In this chapter: transverse and longitudinal waves, the displacement relation for a progressive wave, amplitude, phase, wavelength, wave number and frequency, speed of waves on a string and of sound (Newton's formula and Laplace's correction), the principle of superposition, reflection of waves, standing waves and normal modes in strings and pipes, and beats.Transverse and longitudinal waves
Mechanical waves need a medium, and they travel because each part of the medium is coupled to its neighbours by elastic forces. The particles of the medium only oscillate about their mean positions; it is the disturbance that travels.
- Transverse wave: the particles oscillate perpendicular to the direction of travel, as on a stretched string. Transverse mechanical waves need a medium that resists shear, so they travel in solids and along strings but not through the bulk of a fluid.
- Longitudinal wave: the particles oscillate along the direction of travel, forming compressions and rarefactions. Sound in air is longitudinal. Longitudinal waves can travel in solids, liquids and gases.
Electromagnetic waves (light, radio waves) do not need any medium; they are covered in Class 12.
Displacement relation for a progressive wave
A sinusoidal wave travelling along the +x direction is described by
| Quantity | Meaning | Relation |
|---|---|---|
| Amplitude a | Maximum displacement of a particle from its mean position | always positive |
| Wavelength λ | Distance between two successive points in the same phase (crest to crest) | k = 2π/λ (angular wave number, rad m−1) |
| Period T | Time for one complete oscillation of a particle | ω = 2π/T (angular frequency) |
| Frequency ν | Oscillations per second (Hz) | ν = 1/T = ω/2π |
| Phase | (kx − ωt + φ); φ is the initial phase angle | fixes the state of motion of each particle |
Worked example: A wave on a string is described by y(x, t) = 0.005 sin(80.0x − 3.0t), with x and y in metres and t in seconds. Find the amplitude, wavelength, period, frequency and speed.Solution: a = 0.005 m = 5 mm. k = 80.0 m−1, so λ = 2π/80 ≈ 7.85 cm. ω = 3.0 s−1, so T = 2π/3 ≈ 2.09 s and ν = 1/T ≈ 0.48 Hz. v = ω/k = 3.0/80 ≈ 3.75 cm s−1.
Speed of a travelling wave
The speed of a mechanical wave depends on the elastic (restoring) property and the inertial (mass) property of the medium, not on the frequency or amplitude of the source.
Worked example: A steel wire 12.0 m long has a mass of 2.10 kg. What tension is needed so that the speed of a transverse wave on it equals the speed of sound in dry air at 20 °C, 343 m s−1?Solution: μ = 2.10/12.0 = 0.175 kg m−1. T = v2μ = (343)2 × 0.175 ≈ 2.06 × 104 N.
Speed of sound in a gas
Newton assumed the compressions and rarefactions in sound are isothermal, so B = P, giving v = √(P/ρ). For air at STP this is about 280 m s−1, about 15% less than the measured value. Laplace pointed out that the changes are so fast that there is no time for heat to flow, so they are adiabatic, and B = γP:
Since P/ρ = RT/M for an ideal gas, the speed of sound does not change with pressure at constant temperature, but increases with temperature as v ∝ √T. It is slightly higher in humid air, because moist air is less dense than dry air. Sound travels faster in liquids and solids than in gases, because their elastic moduli are much larger.
Principle of superposition
When two or more waves overlap, the resultant displacement at any point is the algebraic sum of the displacements due to each wave: y = y1 + y2 + ... The waves pass through each other and continue unchanged. For two waves of the same amplitude a and frequency with a phase difference φ, the resultant amplitude is 2a cos(φ/2): 2a when they are in phase (constructive interference) and zero when they are π out of phase (destructive interference).
Reflection of waves and standing waves
A wave reaching a rigid boundary (a string tied to a wall) is reflected with a phase change of π, so a crest comes back as a trough. At an open boundary (a string end free to move, or the open end of a pipe) the wave is reflected without any phase change.
