Class 11 · Coaching volume 4 · module pp. 1–104

Waves

Follow a disturbance travelling through a medium while its particles oscillate locally. Connect wave equations, superposition, standing modes, beats and the Doppler effect.

9 concepts33 worked examples3 visual experiments

Understand the idea → see it move → solve a question. Extra theory is there when you need it.

Textbook alignment: NCERT Class 11 Physics · chapter 14. Coaching order and separate foundation modules follow your uploaded index.

1 Basics

Travelling waves and particle motion

A mechanical wave transfers energy through coupled particles; the particles generally oscillate about their own positions. Transverse displacement is perpendicular to propagation, while longitudinal displacement is parallel, as in ordinary sound in air.

The phase identifies equivalent points on successive oscillations. In y=A sin(kx−ωt+φ), a fixed phase moves toward increasing x. Changing −ωt to +ωt reverses propagation. Particle velocity is a time derivative, not the wave speed. At a stationary boundary between media the source frequency remains the same; changed propagation speed changes wavelength according to λ=v_w/f.

Travelling waves and particle motion
\[ y=A\sin(kx-\omega t+\phi),\quad k=\frac{2\pi}{\lambda},\quad \omega=2\pi f,\quad v_w=\frac{\omega}{k}=f\lambda \]
Symbols and assumptions: y particle displacement and A amplitude (m); x position (m); t time (s); k wave number (rad/m); λ wavelength (m); ω angular frequency (rad/s); f frequency (Hz); v_w propagation speed (m/s); φ phase (rad).
Other useful formulas
Phase difference from path and time
\[ \Delta\phi=\frac{2\pi}{\lambda}\Delta x,\qquad \Delta\phi=\frac{2\pi}{T}\Delta t \]
Symbols and assumptions: Δφ phase difference (rad); λ wavelength (m); Δx separation along the direction of travel (m); T period (s); Δt time interval (s).
Particle velocity and wave slope
\[ v_p=-v\,\frac{\partial y}{\partial x},\qquad \frac{v_{p,\max}}{v}=Ak=\frac{2\pi A}{\lambda} \]
Symbols and assumptions: v_p particle velocity (m/s); v wave speed (m/s), positive for travel toward +x; ∂y/∂x slope of the snapshot; A amplitude (m); k wave number (rad/m); λ wavelength (m).
Visual experiment

The wave travels; the particles stay local

Play a transverse wave and follow a highlighted particle. The pattern moves right while the marked particle only moves up and down.

The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.

Predict before you change an input

The highlighted medium particle…

Use this in a question

A travelling-wave equation or speed–frequency relation is given.

  1. Read amplitude, k and ω from the equation with their units.
  2. Use λ = 2π/k, f = ω/(2π), and v = ω/k.
  3. Keep particle motion separate from propagation direction.
Why the equation works · optional
  1. 1

    Describe one particle

    \[ y(x,t)=A\sin(kx-\omega t) \]

    x labels a particle’s equilibrium position. A is displacement amplitude, k = 2π/λ, ω = 2πf.

  2. 2

    Follow a fixed phase

    \[ kx-\omega t=\mathrm{constant} \]

    To keep the same phase as time increases, x must increase. The minus sign denotes rightward propagation.

  3. 3

    Connect pattern speed to wavelength

    \[ v=\frac\omega k=f\lambda \]

    Particle velocity is ∂y/∂t and is vertical here. It is different from the horizontal propagation speed.

Model and symbols: Ideal sinusoidal transverse wave with amplitude A = 0.15 m. Speed v and frequency f set λ = v/f. The medium is represented by discrete sample particles.

Watch the assumption

Wave speed and maximum particle speed Aω are different quantities.

Worked example

A wave is y=0.02 sin(4πx−20πt) in SI units. Find wavelength, frequency and speed.

Show solution

k=4π gives λ=0.5 m. ω=20π gives f=10 Hz. Speed is fλ=5 m/s, toward +x.

Worked example · Numerical

A wave is \(y = 0.03\sin(0.5\pi x - 100\pi t)\) in SI units. Find the wave speed, the maximum particle speed, and the phase difference between two points 1 m apart.

Show solution
  1. \(k = 0.5\pi\) rad/m and \(\omega = 100\pi\) rad/s.
  2. Wave speed: \(v = \omega/k = 100\pi/0.5\pi = 200\) m/s, toward \(+x\).
  3. Maximum particle speed: \(A\omega = 0.03\times100\pi\approx9.42\) m/s, much less than the wave speed.
  4. Phase difference: \(\Delta\phi = k\Delta x = 0.5\pi\times1 = \pi/2\).

Answer: 200 m/s; about 9.4 m/s; π/2

More solved problem patterns
Worked example · Graph

A snapshot shows a transverse wave moving to the right. Point P is on the rising slope just in front of a crest (between a trough and the next crest ahead of it, on the side facing the direction of travel). Is P moving up or down?

Show solution
  1. Use \(v_p = -v\,\partial y/\partial x\) with \(v > 0\).
  2. Just in front of a crest, the shape slopes downward as \(x\) increases, so \(\partial y/\partial x < 0\) and \(v_p > 0\): moving up.
  3. A quicker way: shift the whole snapshot slightly to the right. The crest arrives at P a moment later, so P must be rising toward the crest.

Answer: P is moving up.

Worked example · Ratio

For a wave \(y = A\sin(kx - \omega t)\), the maximum particle speed is four times the wave speed. Find \(\lambda\) in terms of \(A\).

Show solution
  1. \(A\omega = 4\,\omega/k\), so \(Ak = 4\).
  2. \(k = 2\pi/\lambda\), so \(2\pi A/\lambda = 4\).
  3. \(\lambda = \pi A/2\).

Answer: \(\lambda = \pi A/2\)

Go deeper · extra explanation and derivations
wave movesArrows: particle velocity. Dots at crests and troughs are momentarily at rest.cresttrough
A snapshot of a wave moving to the right. Particles just ahead of a crest are moving up, and those just behind are moving down. Crests and troughs are momentarily at rest; particles on the mean line move fastest.

Reading a wave equation

Any function of the form \(y = f(x - vt)\) is a wave moving toward \(+x\) with speed \(v\); \(f(x + vt)\) moves toward \(-x\). The shape stays the same and slides along. For example, \(y = \dfrac{1}{1+(x-2t)^2}\) is a pulse moving toward \(+x\) at 2 m/s.

For the harmonic wave \(y = A\sin(kx - \omega t + \phi)\):

  • the coefficient of \(x\) is \(k = 2\pi/\lambda\); the coefficient of \(t\) is \(\omega = 2\pi f\);
  • the wave speed is \(v = \omega/k\) (ratio of the coefficients, with signs dropped);
  • opposite signs of the \(x\) and \(t\) terms mean motion toward \(+x\); the same sign means \(-x\).

If the equation is written as \(y = A\sin2\pi(x/\lambda - t/T)\), read \(\lambda\) and \(T\) directly.

Particle velocity is not wave velocity

Each particle oscillates up and down; the wave pattern moves sideways. Particle velocity is \(v_p = \partial y/\partial t = -A\omega\cos(kx - \omega t)\), with maximum \(A\omega\). The wave speed is \(v = \omega/k\), fixed by the medium.

  • A useful link: \(v_p = -v\,\dfrac{\partial y}{\partial x}\). Where the snapshot slopes upward (in the direction of travel), the particle is moving down, and vice versa.
  • \(v_{p,\max}/v = Ak = 2\pi A/\lambda\). The two are equal only if \(\lambda = 2\pi A\).
  • Particle acceleration is \(-\omega^2y\): each particle does SHM.

Phase difference and path difference

Two points a distance \(\Delta x\) apart along the wave differ in phase by \(\Delta\phi = \dfrac{2\pi}{\lambda}\Delta x\). One point at two times \(\Delta t\) apart differs by \(\Delta\phi = \dfrac{2\pi}{T}\Delta t\).

Points \(\lambda\) apart are in phase. Points \(\lambda/2\) apart are in opposite phase. Points \(\lambda/4\) apart are a quarter cycle apart.

