YOU CAN

  • Read travelling-wave equations and distinguish medium motion from phase speed.
  • Derive string speed, power scaling and standing-wave mode conditions.
  • Analyse open/closed air columns and beats with correct harmonic language.
  • Apply sound speed, intensity and Doppler relations with clear frames and assumptions.

BEFORE YOU START

  • Simple harmonic motion, trigonometry and derivatives.
  • Force, mass density, energy and logarithms.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. A travelling pattern is not a travelling piece of matter

A mechanical wave transfers a disturbance through a medium as neighbouring parts interact. In a small-amplitude wave, material elements typically oscillate about equilibrium while the pattern travels. On a string, a marked point moves transversely even though the wave travels along the string. In sound, compressions move through the air while individual molecules execute small oscillatory motions superposed on thermal motion.

A right-travelling one-dimensional shape can be written y(x,t) = f(x − vt); holding its argument constant gives dx/dt = v. A left-travelling shape uses f(x + vt). For a sinusoid y = A sin(kx − ωt + φ), A is amplitude, k = 2π/λ, ω = 2πf and phase speed is ω/k = fλ. The signs in the argument determine propagation direction, not the sign of displacement at one point.

At a fixed location, time variation describes the medium particle’s oscillation. At a fixed time, position variation describes the wave profile. Particle transverse velocity is ∂y/∂t and particle acceleration is ∂²y/∂t²; these are not the wave speed v. Their maxima for a sinusoid are Aω and Aω², while the phase speed is ω/k.

Phase difference can arise from separation in space or time. At one instant, two points separated by Δx have phase difference kΔx. At one point, two instants separated by Δt differ by ωΔt in magnitude. Points one wavelength apart are in phase; points half a wavelength apart are opposite in phase for a sinusoidal travelling wave.

y = A sin(kx − ωt + φ); vphase = ω/k = fλ

Right-travelling sinusoid for positive k and ω.

vparticle = ∂y/∂t; aparticle = ∂²y/∂t²

These describe local medium motion, not propagation speed.

ORIGINAL WORKED EXAMPLE

Read a wave equation

A string wave is y = 0.020 sin(4πx − 100πt) m with x in metres and t in seconds. Find wavelength, frequency, wave speed and maximum transverse particle speed.

  1. k = 4π rad m⁻¹, so λ = 2π/k = 0.50 m.
  2. ω = 100π rad s⁻¹, so f = 50 Hz and wave speed is fλ = 25 m s⁻¹ in the +x direction.
  3. Maximum transverse particle speed is Aω = 0.020 × 100π = 2π m s⁻¹, different from 25 m s⁻¹.

Answer λ = 0.50 m; f = 50 Hz; v = 25 m s⁻¹; particle maximum speed 2π m s⁻¹.

02 / UNDERSTAND & APPLY

2. Tension and inertia determine string-wave speed

For a taut ideal flexible string with tension T and linear mass density μ, transverse small-amplitude wave speed is √(T/μ). Tension supplies the restoring effect and mass per length supplies inertia. This explains why tighter strings transmit disturbances faster and heavier strings transmit them more slowly at the same tension.

The small-slope approximation allows tension to be treated as nearly constant and longitudinal motion to be neglected. A short string element has a net transverse force from the difference in slopes at its ends, leading to μ∂²y/∂t² = T∂²y/∂x². Comparing with the wave equation identifies v² = T/μ. Large amplitude or substantial stiffness requires a more detailed model.

For a uniform string of material density ρ and cross-section A, μ = ρA. If the string length changes at fixed mass, μ changes too. When a hanging mass provides tension, use the actual force and any acceleration conditions rather than automatically setting T equal to the hanging weight in every moving setup.

For a sinusoidal travelling wave on the ideal string, average transported power is (1/2)μω²A²v. It grows as amplitude squared and frequency squared at fixed medium. Doubling amplitude quadruples average power without changing the ideal linear wave speed. Energy transport speed equals this nondispersive phase speed in the elementary string model.