When a wave and its reflection travel in opposite directions and overlap, they form a standing (stationary) wave:
- Nodes (zero amplitude) occur at x = nλ/2; antinodes (maximum amplitude) at x = (n + ½)λ/2.
- Adjacent nodes (or antinodes) are λ/2 apart; a node and the next antinode are λ/4 apart.
String fixed at both ends
Both ends must be nodes, so the length holds a whole number of half-wavelengths: L = nλ/2.
Air columns
- Pipe closed at one end: node at the closed end, antinode at the open end, so L = (2n − 1)λ/4 and ν = (2n − 1)v/4L. Only odd harmonics are present (ν, 3ν, 5ν, ...).
- Pipe open at both ends: antinodes at both ends, so L = nλ/2 and ν = nv/2L. All harmonics are present, as for a string.
For a pipe 30 cm long with v = 340 m s−1, the fundamental is 340/(2 × 0.30) ≈ 567 Hz if both ends are open, and 340/(4 × 0.30) ≈ 283 Hz if one end is closed. The closed pipe has half the fundamental frequency of an open pipe of the same length.
Beats
When two sound waves of slightly different frequencies ν1 and ν2 are heard together, the loudness rises and falls periodically. These waxings and wanings are beats:
Musicians use beats to tune instruments: they adjust the tension of a string until the beats with a standard fork disappear. In problems, remember that loading a tuning fork with wax lowers its frequency, while filing its prongs raises it.
Worked example: A tuning fork of unknown frequency gives 4 beats per second with a 256 Hz fork. When the unknown fork is loaded with a little wax, the beat frequency decreases. Find its original frequency.Solution: The unknown frequency is 256 ± 4, that is, 252 Hz or 260 Hz. Wax lowers its frequency. If it were 252 Hz, lowering it would increase the difference from 256 Hz; if it were 260 Hz, lowering it would reduce the difference. Since the beats decreased, the frequency was 260 Hz.
Common mistakes: (1) Thinking the particles of the medium travel with the wave; they only oscillate. (2) Believing the speed of sound depends on pressure at constant temperature. (3) Including even harmonics for a pipe closed at one end. (4) Taking the distance between a node and the next antinode as λ/2 instead of λ/4. (5) Mixing up the phase change on reflection: π at a rigid end, none at an open end.JEE and NEET focus
- Reading a, λ, ν, k, ω and v from a given wave equation, and the direction of travel from the sign.
- Speed of a wave on a string, and how it changes with tension and mass per unit length.
- Newton's formula, Laplace's correction and the dependence of the speed of sound on temperature.
- Harmonics in strings, open pipes and closed pipes, including the ratio of their frequencies.
- Beats, including wax-loading and filing problems.
Practice questions
Sound waves in air are:
- Transverse
- Longitudinal
- Electromagnetic
- Neither transverse nor longitudinal
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The tension in a stretched string is made four times. The speed of transverse waves on it becomes:
- Four times
- Twice
- Half
- Unchanged
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Laplace's correction to Newton's formula for the speed of sound assumes that the compressions and rarefactions are:
- Isothermal
- Adiabatic
- Isobaric
- Isochoric
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In a standing wave, the distance between a node and the nearest antinode is:
- λ/4
- λ/2
- λ
- 2λ
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The fundamental frequency of a pipe closed at one end is 200 Hz. Which of these is also a natural frequency of the pipe?
- 400 Hz
- 600 Hz
- 800 Hz
- 100 Hz
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Two tuning forks of frequencies 256 Hz and 260 Hz are sounded together. The number of beats heard per second is:
- 2
- 4
- 8
- 258
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At constant temperature, if the pressure of air is doubled, the speed of sound in it:
- Doubles
- Increases by √2
- Halves
- Does not change
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A string 1 m long, fixed at both ends, carries waves at 200 m s−1. The frequency of its third harmonic is:
- 100 Hz
- 200 Hz
- 300 Hz
- 600 Hz