Question patterns
  • “For y = 0.03 sin(0.5πx − 100πt), find the wave speed, the maximum particle speed and the phase difference between points 1 m apart.”
  • “Which of these functions represents a travelling wave?” (only f(x ± vt) forms)
  • “Find the phase difference between two points 15 cm apart when f = 500 Hz and v = 300 m/s.”
  • Graph: “In this snapshot of a wave moving to the right, what is the direction of motion of particle P?”
  • “If the maximum particle speed is four times the wave speed, find λ in terms of A.”
Another common mistake

In a transverse wave the particles do not travel with the wave. They oscillate about fixed mean positions; only energy and the pattern move.

Another common mistake

In \(y = A\sin(\omega t - kx)\) the wave still moves toward \(+x\). What matters is that \(x\) and \(t\) have opposite signs, not which comes first.

Check the idea

2 Core

Wave speed on strings and in gases

A stretched string transmits transverse disturbances through its tension. More tension strengthens the restoring interaction; more mass per length increases inertia. Sound in a gas depends on compressional stiffness and density.

Sound compressions are approximately adiabatic for ordinary rapid acoustic oscillations, so the gas bulk stiffness is γp rather than p. At a fixed gas composition, sound speed scales with the square root of absolute temperature.

Wave speed on strings and in gases
\[ v_{\rm string}=\sqrt{\frac{F_T}{\mu}},\qquad v_{\rm sound}=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma RT}{M}} \]
Symbols and assumptions: F_T string tension (N); μ mass per length (kg/m); γ gas heat-capacity ratio; p gas pressure (Pa); ρ density (kg/m³); R gas constant (J/mol K); T absolute temperature (K); M molar mass (kg/mol); speeds in m/s. Ideal-gas sound is assumed.
Other useful formulas
Wire in terms of stress
\[ v=\sqrt{\frac{T}{\rho A}}=\sqrt{\frac{\text{stress}}{\rho}} \]
Symbols and assumptions: v transverse wave speed (m/s); T tension (N); ρ density of the wire material (kg/m³); A cross-sectional area (m²); stress = T/A (Pa).
Hanging rope
\[ v=\sqrt{gx},\qquad t_{\rm total}=2\sqrt{\frac Lg} \]
Symbols and assumptions: v wave speed at height x above the free lower end (m/s); g gravity (m/s²); x height above the lower end (m); t_total time for a pulse to run the whole rope (s); L rope length (m). Uniform rope.
Elastic waves in a rod and a fluid
\[ v_{\rm rod}=\sqrt{\frac{Y}{\rho}},\qquad v_{\rm fluid}=\sqrt{\frac{B}{\rho}} \]
Symbols and assumptions: v speeds (m/s); Y Young’s modulus (Pa); B bulk modulus (Pa); ρ density (kg/m³). Rod formula is for longitudinal waves in a thin rod.
Watch the assumption

Use linear density for a string, not its bulk mass density.

Worked example

A string has tension 100 N and μ=0.01 kg/m. Find wave speed.

Show solution

v=√(100/0.01)=100 m/s. Quadrupling tension at fixed μ would double the speed.

Worked example · Numerical

A steel wire has density 8000 kg/m³ and cross-section 1 mm². It is stretched with a tension of 80 N. Find the speed of transverse waves on it.

Show solution
  1. \(\mu = \rho A = 8000\times1\times10^{-6} = 8\times10^{-3}\) kg/m.
  2. \(v = \sqrt{T/\mu} = \sqrt{80/(8\times10^{-3})} = \sqrt{10^4}\).
  3. \(v = 100\) m/s.

Answer: 100 m/s

More solved problem patterns
Worked example · Ratio

Two wires of the same material carry the same tension. Their radii are \(r\) and \(2r\). Compare the wave speeds on them.

Show solution
  1. \(\mu = \rho\pi r^2\), so \(v = \sqrt{T/(\rho\pi r^2)}\propto1/r\).
  2. \(v_1/v_2 = 2r/r = 2\).
  3. The thinner wire carries waves twice as fast.

Answer: 2 : 1

Worked example · Numerical

A uniform rope 10 m long hangs freely from a ceiling. How long does a small pulse take to travel from its lower end to the top? Take \(g = 10\) m/s².

Show solution
  1. At height \(x\) above the lower end, \(v = \sqrt{gx}\), so the pulse speeds up as it climbs.
  2. \(t = \int_0^L dx/\sqrt{gx} = 2\sqrt{L/g}\).
  3. \(t = 2\sqrt{10/10} = 2\) s.

Answer: 2 s

Go deeper · extra explanation and derivations

Why v = √(T/μ) for a string

Dimensional check. Tension has units N = kg m s⁻²; linear density has kg m⁻¹. Their ratio has units m² s⁻², so \(\sqrt{T/\mu}\) is a speed. No other combination of \(T\) and \(\mu\) gives a speed.

Physical argument. Ride along with a pulse. A small piece of string of length \(\Delta l\) at the top of the pulse moves on a circular arc of radius \(R\) at speed \(v\). The two tension forces, each at a small angle, give a net inward force \(T\Delta l/R\). This must equal \(\mu\Delta l\,v^2/R\). So \(v^2 = T/\mu\).

For a wire of density \(\rho\) and cross-section \(A\): \(\mu = \rho A\), so \(v = \sqrt{T/(\rho A)} = \sqrt{\text{stress}/\rho}\).

Variable tension: a hanging rope

A uniform rope of length \(L\) hangs from the ceiling. At height \(x\) above its lower end, it supports the rope below it: \(T = \mu gx\). So \(v = \sqrt{gx}\), zero at the bottom and \(\sqrt{gL}\) at the top.

The time for a pulse to travel the full length is \(t = \int_0^L dx/\sqrt{gx} = 2\sqrt{L/g}\). It does not depend on the rope’s mass.

If a block hangs from a string and is then dipped in a liquid, buoyancy lowers the tension, which lowers the wave speed and the string’s frequencies.

Waves in solids, liquids and gases

Wave speed always has the form \(\sqrt{\text{elastic property}/\text{inertial property}}\).

  • Longitudinal waves in a thin rod: \(v = \sqrt{Y/\rho}\).
  • Sound in a liquid or gas: \(v = \sqrt{B/\rho}\), with \(B\) the bulk modulus.
  • Solids are far stiffer than gases, so sound typically travels fastest in solids (steel about 5000 m/s), then liquids (water about 1500 m/s), then gases (air about 340 m/s).
  • Transverse waves need a shear stiffness, so they cannot travel through the bulk of a liquid or gas.

The details for gases (Newton’s formula and Laplace’s correction) are in the next section.

Question patterns
  • “Find the wave speed on a wire of density ρ and radius r under tension T.”
  • Ratio: “Two wires of the same material, radii in the ratio 1 : 2, same tension. Compare wave speeds.”
  • “A block hanging from a string is immersed in water. How does the wave speed change?”
  • “A rope hangs from a ceiling. Find the time a pulse takes to travel from bottom to top.” (2√(L/g))
  • Concept: “Can transverse waves travel through air?”
Another common mistake

Doubling the tension increases the speed by \(\sqrt2\), not 2. Speed goes as \(\sqrt T\).

Another common mistake

Two wires of the same material under the same tension can have different speeds: the thinner wire has smaller \(\mu\) and carries waves faster.

Check the idea

3 Core

Speed of sound: Newton, Laplace and what changes it

Sound in air is a train of squeezes and stretches. How fast it moves depends on how hard the air pushes back when squeezed and how heavy it is. Newton first assumed the air stays at the same temperature while it is squeezed. His answer came out about 15% too low. Laplace fixed it by noting that the squeezes happen so fast that heat has no time to flow out.

Choose the speed law for the medium: strings, solids, and gases use different stiffness and density quantities. In an ideal gas, temperature must be in kelvin. A change in the source frequency does not by itself change sound speed in the same medium.