ORIGINAL WORKED EXAMPLE

Tension and a mass-per-length conversion

A 2.0 m string has mass 8.0 g and tension 100 N. Find wave speed and the effect of reducing tension to 25 N.

  1. Linear density is 0.008/2.0 = 0.004 kg m⁻¹.
  2. Initial speed is √(100/0.004) = √25000 ≈ 158 m s⁻¹.
  3. Quartering tension halves speed, giving about 79.1 m s⁻¹.

Answer Approximately 158 m s⁻¹ initially and 79.1 m s⁻¹ afterward.

03 / UNDERSTAND & APPLY

3. Boundaries change phase; overlapping waves add displacement

In a linear medium, the net displacement is the algebraic sum of component wave displacements. Waves do not permanently annihilate one another at a moment of destructive interference; they continue propagating and can later reinforce elsewhere. Energy redistribution must be considered over the complete field and boundary interactions.

A pulse reflected from a fixed end of a string is inverted because the endpoint must maintain zero displacement. A pulse reflected from an ideal free end is not inverted because the boundary instead imposes a zero transverse-force or slope condition. The transmitted and reflected amplitudes at a junction depend on the mechanical impedances, not only on the words “heavy” and “light”.

For two coherent sinusoidal waves of the same frequency with amplitudes A₁ and A₂ and phase difference δ, resultant amplitude satisfies A² = A₁² + A₂² + 2A₁A₂ cosδ. Equal amplitudes can cancel completely at δ = π; unequal amplitudes leave a minimum |A₁ − A₂|. Coherence means a stable relative phase over the observation time.

On a string, two equal-amplitude opposite-travelling waves of the same frequency can produce a standing pattern. The pattern has fixed nodes and antinodes instead of moving peaks. Calling it “standing” does not mean every string point is at rest; only nodes remain at zero displacement.

Aresult² = A₁² + A₂² + 2A₁A₂ cosδ

Same-frequency coherent scalar displacement components at the point considered.

04 / UNDERSTAND & APPLY

4. Standing waves are allowed shapes selected by boundaries

Adding A sin(kx − ωt) and A sin(kx + ωt) gives y = 2A sin(kx)cos(ωt). The position-dependent amplitude is 2A|sin kx|. Nodes occur where sin kx = 0, separated by λ/2. Antinodes lie halfway between adjacent nodes, so neighbouring node–antinode separation is λ/4.

For a string fixed at both ends separated by L, both endpoints must be nodes. The length therefore contains an integer number n of half-wavelengths: L = nλn/2. Allowed frequencies are fn = nv/(2L), with n = 1,2,3… . The fundamental is n = 1; the second harmonic is also the first overtone for this all-harmonics system.

All points within one segment between neighbouring nodes oscillate in phase, while adjacent segments oscillate oppositely. Amplitude varies with position. The ideal standing wave is a superposition of equal counter-propagating powers and has zero net time-averaged energy transport, though energy still exchanges locally between kinetic and elastic forms.

Resonance occurs when a drive efficiently excites an allowed mode. Finite damping limits amplitude and broadens the response. End conditions determine the mode frequencies; drive strength primarily affects amplitude in the linear regime. A missing harmonic can result from how the string is plucked or driven, even when that mode is mathematically allowed.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Allowed standing waves fit integer half-wavelengthsFirst three fixed–fixed string modes have one, two and three loops. Nodes occur at both ends and at equally spaced interior positions for higher modes. Opposite-phase snapshots show fixed nodes and oscillating loops.n = 11f₁n = 22f₁n = 33f₁Fixed–fixed string: nodes at both endsx = 0x = LSolid and dashed curves are opposite-phase snapshots
Normalized position scale. fn = nv/(2L), n=1,2,3. Curves show displacement snapshots, not trajectories.

ORIGINAL WORKED EXAMPLE

Find tension from a harmonic

A 1.0 m string with μ = 0.010 kg m⁻¹ is fixed at both ends. Its third harmonic is 150 Hz. Find fundamental frequency, wave speed and tension.