Speed of sound: Newton, Laplace and what changes it
\[ v_{\rm Newton}=\sqrt{\frac p\rho},\qquad v_{\rm Laplace}=\sqrt{\frac{\gamma p}{\rho}}=\sqrt{\frac{\gamma RT}{M}} \]
Symbols and assumptions: v speed of sound (m/s); p gas pressure (Pa); ρ gas density (kg/m³); γ = C_p/C_v (1.4 for air); R = 8.314 J/(mol K); T absolute temperature (K); M molar mass (kg/mol). Ideal gas, small-amplitude sound.
Other useful formulas
Temperature dependence near room temperature
\[ v_t\approx v_0+0.61\,t \]
Symbols and assumptions: v_t speed at t °C (m/s); v₀ ≈ 332 m/s at 0 °C; t Celsius temperature. Linear approximation of v ∝ √T for air, good for ordinary temperatures.
Comparing two gases or temperatures
\[ \frac{v_1}{v_2}=\sqrt{\frac{\gamma_1T_1M_2}{\gamma_2T_2M_1}} \]
Symbols and assumptions: v₁, v₂ sound speeds (m/s); γ₁, γ₂ heat-capacity ratios; T₁, T₂ absolute temperatures (K); M₁, M₂ molar masses (any consistent unit).
Sound speed and rms speed
\[ \frac{v_{\rm sound}}{v_{\rm rms}}=\sqrt{\frac\gamma3} \]
Symbols and assumptions: v_sound speed of sound (m/s); v_rms root-mean-square molecular speed (m/s); γ = C_p/C_v of the gas. Same gas and temperature.
Watch the assumption

Doubling the pressure of a gas at constant temperature does not change the speed of sound. Density doubles too, so \(p/\rho\) is unchanged.

Worked example

Find the speed of sound in air at STP from Newton’s formula and from Laplace’s formula. Take \(p = 1.013\times10^5\) Pa, \(\rho = 1.293\) kg/m³ and \(\gamma = 1.4\).

Show solution

Newton: \(v = \sqrt{1.013\times10^5/1.293}\approx280\) m/s. Laplace: \(v = \sqrt{1.4}\times280\approx331\) m/s, which matches the measured value of about 332 m/s.

Worked example · Numerical

The speed of sound in air is 332 m/s at 0 °C. Find it at 27 °C.

Show solution
  1. \(v\propto\sqrt T\): \(v = 332\sqrt{300/273}\).
  2. \(\sqrt{300/273} = \sqrt{1.0989}\approx1.0483\).
  3. \(v\approx348\) m/s. The rule \(v_0 + 0.61t\) gives \(332 + 16.5 = 348.5\) m/s, in close agreement.

Answer: About 348 m/s

More solved problem patterns
Worked example · Ratio

At what temperature is the speed of sound in hydrogen equal to its speed in oxygen at 100 °C? Both are diatomic.

Show solution
  1. Same \(\gamma\), so \(v\propto\sqrt{T/M}\). Set \(T_H/M_H = T_O/M_O\).
  2. \(T_H = 373\times2/32\approx23.3\) K.
  3. In Celsius: \(23.3 - 273\approx-250\) °C. Hydrogen must be very cold to slow sound down to oxygen’s level.

Answer: About 23 K (about −250 °C)

Go deeper · extra explanation and derivations

Newton’s formula and why it failed

Newton treated the compressions in a sound wave as isothermal. For an isothermal ideal gas, \(pV = \text{constant}\), so \(B = -V\,dp/dV = p\). That gives \(v = \sqrt{p/\rho}\approx280\) m/s at STP. The measured value is about 332 m/s, roughly 18% higher.

Laplace pointed out that compressions and rarefactions alternate very fast, and air is a poor conductor of heat. A compressed region warms up and cannot lose that heat before it expands again. The process is adiabatic: \(pV^\gamma = \text{constant}\), so \(B = \gamma p\). The speed rises by \(\sqrt\gamma = \sqrt{1.4}\approx1.18\), giving about 331 m/s.

What changes the speed of sound in a gas

  • Temperature: \(v\propto\sqrt T\) (absolute). Near room temperature, \(v_t\approx v_0 + 0.61\,t\) m/s, with \(t\) in °C and \(v_0\approx332\) m/s.
  • Pressure: no effect at constant temperature, because \(p/\rho = RT/M\).
  • Humidity: moist air is less dense (water vapour, \(M = 18\) g/mol, replaces heavier N₂ and O₂), so sound travels slightly faster on a humid day.
  • Nature of the gas: \(v\propto\sqrt{\gamma/M}\) at the same temperature. Sound in hydrogen is 4 times as fast as in oxygen (same \(\gamma\), \(M\) ratio 1 : 16).
  • Wind: adds its component along the direction of travel.
  • Frequency, wavelength and amplitude: no effect. All audible frequencies travel together, which is why music from far away still sounds in tune.

Link to molecular speeds

Kinetic theory gives \(v_{\rm rms} = \sqrt{3RT/M}\). Dividing, \(v_{\rm sound}/v_{\rm rms} = \sqrt{\gamma/3}\), about 0.68 for air. Sound cannot outrun the molecules that carry it, and both scale as \(\sqrt{T/M}\).

Question patterns
  • “Find the speed of sound at STP by Newton’s formula. Why is it wrong? What correction did Laplace make?”
  • “At what temperature is the speed of sound double (or 1.5 times) its value at 0 °C / 27 °C?”
  • “How does the speed of sound change with pressure, humidity, temperature?”
  • “Compare the speed of sound in hydrogen and oxygen at the same temperature.” (4 : 1)
  • Ratio of v_sound to v_rms.
Another common mistake

Use absolute temperature in \(v\propto\sqrt T\). Going from 27 °C to 54 °C does not double \(T\); it changes it from 300 K to 327 K.

Another common mistake

Humid air carries sound faster, not slower. Water vapour is lighter than the nitrogen and oxygen it replaces.

Check the idea

4 Core

Intensity, energy and loudness

A wave carries energy without carrying matter. Stand close to a loudspeaker and you can feel the air push on you; walk away and the sound fades. The energy a wave delivers each second to each square metre is its intensity. It grows with the square of the amplitude and the square of the frequency.

At fixed medium and frequency, intensity is proportional to amplitude squared. For a spreading point source, also account for the area over which the power spreads. Add displacements for coherent interference before squaring to find intensity.

Intensity, energy and loudness
\[ I=\tfrac12\rho v\omega^2A^2=2\pi^2\rho vf^2A^2 \]
Symbols and assumptions: I intensity (W/m²); ρ density of the medium (kg/m³); v wave speed (m/s); ω angular frequency (rad/s); f frequency (Hz); A displacement amplitude (m). Harmonic wave in a uniform medium.
Other useful formulas
Spreading from point and line sources
\[ I_{\rm point}=\frac{P}{4\pi r^2}\propto\frac1{r^2},\qquad I_{\rm line}\propto\frac1r \]
Symbols and assumptions: I intensity (W/m²); P power of the source (W); r distance from the source (m). No absorption; amplitude goes as 1/r for a point source and 1/√r for a line source.
Sound level in decibels
\[ \beta=10\log_{10}\frac{I}{I_0},\qquad \beta_2-\beta_1=10\log_{10}\frac{I_2}{I_1} \]
Symbols and assumptions: β sound level (dB); I intensity (W/m²); I₀ = 10⁻¹² W/m², the reference threshold of hearing; I₁, I₂ two intensities being compared.
Watch the assumption

Intensity depends on both amplitude and frequency. Two waves with the same amplitude but different frequencies do not have the same intensity.

Worked example

Two sound waves in the same air have amplitudes in the ratio 2 : 1 and frequencies in the ratio 1 : 2. Compare their intensities.

Show solution

\(I\propto f^2A^2\), so \(I_1/I_2 = (1/2)^2\times2^2 = 1\). The intensities are equal.

Worked example · Numerical

A conversation has a sound level of 60 dB and a whisper 40 dB. How many times more intense is the conversation?

Show solution
  1. \(\beta_2 - \beta_1 = 10\log_{10}(I_2/I_1)\), so \(20 = 10\log_{10}(I_2/I_1)\).
  2. \(\log_{10}(I_2/I_1) = 2\), so \(I_2/I_1 = 100\).
  3. The conversation is 100 times as intense, though it does not sound 100 times as loud. The ear compresses the range.

Answer: 100 times

More solved problem patterns
Worked example · Ratio

A listener moves from 5 m to 10 m away from a small loudspeaker (no echoes or absorption). Find the change in intensity, amplitude and sound level.

Show solution
  1. Point source: \(I\propto1/r^2\), so \(I\) becomes \((5/10)^2 = 1/4\) of its value.
  2. \(A\propto1/r\), so the amplitude halves.
  3. Level change \(= 10\log_{10}(1/4)\approx-6\) dB.

Answer: Intensity ÷ 4, amplitude ÷ 2, level drops by about 6 dB

Worked example · Conceptual

A violin and a flute play the same note at the same loudness. Why can a listener still tell them apart?

Show solution
  1. Same note means the same fundamental frequency, so the pitch is the same.
  2. Same loudness means similar intensity.
  3. The instruments produce different sets of overtones with different strengths, so the waveforms differ in shape.
  4. This property is called quality or timbre.

Answer: They differ in quality (timbre), set by their overtones.

Go deeper · extra explanation and derivations

Derivation: intensity of a wave

  1. A particle of mass \(\Delta m\) in SHM has total energy \(\tfrac12\Delta m\,\omega^2A^2\).
  2. Energy per unit volume (energy density): \(u = \tfrac12\rho\omega^2A^2\).
  3. In time \(\Delta t\), the energy in a column of length \(v\Delta t\) and area \(S\) crosses the area \(S\): energy \(= u\,Sv\Delta t\).
  4. Intensity \(= \dfrac{\text{energy}}{S\Delta t} = uv = \tfrac12\rho v\omega^2A^2\).

At fixed medium, \(I\propto A^2\omega^2\), or \(I\propto A^2f^2\).

How intensity falls with distance

  • Point source (spherical waves): power \(P\) spreads over \(4\pi r^2\). \(I = P/4\pi r^2\propto1/r^2\), and amplitude \(A\propto1/r\).
  • Line source (cylindrical waves, such as a long busy road): \(I\propto1/r\) and \(A\propto1/\sqrt r\).
  • Plane waves (far from the source, ideal): \(I\) and \(A\) stay constant.

Absorption in the medium makes real sound fade faster than these ideal laws.

Loudness, decibels, pitch and quality

The ear responds to an enormous range of intensities, from about \(10^{-12}\) W/m² (threshold of hearing) to about 1 W/m² (threshold of pain). Its sense of loudness grows roughly with the logarithm of intensity. So we use the sound level

\(\beta = 10\log_{10}(I/I_0)\) dB, with \(I_0 = 10^{-12}\) W/m².

  • 10 times the intensity adds 10 dB; 100 times adds 20 dB; doubling adds about 3 dB.
  • Typical levels: whisper about 20–30 dB, conversation about 60 dB, busy traffic about 80 dB, pain about 120 dB.
  • Loudness is the listener’s sensation and depends mainly on intensity (and also on frequency).
  • Pitch depends on frequency. A female voice usually has a higher pitch than a male voice.
  • Quality (timbre) depends on which overtones are present and how strong they are. It lets you tell a violin from a flute playing the same note.
  • Humans hear about 20 Hz to 20 kHz. Below is infrasound; above is ultrasound.
Question patterns
  • “Two waves have amplitude ratio 2 : 1 and frequency ratio 1 : 2. Compare intensities.”
  • “The distance from a point source is tripled. Find the new intensity / amplitude.”
  • “The intensity of a sound increases 100 times. By how many decibels does the level rise?” (20 dB)
  • Match the column: loudness–intensity, pitch–frequency, quality–overtones.
  • “What is the audible range? What are infrasonic and ultrasonic waves?”
Another common mistake

Doubling the distance from a point source halves the amplitude but quarters the intensity. Keep \(A\propto1/r\) and \(I\propto1/r^2\) apart.

Another common mistake

Loudness and pitch are different. Turning up the volume raises loudness (intensity) without changing pitch (frequency).

Check the idea

5 Core

Superposition, interference and intensity

In a linear medium, overlapping displacements add algebraically. Waves reinforce when their phases match and cancel when opposite. The particles respond to the total displacement; the waves continue through each other in the linear model.

Intensity measures average energy flow per unit area and scales with amplitude squared for the same medium and frequency. A fixed phase difference requires coherent waves for a stable interference pattern.

Superposition, interference and intensity
\[ y_{\rm net}=y_1+y_2,\quad A_R^2=A_1^2+A_2^2+2A_1A_2\cos\delta,\quad I\propto A^2 \]
Symbols and assumptions: y displacements (m); A₁/A₂ component amplitudes (m); A_R resultant amplitude; δ phase difference (rad); I intensity (W/m²). Amplitude formula assumes same frequency and propagation direction at the point considered; intensity scaling holds for fixed medium and frequency.
Other useful formulas
Resultant intensity
\[ I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\delta \]
Symbols and assumptions: I_R resultant intensity (W/m²); I₁, I₂ intensities of the two waves (W/m²); δ phase difference at the point (rad). Coherent waves of the same frequency in the same medium.
Ratio of maximum to minimum intensity
\[ \frac{I_{\max}}{I_{\min}}=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2 \]
Symbols and assumptions: I_max, I_min extreme intensities (W/m²); I₁, I₂ individual intensities; A₁, A₂ amplitudes (m), with A₁ > A₂.
Interference conditions (sources in phase)
\[ \Delta x=n\lambda\ \text{(max)},\qquad \Delta x=\left(n+\tfrac12\right)\lambda\ \text{(min)} \]
Symbols and assumptions: Δx path difference (m); λ wavelength (m); n = 0, 1, 2, … Swap the conditions if the sources are in antiphase.
Watch the assumption

Intensity is not generally the sum of individual intensities where coherent waves overlap.

Worked example

Two equal-amplitude waves A meet in phase. Find resultant amplitude and intensity relative to one wave.

Show solution

A_R=2A. Intensity scales as (2A)², so it is four times that of one wave, not twice.

Worked example · Ratio

Two coherent waves with intensities in the ratio 9 : 4 interfere. Find the ratio of maximum to minimum intensity.

Show solution
  1. Amplitude ratio \(= \sqrt{9/4} = 3/2\).
  2. \(A_{\max} : A_{\min} = (3+2) : (3-2) = 5 : 1\).
  3. \(I_{\max}/I_{\min} = 5^2 = 25\).

Answer: 25 : 1

More solved problem patterns
Worked example · Numerical

Two loudspeakers driven in phase emit sound of 680 Hz (speed 340 m/s). A listener is 4.25 m from one speaker and 5.00 m from the other. Is the sound loud or quiet there?

Show solution
  1. \(\lambda = 340/680 = 0.5\) m.
  2. Path difference \(= 5.00 - 4.25 = 0.75\) m \(= 1.5\lambda\).
  3. An odd number of half-wavelengths with sources in phase means destructive interference.
  4. If the speakers are equal and the distances similar, the sound there is very weak.

Answer: Quiet (destructive interference)

Worked example · Numerical

Two waves of equal intensity \(I_0\) meet with a phase difference of \(2\pi/3\). Find the resultant intensity.

Show solution
  1. \(I_R = I_0 + I_0 + 2I_0\cos(2\pi/3)\).
  2. \(\cos(2\pi/3) = -1/2\), so \(I_R = 2I_0 - I_0 = I_0\).
  3. The resultant amplitude is also \(A_0\), matching \(2A_0\cos(\delta/2) = 2A_0\times\tfrac12\).

Answer: \(I_0\)

Go deeper · extra explanation and derivations

Resultant amplitude and intensity

Two waves of the same frequency meeting with phase difference \(\delta\) give \(A_R^2 = A_1^2 + A_2^2 + 2A_1A_2\cos\delta\). Since \(I\propto A^2\):

\(I_R = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\delta\).