  1. For a fixed–fixed string, f₃ = 3f₁, so f₁ = 50 Hz.
  2. Wave speed is 2Lf₁ = 100 m s⁻¹.
  3. Tension follows T = μv² = 0.010 × 10000 = 100 N.

Answer 50 Hz; 100 m s⁻¹; 100 N.

05 / UNDERSTAND & APPLY

5. Sound displacement and pressure have opposite node patterns

Sound in a fluid is longitudinal. At an ideal rigid closed end of an air column, air displacement has a node because the boundary cannot move, while pressure variation has an antinode. At an ideal open end, pressure variation is approximately zero relative to ambient, so it is a pressure node and displacement antinode.

An open–open pipe or ideal closed–closed pipe has the same type of boundary at both ends and allows L = nλ/2. An open–closed pipe has a displacement node at one end and antinode at the other, allowing L = (2n − 1)λ/4. Its ideal resonant frequencies are (2n − 1)v/(4L): only odd harmonics.

For an open–closed pipe, the first overtone is the third harmonic, not the second. This vocabulary difference frequently causes errors. Draw boundary conditions before counting loops or applying an overtone label. Also distinguish displacement diagrams from pressure diagrams, which interchange nodes and antinodes.

Real open ends require an end correction because the oscillating air extends slightly beyond the physical tube. The effective length is somewhat longer; the correction depends on geometry and opening conditions. Use a specified correction when supplied and do not silently apply one universal number to every pipe. Temperature also changes the speed of sound and therefore the resonance frequencies.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Closed/open ends swap pressure and displacement nodesFor the fundamental of a pipe closed on the left and open on the right, displacement amplitude rises from zero at the wall to maximum at the open end. Pressure-variation amplitude falls from maximum at the wall to zero at the open end.Displacement amplitudePressure-variation amplitudeClosed endOpen endNodeAntinodeAntinodeNodeFundamental: L = λ/4; pressure and displacement swap nodes
Normalized spatial amplitude profiles, not bent pipe shapes or simultaneous signed pressure/displacement snapshots. End correction neglected; L=λ/4.

ORIGINAL WORKED EXAMPLE

An open–closed air column

An ideal pipe of length 0.85 m is closed at one end. With sound speed 340 m s⁻¹ and negligible end correction, find its fundamental and first overtone.

  1. Fundamental frequency is v/(4L) = 340/(3.4) = 100 Hz.
  2. Only odd harmonics occur, so the next resonance is 3f₁ = 300 Hz.
  3. The first overtone is therefore the third harmonic.

Answer 100 Hz and 300 Hz.

06 / UNDERSTAND & APPLY

6. Nearby frequencies create a slowly varying envelope

When two waves of close frequencies overlap at a point, their relative phase changes slowly. For equal amplitudes, adding cos(2πf₁t) and cos(2πf₂t) gives a rapid oscillation at their mean frequency multiplied by an envelope involving half their frequency difference. Loudness maxima occur at the beat frequency |f₁ − f₂|.

The absolute value matters because the audible beat count does not reveal which source has higher frequency. If one source is adjusted and the beat rate decreases, its frequency is moving toward the other’s. Use the direction of a known adjustment, such as increasing string tension, to resolve the ambiguity.

Equal source amplitudes can make the resultant amplitude momentarily zero at a beat minimum; unequal amplitudes leave a nonzero minimum. Beat frequency concerns envelope recurrence, not the frequency of rapid pressure oscillation. The mathematical cosine factor changes sign after half its full period, but amplitude magnitude and loudness repeat twice in that interval.