  • Maximum (\(\delta = 0, 2\pi, \ldots\)): \(A_{\max} = A_1 + A_2\), \(I_{\max} = (\sqrt{I_1}+\sqrt{I_2})^2\).
  • Minimum (\(\delta = \pi, 3\pi, \ldots\)): \(A_{\min} = |A_1 - A_2|\), \(I_{\min} = (\sqrt{I_1}-\sqrt{I_2})^2\).
  • \(\dfrac{I_{\max}}{I_{\min}} = \left(\dfrac{A_1+A_2}{A_1-A_2}\right)^2\).
  • For equal intensities \(I_0\): \(I_{\max} = 4I_0\), \(I_{\min} = 0\), and the average is \(2I_0\). Energy is redistributed, not created.

Path difference decides the result

Two sources vibrating in phase send waves to a point P. If the path difference is \(\Delta x\), the phase difference is \(\delta = 2\pi\Delta x/\lambda\).

  • Constructive at P: \(\Delta x = n\lambda\) (\(n = 0, 1, 2, \ldots\)).
  • Destructive at P: \(\Delta x = (2n+1)\lambda/2\).

If the sources themselves are in opposite phase, swap the two conditions.

Coherence

A steady pattern of loud and quiet spots needs a constant phase difference, so the sources must have the same frequency and a fixed phase relation (coherent). Two independent sources change their phase randomly; the \(\cos\delta\) term averages to zero, and intensities simply add: \(I = I_1 + I_2\).

Question patterns
  • “Intensities in the ratio 9 : 4 (or amplitudes 3 : 1) interfere. Find I_max/I_min.”
  • “Two speakers in phase, λ = 0.5 m. Is a point with path difference 0.75 m loud or quiet?”
  • “Two equal waves interfere with phase difference π/2 (or 2π/3). Find the resultant intensity in terms of I₀.”
  • Concept: “Why do two independent sources not give a steady interference pattern?”
Another common mistake

Given an intensity ratio, take square roots before using the amplitude formula. A 9 : 4 intensity ratio means amplitudes 3 : 2, so \(I_{\max}/I_{\min} = (5/1)^2 = 25\), not \(13/5\).

Another common mistake

At a point of constructive interference, two equal waves give 4 times one intensity, not 2 times. The average over the pattern is still 2 times.

Check the idea

6 Core

Reflection and standing waves

A wave reflecting at a fixed string end reverses displacement phase; at a free end it does not. Two equal opposite-travelling waves of the same frequency can form a standing pattern with fixed nodes and antinodes.

At a displacement node the medium never moves; at an antinode its amplitude is maximal. Adjacent nodes are λ/2 apart and a node-to-neighbouring-antinode distance is λ/4. In an ideal standing wave the average net energy flow is zero.

Reflection and standing waves
\[ y=2A\sin(kx)\cos(\omega t),\quad \Delta x_{NN}=\frac\lambda2,\quad \Delta x_{NA}=\frac\lambda4 \]
Symbols and assumptions: y displacement (m); A amplitude of each travelling component (m); k wave number (rad/m); ω angular frequency (rad/s); λ wavelength (m); Δx_NN adjacent-node spacing; Δx_NA adjacent node–antinode spacing. Origin is a node.
Other useful formulas
Positions of nodes and antinodes
\[ x_{\rm N}=\frac{n\lambda}{2},\qquad x_{\rm A}=\frac{(2n+1)\lambda}{4}\qquad(n=0,1,2,\ldots) \]
Symbols and assumptions: x_N node positions and x_A antinode positions (m), measured from a node at x = 0; λ wavelength (m). For y = 2A sin kx cos ωt.
Amplitude at a point
\[ A(x)=2A|\sin kx| \]
Symbols and assumptions: A(x) amplitude of the particle at position x (m); A amplitude of each travelling wave (m); k wave number (rad/m). Origin at a node.
An = 1: fundamentalλ = 2L, f₁ = v/2LAAn = 2: 2nd harmonicλ = L, f = 2f₁AAAn = 3: 3rd harmonicλ = 2L/3, f = 3f₁● nodes (including both fixed ends)A antinodes
The first three modes of a string fixed at both ends. Solid and dashed curves show the string at its two extremes. The nth harmonic has n loops and n + 1 nodes, with λ = 2L/n.
Watch the assumption

For sound, a displacement node corresponds to a pressure antinode.

Worked example

Adjacent nodes are 0.30 m apart. Find wavelength and nearest node–antinode distance.

Show solution

λ=0.60 m. The nearest node–antinode distance is λ/4=0.15 m.

Worked example · Numerical

A standing wave on a string is \(y = 4\sin(\pi x/15)\cos(96\pi t)\), with \(x\) and \(y\) in cm and \(t\) in s. Find (a) the amplitude at \(x = 5\) cm, (b) the node positions, (c) the amplitude and speed of the component travelling waves.

Show solution
  1. (a) \(A(5) = 4\sin(\pi/3) = 4\times0.866\approx3.46\) cm.
  2. (b) Nodes where \(\sin(\pi x/15) = 0\): \(x = 0, 15, 30, \ldots\) cm. So \(\lambda/2 = 15\) cm and \(\lambda = 30\) cm.
  3. (c) The standing-wave amplitude \(2A = 4\) cm, so each travelling wave has \(A = 2\) cm.
  4. Speed \(= \omega/k = 96\pi/(\pi/15) = 1440\) cm/s \(= 14.4\) m/s. The frequency is 48 Hz.

Answer: (a) about 3.46 cm; (b) x = 0, 15, 30 cm …; (c) 2 cm each, 14.4 m/s

More solved problem patterns
Worked example · Conceptual

A wave pulse with a crest travels along a light string tied to a heavy rope. Describe the reflected and transmitted pulses.

Show solution
  1. The heavy rope is the denser medium, so it behaves partly like a fixed end.
  2. The reflected pulse returns inverted (as a trough), with phase change \(\pi\).
  3. The transmitted pulse continues as a crest (never inverted), but moves more slowly in the heavy rope.
  4. Both pulses keep the original frequency content; the transmitted pulse is shorter because its speed is lower.

Answer: Reflected: inverted. Transmitted: upright, slower and shorter.

Go deeper · extra explanation and derivations

Derivation: the standing-wave equation

Add two equal waves travelling in opposite directions:

\(y = A\sin(kx - \omega t) + A\sin(kx + \omega t) = 2A\sin kx\cos\omega t\).

  • Every point oscillates at the same \(\omega\), but with its own amplitude \(2A|\sin kx|\).
  • Nodes (amplitude zero): \(\sin kx = 0\), \(x = 0, \lambda/2, \lambda, \ldots\)
  • Antinodes (amplitude \(2A\)): \(x = \lambda/4, 3\lambda/4, \ldots\)
  • All points between two neighbouring nodes move in phase. Points on opposite sides of a node are in antiphase.
  • No energy flows past a node, so the wave does not transport energy; energy sloshes between kinetic and potential forms within each loop.

If the waves are \(A\sin(kx - \omega t)\) and \(-A\sin(kx+\omega t)\) instead, the result is \(-2A\cos kx\sin\omega t\), with an antinode at \(x = 0\). The origin’s nature depends on the boundary.

Reflection rules

  • Fixed end or denser medium: the reflected wave is inverted (phase change \(\pi\)). A crest returns as a trough. The end is a node.
  • Free end or rarer medium: no phase change. A crest returns as a crest. The end is an antinode.
  • Frequency never changes on reflection or transmission. Wavelength changes only in the transmitted wave if the speed changes.

Pressure and displacement in sound

In a sound standing wave, a displacement node is a pressure antinode, and vice versa. Air molecules on both sides of a displacement node move toward it and away from it together, so the pressure there swings the most. At the closed end of a pipe, the air cannot move (displacement node), but the pressure varies most. At an open end, the pressure stays close to atmospheric (pressure node), while the air moves most.

Question patterns
  • “For y = 4 sin(πx/15) cos(96πt) cm, find the amplitude at x = 5 cm, the node positions, and the speed of the component waves.”
  • “Distance between a node and the next antinode is 10 cm. Find λ.” (40 cm)
  • “What is the phase difference between particles in adjacent loops?” (π)
  • Reflection: “A pulse reflects from a fixed end / free end. Draw the reflected pulse.”
  • Statement-type: “Energy is not transported in a standing wave.”
Another common mistake

Adjacent nodes are \(\lambda/2\) apart, not \(\lambda\). One full wavelength holds two loops.

Another common mistake

Particles in a standing wave do not all have the same amplitude, but those within one loop are in the same phase. In a travelling wave it is the reverse: same amplitude, different phases.

Check the idea

7 Exam

String and air-column harmonics

Boundary conditions select wavelengths that fit the resonator. A string fixed at both ends has displacement nodes at both ends; an open air-column end is approximately a displacement antinode, while a closed end is a node.

An open–open pipe supports all integer harmonics, like a fixed–fixed string. An open–closed pipe fits odd quarter-wavelengths and supports odd harmonics in the ideal model. End corrections slightly increase the effective length of real pipes.

String and air-column harmonics
\[ f_n=\frac{nv}{2L}\ \text{(fixed string or open pipe)},\qquad f_n=\frac{(2n-1)v}{4L}\ \text{(one closed end)} \]
Symbols and assumptions: f_n mode frequency (Hz); n=1,2,… is mode index; v appropriate wave speed (m/s); L effective resonator length (m). Closed-pipe harmonic numbers are 1,3,5,… rather than n itself. Ideal boundary conditions are assumed.
Other useful formulas
End correction
\[ f_{\rm closed}=\frac{v}{4(L+e)},\qquad f_{\rm open}=\frac{v}{2(L+2e)},\qquad e\approx0.6r \]
Symbols and assumptions: f fundamental frequencies (Hz); v speed of sound (m/s); L physical pipe length (m); e end correction at each open end (m); r inner radius of the pipe (m).
Resonance tube
\[ v=2f(l_2-l_1),\qquad e=\frac{l_2-3l_1}{2} \]
Symbols and assumptions: v speed of sound (m/s); f tuning-fork frequency (Hz); l₁, l₂ first and second resonating lengths of the air column (m); e end correction (m).
Fundamental from two consecutive resonances
\[ f_1=f_{n+1}-f_n\ \text{(string or open pipe)},\qquad f_1=\frac{f_{\rm next}-f_{\rm prev}}{2}\ \text{(closed pipe)} \]
Symbols and assumptions: f₁ fundamental (Hz); f_n, f_(n+1) consecutive resonant frequencies (Hz). In a closed pipe consecutive resonances are odd harmonics, so they differ by 2f₁.
Visual experiment

Nodes do not move when the string vibrates

Choose a harmonic of a string fixed at both ends. Watch stationary nodes divide the string into loops.

The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.

Predict before you change an input

In the third harmonic, the number of internal nodes is…

Use this in a question

A string fixed at both ends vibrates in a harmonic.

  1. Count the loops to identify harmonic n.
  2. Use λ = 2L/n and f = nv/(2L).
  3. Count internal nodes as n − 1; include endpoints only if asked.
Why the equation works · optional
  1. 1

    Fit whole half-wavelengths

    \[ L=n\frac\lambda2 \]

    Each loop contains one half-wavelength. The two fixed endpoints must be nodes.

  2. 2

    Find the allowed frequency

    \[ \lambda_n=\frac{2L}{n},\qquad f_n=\frac{nv}{2L} \]

    Frequency changes discretely with harmonic number; n is not a continuous slider value.

  3. 3

    Combine counter-propagating waves

    \[ y=2A\sin\left(\frac{n\pi x}{L}\right)\cos(2\pi f_nt) \]

    The spatial factor fixes nodes where it vanishes. All particles between adjacent nodes oscillate in phase; the neighbouring loop has opposite phase.

Model and symbols: String length L = 2 m; wave speed v = 4 m/s; fixed ends; harmonic n is a positive integer. Damping and driving are omitted.

Watch the assumption

The first overtone of a closed pipe is the third harmonic, not the second harmonic.

Worked example

For a closed pipe L=0.85 m with sound speed 340 m/s, find fundamental and first overtone.

Show solution

f₁=340/(4×0.85)=100 Hz. The next allowed mode is 3f₁=300 Hz.

Worked example · Numerical

In a resonance-tube experiment with a 500 Hz tuning fork, the first and second resonances occur at air-column lengths 16 cm and 50 cm. Find the speed of sound and the end correction.

Show solution
  1. \(\lambda/2 = l_2 - l_1 = 34\) cm, so \(\lambda = 68\) cm.
  2. \(v = f\lambda = 500\times0.68 = 340\) m/s.
  3. \(e = (l_2 - 3l_1)/2 = (50 - 48)/2 = 1\) cm.
  4. Check: \(l_1 + e = 17\) cm \(= \lambda/4\). Correct.

Answer: 340 m/s; 1 cm

More solved problem patterns
Worked example · Ratio

The third harmonic of a pipe closed at one end has the same frequency as the fundamental of an open pipe. Find the ratio of the open pipe’s length to the closed pipe’s length.

Show solution
  1. Closed pipe third harmonic: \(3v/(4L_c)\).
  2. Open pipe fundamental: \(v/(2L_o)\).
  3. Equate: \(3/(4L_c) = 1/(2L_o)\), so \(L_o = 2L_c/3\).
  4. Ratio \(L_o : L_c = 2 : 3\).

Answer: 2 : 3

Worked example · Numerical

A pipe has consecutive resonant frequencies 425 Hz and 595 Hz. Is it open or closed at one end? Find its fundamental and, if \(v = 340\) m/s, its length.

Show solution
  1. Difference \(= 170\) Hz.
  2. If open, \(f_1 = 170\) Hz and 425 would be \(2.5f_1\), not a whole number. So it is not open.
  3. If closed, \(f_1 = 85\) Hz: \(425 = 5f_1\) and \(595 = 7f_1\), both odd. It is closed at one end.
  4. \(L = v/(4f_1) = 340/340 = 1\) m (ignoring end correction).

Answer: Closed at one end; f₁ = 85 Hz; L = 1 m

Go deeper · extra explanation and derivations
Closed at one endf₁ = v/4Lf₃ = 3v/4L = 3f₁Open at both endsf₁ = v/2Lf₂ = 2v/2L = 2f₁Closed end: displacement node. Open end: displacement antinode.
Displacement envelopes of air in pipes. A closed end is a displacement node and an open end an antinode. The closed pipe fits odd quarter-wavelengths (f₁, 3f₁, 5f₁ …); the open pipe fits whole half-wavelengths (f₁, 2f₁, 3f₁ …).

String fixed at both ends

Both ends must be nodes, so a whole number of half-wavelengths fits: \(L = n\lambda/2\).

  • \(\lambda_n = 2L/n\), \(f_n = \dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}\), with \(n = 1, 2, 3, \ldots\). All harmonics are present.
  • The nth harmonic is the \((n-1)\)th overtone. It has \(n\) loops, \(n+1\) nodes and \(n\) antinodes.
  • Laws of a sonometer: \(f\propto1/L\) (fixed \(T, \mu\)), \(f\propto\sqrt T\) (fixed \(L, \mu\)), \(f\propto1/\sqrt\mu\) (fixed \(L, T\)).
  • Consecutive harmonics differ by \(f_1\). If two consecutive resonances are known, their difference gives the fundamental.

Pipes: closed and open

Closed at one end (node at the closed end, antinode at the open end): \(L = (2n-1)\lambda/4\), so \(f = (2n-1)\dfrac{v}{4L}\). Only odd harmonics: \(f_1, 3f_1, 5f_1, \ldots\) Consecutive resonances differ by \(2f_1\).

Open at both ends (antinodes at both ends): \(L = n\lambda/2\), so \(f = n\dfrac{v}{2L}\). All harmonics.