Beats are most readily perceived as separate loudness changes when frequencies are close enough for the auditory system to resolve the modulation. The formula itself is a superposition result; whether a listener hears beats, roughness or separate tones depends on perception and frequency range.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Close frequencies produce a slow amplitude envelopeThe sum of equal-amplitude 20 Hz and 22 Hz cosines has a rapid oscillation with an envelope whose magnitude peaks at 0,0.50 and1.00 seconds. Beat frequency is2 Hz.00.250.500.751.00+2A−2AEqual-amplitude tones: 20 Hz + 22 HzTime t (s)Envelope maxima every 0.50 s → beat frequency 2 Hz
Quantitative illustrative frequencies chosen to make the graph readable. The red dashed curves are ±2A|cos(2πt)|. Beat frequency is the difference, not the average carrier frequency.

ORIGINAL WORKED EXAMPLE

Use an adjustment to identify the unknown frequency

A 256 Hz fork gives 4 beats per second with a string. Increasing string tension slightly reduces the beat rate. Was the original string frequency 252 or 260 Hz?

  1. A beat rate of 4 allows either 256 − 4 = 252 Hz or 256 + 4 = 260 Hz.
  2. Increasing tension raises string frequency because f ∝ √T at fixed length and linear density.
  3. Since beats decrease, the string is moving toward 256 Hz from below.

Answer 252 Hz.

07 / UNDERSTAND & APPLY

7. Source and observer motion have different roles for sound

For sound in a stationary medium, source motion changes the spacing of wavefronts in the medium, while observer motion changes the rate at which those fronts are encountered. Consequently source and observer speeds enter different places in the formula. The medium provides the reference frame; relative source–observer speed alone is not sufficient for an exact classical sound calculation.

For collinear motion with sound speed v in the medium, define vo positive when the observer moves toward the source and vs positive when the source moves toward the observer. Then observed frequency is f′ = f(v + vo)/(v − vs), for the ordinary subsonic approach situation and corresponding signed extensions. Draw the propagation direction and define signs rather than memorising plus/minus slogans.

Approach raises frequency and recession lowers it. A moving source at rest-relative emission frequency f produces wavelength (v − vs)/f ahead when moving toward the observer. An approaching observer meets those fronts at rate (v + vo)/λ. This derivation explains the numerator and denominator rather than treating them as interchangeable.

Wind changes propagation relative to the ground; source and observer velocities must be expressed consistently relative to the medium when using this formula. Light uses a relativistic Doppler relation without a material rest medium, so the sound formula cannot be transplanted to electromagnetic waves. Supersonic source motion produces shock geometry beyond the simple subsonic denominator interpretation.

ORIGINAL WORKED EXAMPLE

Both source and observer approach

A source emits 500 Hz while moving toward an observer at 20 m s⁻¹. The observer moves toward the source at 10 m s⁻¹. Sound speed is 340 m s⁻¹ in still air. Find observed frequency.

  1. Use medium-referenced approach speeds: vo = 10 and vs = 20 m s⁻¹.
  2. f′ = 500(340 + 10)/(340 − 20) = 500 × 350/320.
  3. This is 546.875 Hz, about 547 Hz. Both motions raise the encounter rate.

Answer Approximately 547 Hz.

08 / UNDERSTAND & APPLY

8. Elasticity, inertia and spreading control sound

For longitudinal sound in a fluid, speed is √(B/ρ), where B is the relevant bulk modulus and ρ the equilibrium density. In an ideal gas undergoing small rapid adiabatic compressions, v = √(γRT/M), where M is molar mass. At fixed composition this scales as √T with absolute temperature. The pressure cancels in the ideal-gas relation at fixed temperature.

For an isotropic point source in a lossless far field, power spreads over spheres and intensity is P/(4πr²). Doubling distance reduces intensity by four while pressure/displacement amplitudes scale approximately as 1/r under the same conditions. Reflections, absorption and directional sources can invalidate the simple spreading model.

Sound level in decibels is β = 10 log₁₀(I/Iref), with a specified reference intensity, conventionally 10⁻¹² W m⁻² for the elementary sound-intensity scale. A tenfold intensity increase adds 10 dB; a hundredfold increase adds 20 dB. Do not add decibel levels directly when combining independent sources; add intensities first.