For the same length, the open pipe’s fundamental is twice the closed pipe’s. If an open pipe is closed at one end, its fundamental halves. An open pipe sounds richer because it has all harmonics.

End correction and the resonance tube

The antinode at an open end forms slightly outside the pipe, at a distance \(e\approx0.6r\) (\(r\) is the pipe’s inner radius). So the effective length is \(L + e\) for a closed pipe and \(L + 2e\) for an open pipe.

In a resonance-tube experiment with a tuning fork, the first two resonating lengths satisfy \(l_1 + e = \lambda/4\) and \(l_2 + e = 3\lambda/4\). Subtracting removes \(e\):

\(\lambda = 2(l_2 - l_1)\), \(v = 2f(l_2 - l_1)\), and \(e = \dfrac{l_2 - 3l_1}{2}\).

Question patterns
  • “A string vibrates at 420 Hz and 490 Hz in two consecutive modes. Find the fundamental.” (70 Hz)
  • “Consecutive resonances of a pipe are 425 Hz and 595 Hz. Is the pipe open or closed? Find its fundamental.”
  • “The third harmonic of a closed pipe equals the fundamental of an open pipe. Find the ratio of their lengths.”
  • “In a resonance tube, l₁ = 16 cm and l₂ = 50 cm with a 500 Hz fork. Find v and the end correction.”
  • “An open pipe is suddenly closed at one end. How does its fundamental change?” (halves)
Another common mistake

Do not count the closed pipe’s overtones as 2f₁, 3f₁. Its first overtone is the 3rd harmonic, its second overtone the 5th harmonic.

Another common mistake

If two consecutive resonances of an unknown pipe differ by \(\Delta f\), check whether the lower one is an odd multiple of \(\Delta f/2\). If yes, it may be a closed pipe with \(f_1 = \Delta f/2\).

Check the idea

8 Core

Beats and nearby frequencies

Two nearby frequencies produce an amplitude that grows and shrinks slowly. These loud–soft cycles are beats. Their rate equals the frequency difference, while the rapid oscillation is near the average frequency.

Beats help tune instruments: changing one frequency until the beats slow indicates the frequencies are getting closer. A beat rate alone gives the absolute difference and cannot tell which source is higher without another observation.

Beats and nearby frequencies
\[ f_b=|f_1-f_2| \]
Symbols and assumptions: f_b is beat frequency (beats/s or Hz); f₁ and f₂ are nearby source frequencies (Hz). Audible beats require suitable overlap and a small enough difference for the modulation to be perceived.
Other useful formulas
Resultant of two close frequencies
\[ y=2A\cos\left(\pi\Delta f\,t\right)\sin\left(2\pi\bar f\,t\right),\quad \Delta f=f_1-f_2,\ \bar f=\frac{f_1+f_2}{2} \]
Symbols and assumptions: y displacement (m); A amplitude of each wave (m); Δf frequency difference (Hz); f̄ average frequency, the pitch heard (Hz); t time (s).
Beat period
\[ T_{\rm beat}=\frac{1}{|f_1-f_2|} \]
Symbols and assumptions: T_beat time between successive loudest sounds (s); f₁, f₂ the two frequencies (Hz).
Visual experiment

Two nearby tones make a slow envelope

Vary the small frequency difference. Compare the rapid oscillation with the slowly changing combined amplitude.

The equation walkthrough below explains this experiment. Interactive controls load when JavaScript is available.

Predict before you change an input

Two equal-amplitude tones with exactly equal frequency and phase produce…

Use this in a question

Two nearby tones produce beats.

  1. Use beat frequency as the absolute difference of tone frequencies.
  2. If a fork is loaded, its frequency falls.
  3. Use whether beats increase or decrease to choose the correct unknown frequency.
Why the equation works · optional
  1. 1

    Add both disturbances

    \[ y=A\cos(2\pi f_1t)+A\cos(2\pi f_2t) \]

    Superposition adds displacements, not intensities.

  2. 2

    Separate carrier and envelope

    \[ y=2A\cos[\pi(f_2-f_1)t]\cos[\pi(f_1+f_2)t] \]

    The average frequency gives rapid oscillation; the difference controls the slow modulation.

  3. 3

    Count loudness maxima

    \[ f_{\rm beat}=|f_2-f_1| \]

    Intensity follows amplitude squared, so both positive and negative envelope peaks give loudness maxima. This removes the apparent factor-of-two ambiguity.

Model and symbols: Two equal-amplitude waves at the same point with frequencies f₁ = 20 Hz and f₂ close to it. These low illustrative frequencies make the oscillations visible; no audio is generated.

Watch the assumption

Five beats per second could mean a source is either 5 Hz above or 5 Hz below the reference.

Worked example

A 256 Hz fork and another fork produce 4 beats/s. What frequencies are possible for the other fork?

Show solution

It can be 252 or 260 Hz. A change that raises its frequency and reduces beats would identify 252 Hz as the starting value.

Worked example · Numerical

A tuning fork of 256 Hz and an unknown fork X produce 6 beats per second. When X is loaded with a little wax, the beat rate drops to 4 per second. Find the original frequency of X.

Show solution
  1. Possible values: \(256 \pm 6\), that is 262 Hz or 250 Hz.
  2. Wax lowers the frequency of X.
  3. If X were 262 Hz, lowering it (to, say, 260 Hz) brings it closer to 256 Hz: beats drop to 4. This matches.
  4. If X were 250 Hz, lowering it moves it farther away, so beats would increase. This does not match.

Answer: 262 Hz

More solved problem patterns
Worked example · Ratio

Two identical wires vibrate in unison at 400 Hz. The tension of one is increased by 2%. How many beats per second are heard?

Show solution
  1. \(f\propto\sqrt T\), so for a small change \(\Delta f/f = \tfrac12\,\Delta T/T = 1\%\).
  2. \(\Delta f = 0.01\times400 = 4\) Hz.
  3. An exact calculation gives \(400(\sqrt{1.02} - 1)\approx3.98\) Hz, so about 4 beats per second.

Answer: About 4 beats per second

Go deeper · extra explanation and derivations
one beat: 1/|f₁ − f₂|waxingwaningdashed: slowly varying amplitude
Two waves of 20 Hz and 22 Hz added together over one second. The amplitude (dashed envelope) swells and fades twice, giving 2 beats per second, while the fast oscillation runs at about 21 Hz.

Derivation: why the beat rate is f₁ − f₂

At one place, two sound waves of equal amplitude give

\(y = A\sin2\pi f_1t + A\sin2\pi f_2t = \left[2A\cos\pi(f_1 - f_2)t\right]\sin\pi(f_1 + f_2)t\).

  • The fast factor \(\sin\pi(f_1+f_2)t\) oscillates at the average frequency \((f_1+f_2)/2\). That is the pitch you hear.
  • The slow factor \(2A\cos\pi(f_1-f_2)t\) is the varying amplitude. Loudness depends on its size, not its sign.
  • \(|\cos\pi\Delta f\,t|\) peaks every \(1/\Delta f\) seconds. So there are \(|f_1 - f_2|\) loud moments (beats) per second.

When beats can be heard

The ear holds a sound impression for about 0.1 s. If the loud–soft cycles come faster than about 10 per second, they blur together. So beats are clearly heard only for small frequency differences, up to roughly 10 Hz. Larger differences are heard as a rough or a combined tone.

Finding an unknown frequency

Suppose a known fork of frequency \(f_0\) gives \(b\) beats per second with an unknown fork X. Then \(f_X = f_0\pm b\). To decide the sign, change X slightly:

  • Waxing (loading the prongs) lowers \(f_X\). Filing the prongs raises \(f_X\).
  • If lowering \(f_X\) makes the beats fewer, \(f_X\) was above \(f_0\): \(f_X = f_0 + b\).
  • If lowering \(f_X\) makes the beats more, \(f_X\) was below: \(f_X = f_0 - b\).
  • Reverse the reasoning for filing. For a string, more tension raises the frequency.

If the beats stay the same after the change, the frequency crossed over \(f_0\) (for example, from \(f_0 + b\) to \(f_0 - b\)).

Question patterns
  • “A fork of 256 Hz gives 6 beats with X. On waxing X, the beats become 4. Find the original frequency of X.”
  • “Filing X increases the beats. What was X’s frequency?”
  • “Two identical strings at 400 Hz. The tension in one is raised by 2%. How many beats per second?”
  • “How many beats are heard in 5 s from 300 Hz and 304 Hz sources?”
  • Beats combined with Doppler or with pipes and strings.
Another common mistake

Wax lowers a fork’s frequency; filing raises it. Students often reverse these.

Another common mistake

Beats per second equal the difference \(|f_1 - f_2|\), not half of it, even though the envelope factor contains \(\pi\Delta f\). The envelope has two loud peaks per cycle of the cosine.

Check the idea

9 Exam

Doppler effect for sound

A source moving toward an observer compresses the spacing of emitted wavefronts; an observer moving toward incoming waves encounters them faster. These are distinct effects because sound travels relative to the medium.

Choose signs by whether motion increases or decreases the observed frequency. For the simple collinear case below, both positive speed magnitudes describe approach. Recession reverses the appropriate sign. Wind requires velocities relative to the air.

Doppler effect for sound
\[ f'=f\frac{v+v_o}{v-v_s}\quad\text{(source and observer approaching)} \]
Symbols and assumptions: f emitted and f′ observed frequency (Hz); v sound speed relative to medium (m/s); v_o observer speed toward source and v_s source speed toward observer (m/s). Assume collinear subsonic motion and a stationary uniform medium.
Other useful formulas
General Doppler formula
\[ f'=f\,\frac{v\pm v_o}{v\mp v_s} \]
Symbols and assumptions: f′ observed and f emitted frequency (Hz); v speed of sound in still air (m/s); v_o observer speed and v_s source speed along the line joining them (m/s). Upper signs for motion toward the other; lower signs for motion away. Subsonic speeds.
Echo from a moving reflector
\[ f''=f\,\frac{v+u}{v-u} \]
Symbols and assumptions: f″ frequency of the echo heard by the moving source (Hz); f emitted frequency (Hz); v speed of sound (m/s); u speed of the source approaching a fixed wall, or of a reflector approaching a fixed source (m/s).
Approaching and receding source
\[ \frac{f'_{\rm app}}{f'_{\rm rec}}=\frac{v+v_s}{v-v_s} \]
Symbols and assumptions: f′_app, f′_rec frequencies heard by a stationary observer as the source approaches and recedes (Hz); v speed of sound (m/s); v_s source speed (m/s).
Sv_sahead: shorter λ,higher fbehind: longer λ,lower fGrey dots: where S was when each wavefront left it.
Wavefronts from a source S moving to the right at half the speed of sound. Each circle is centred where S was when it was emitted. Ahead of S the wavefronts crowd together (shorter wavelength, higher frequency); behind it they spread out.
Watch the assumption

A moving observer changes encounter rate; a moving source changes wavelength. Their denominators are therefore different.

Worked example

A stationary observer hears a 500 Hz source approaching at 20 m/s in air with v=340 m/s. Find heard frequency.

Show solution

f′=500×340/(340−20)=531.25 Hz. Approach raises the pitch.

Worked example · Numerical

A train whistle of 640 Hz approaches a platform at 20 m/s and then moves away at the same speed. Find the frequencies heard by a person on the platform. Speed of sound 340 m/s.

Show solution
  1. Approaching: \(f^{\prime} = 640\times340/(340 - 20) = 640\times340/320 = 680\) Hz.
  2. Receding: \(f^{\prime} = 640\times340/(340 + 20) = 640\times340/360\approx604\) Hz.
  3. Ratio: \(680/604.4 = 360/320 = 1.125\), as the formula \((v+v_s)/(v-v_s)\) predicts.

Answer: 680 Hz approaching, about 604 Hz receding

More solved problem patterns
Worked example · Numerical

A car approaches a large wall at 10 m/s, sounding a 500 Hz horn. Speed of sound 340 m/s. Find the frequency of the echo heard by the driver and the beat frequency with the direct sound.

Show solution
  1. Step 1, wall as a stationary observer of a moving source: \(f_1 = 500\times340/330\).
  2. Step 2, wall as a stationary source and driver as a moving observer approaching it: \(f_2 = f_1\times350/340 = 500\times350/330\).
  3. \(f_2\approx530.3\) Hz.
  4. Beats with the 500 Hz horn: about 30 per second (too fast to hear clearly as beats).

Answer: About 530 Hz; about 30 beats per second

Worked example · Ratio

At what speed must a sound source approach a stationary observer for the observed frequency to be double the emitted frequency?

Show solution
  1. \(v/(v - v_s) = 2\).
  2. \(v = 2v - 2v_s\), so \(v_s = v/2\).
  3. For comparison, an observer would have to approach at \(v\) (the speed of sound) to double the frequency: \((v + v_o)/v = 2\).

Answer: Half the speed of sound

Go deeper · extra explanation and derivations

Deriving the two effects separately

Moving source. A source of frequency \(f\) moves toward the observer at \(v_s\). In one period \(1/f\), a wavefront travels \(v/f\) but the source follows it by \(v_s/f\). So the wavelength ahead shrinks to \(\lambda' = (v - v_s)/f\). The observer receives \(f^{\prime} = v/\lambda' = f\dfrac{v}{v - v_s}\).

Moving observer. The wavelength is unchanged, \(\lambda = v/f\), but an observer moving toward the source at \(v_o\) meets the waves at relative speed \(v + v_o\). So \(f^{\prime} = (v + v_o)/\lambda = f\dfrac{v + v_o}{v}\).

Combining: \(f^{\prime} = f\dfrac{v \pm v_o}{v \mp v_s}\). Sign rule: use the upper signs for motion toward the other party (each raises \(f'\)), the lower signs for motion away.

Special cases

  • Same velocity. Source and observer moving together in still air: \(f^{\prime} = f\) (the factors cancel).
  • Wind blowing at \(w\) from source to observer: replace \(v\) by \(v + w\) (or \(v - w\) if it blows the other way). If nothing moves except the air, there is no shift.
  • Motion at an angle. Use only the component of velocity along the line joining source and observer. At the moment of closest approach (motion perpendicular to the line), there is no shift.
  • Reflection from a moving object. Treat it in two steps: the object first acts as a moving observer, then as a moving source. A car approaching a wall at \(u\) hears its own echo at \(f\dfrac{v+u}{v-u}\).
  • Source passing by. The pitch drops suddenly from \(f\dfrac{v}{v - v_s}\) to \(f\dfrac{v}{v + v_s}\).

Why source and observer motion are not symmetric

A source approaching at 100 m/s (with \(v = 340\) m/s) gives \(f^{\prime} = f\times340/240 = 1.42f\). An observer approaching at 100 m/s gives \(f^{\prime} = f\times440/340 = 1.29f\). The results differ because sound moves relative to the air, so “who is moving” relative to the air matters. Only the relative velocity matters for light, which needs no medium.

Question patterns
  • “A train whistle of 640 Hz approaches at 20 m/s and then recedes. Find both frequencies heard on the platform.”
  • “A car sounding its horn approaches a wall. Find the frequency of the echo heard by the driver and the beats with the direct sound.”
  • “At what speed must a source approach for the frequency heard to double?” (v/2)
  • “Source and observer both move.” Choose signs carefully.
  • Assertion–reason on why the formulas for a moving source and a moving observer differ.
Another common mistake

Do not put the observer’s speed in the denominator or the source’s speed in the numerator. A moving observer changes the encounter rate (numerator); a moving source changes the wavelength (denominator).

Another common mistake

When the source and observer move with the same velocity in still air, the frequency does not change, even though both are “moving”.

Check the idea

Chapter check

Explain each formula’s symbols and conditions without looking. Solve the examples before revealing their steps, then use chapter practice to check what needs revision.