These relations connect the wave motion, medium and measurement. Frequency is set by the source for a stationary interface, while speed follows the medium and wavelength adjusts. A wave crossing a stationary interface into a different medium changes wavelength when its speed changes while frequency remains unchanged.

09 / UNDERSTAND & APPLY

9. Original practice and boundary-condition audit

Before calculating a resonant frequency, name the endpoint conditions. Before calculating a Doppler shift, name the medium frame and motion directions.

  • Particle speed and wave speed are different derivatives.
  • String speed comes from tension divided by mass per length.
  • Standing-wave frequencies are selected by boundary conditions.
  • Pressure and displacement nodes interchange in an air column.
  • Beats measure frequency difference magnitude.
  • Sound Doppler shifts distinguish source from observer motion relative to the medium.

ORIGINAL PRACTICE · 1

For y = A sin(kx + ωt), which way does the pattern travel for k,ω>0?

Reveal answer & reasoning

Toward negative x.

Constant phase gives kx + ωt = constant, so dx/dt = −ω/k.

ORIGINAL PRACTICE · 2

Two points in a sinusoidal travelling wave are λ/4 apart. Find phase difference magnitude.

Reveal answer & reasoning

π/2 radians.

Δφ = 2πΔx/λ.

ORIGINAL PRACTICE · 3

A string’s tension quadruples with μ fixed. What happens to wave speed?

Reveal answer & reasoning

It doubles.

v = √(T/μ).

ORIGINAL PRACTICE · 4

At fixed medium and frequency, wave amplitude triples. How does average string-wave power change?

Reveal answer & reasoning

It increases ninefold.

Average power is proportional to A².

ORIGINAL PRACTICE · 5

A standing wave has adjacent nodes 0.30 m apart. Find wavelength.

Reveal answer & reasoning

0.60 m.

Adjacent node spacing is λ/2.

ORIGINAL PRACTICE · 6

A fixed–fixed string has fundamental 80 Hz. Find its second overtone.

Reveal answer & reasoning

240 Hz.

Second overtone is third harmonic in an all-harmonics fixed–fixed string.

ORIGINAL PRACTICE · 7

An ideal open–closed pipe has fundamental 120 Hz. Find its second overtone.

Reveal answer & reasoning

600 Hz.

Its resonances are 120, 360, 600…; second overtone is fifth harmonic.

ORIGINAL PRACTICE · 8

Tones at 440 and 446 Hz overlap. Find beat frequency.

Reveal answer & reasoning

6 Hz.

Beat frequency is the absolute difference.

ORIGINAL PRACTICE · 9

A stationary 600 Hz source is approached by an observer at 34 m s⁻¹ in still air with v=340. Find f′.

Reveal answer & reasoning

660 Hz.

f′ = 600(340+34)/340 = 600×1.1.

ORIGINAL PRACTICE · 10

A source recedes at 20 m s⁻¹ from a stationary observer, emitting 500 Hz with v=340. Find f′.

Reveal answer & reasoning

Approximately 472 Hz.

For recession use denominator v+20: 500×340/360 ≈472.2.

ORIGINAL PRACTICE · 11

Gas temperature rises from 300 K to 675 K with composition unchanged. How does sound speed change?

Reveal answer & reasoning

It increases by a factor of 1.5.

v₂/v₁ = √(675/300) = √2.25.

ORIGINAL PRACTICE · 12

Intensity increases by a factor of 100. What is the sound-level change?

Reveal answer & reasoning

+20 dB.

Δβ = 10log₁₀100 =20.

ORIGINAL PRACTICE · 13

At a rigid closed pipe end, what are the displacement and pressure conditions?

Reveal answer & reasoning

Displacement node; pressure antinode.

Air cannot move through the wall, but pressure variation can be maximal there.

KEEP THE BIG IDEAS

Quick revision

  • A wave transports a pattern and energy, not necessarily bulk matter.
  • Boundary conditions decide which wavelengths fit.
  • Superposition creates fixed nodes or beats depending on wave directions and frequencies.
  • The medium frame is essential for classical sound Doppler calculations.
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