Waves

What this chapter is about

This chapter introduces you to wave motion, one of the most important ideas in physics. A wave carries energy and information from one place to another without carrying matter along with it. When you throw a stone into a still pond, ripples spread outward, yet the water itself does not travel across the pond; each bit of water simply moves up and down. This distinction between matter transport and energy transport is central to understanding waves.

You will study two main types of mechanical waves: transverse waves, where particles oscillate perpendicular to the wave direction (like a vibrating string), and longitudinal waves, where particles oscillate along the wave direction (like sound in air). The chapter develops the mathematical description of a sinusoidal travelling wave, relates wave speed to the medium's properties, and explains superposition, interference, standing waves, beats, and the Doppler effect.

By the end, you should be able to write the equation of a travelling wave, calculate its speed in strings and air columns, analyse standing waves on strings and in pipes, and solve problems on beats and the change of frequency due to motion of source or observer.

Key ideas

• A mechanical wave requires a material medium and arises because of the medium's elasticity and inertia; electromagnetic waves need no medium.
• In a transverse wave the particle displacement is perpendicular to the wave velocity; in a longitudinal wave it is parallel.
• The wave equation y(x, t) = A sin(kx − ωt + φ) describes a sinusoidal wave travelling in the +x direction with amplitude A, angular wave number k = 2π/λ, angular frequency ω = 2πf, and initial phase φ.
• Wave speed v = fλ = ω/k; for a string under tension T with linear mass density μ, v = √(T/μ).
• When two or more waves meet, the resultant displacement is the algebraic sum of individual displacements (principle of superposition).
• Standing waves form when two identical waves travelling in opposite directions superpose, producing nodes (zero displacement) and antinodes (maximum displacement) at fixed positions.
• Beats occur when two waves of slightly different frequencies superpose; the beat frequency equals the magnitude of the difference of the two frequencies.
• The Doppler effect is the apparent change in frequency when source or observer or both move relative to the medium.

Formulas and facts to remember

1. Travelling wave equation: y = A sin(kx − ωt) for motion in +x; y = A sin(kx + ωt) for motion in −x.
2. Wave number and wavelength: k = 2π/λ.
3. Angular frequency and period: ω = 2π/T = 2πf.
4. Wave speed: v = fλ = ω/k.
5. Speed on a stretched string: v = √(T/μ), where T is tension (N) and μ is mass per unit length (kg/m).
6. Speed of sound in a gas (Newton-Laplace): v = √(γP/ρ), where γ = Cp/Cv, P is pressure, ρ is density.
7. Standing wave on a string fixed at both ends: frequencies fₙ = n v/(2L), n = 1, 2, 3, …; first harmonic (fundamental) n = 1.
8. Open pipe (both ends open): fₙ = n v/(2L), n = 1, 2, 3, …; closed pipe (one end closed): fₙ = n v/(4L), n = 1, 3, 5, … (odd harmonics only).
9. Beat frequency: f_beat = |f₁ − f₂|.
10. Doppler effect for sound (observer and source along the line of propagation): f′ = f (v + vₒ)/(v − vₛ), taking signs so that motion towards each other increases frequency.

Worked examples

Example 1 – Speed of a wave on a wire

A steel wire of length 80 cm has mass 4.0 g and is stretched with a tension of 100 N. Find the speed of transverse waves on the wire.

Solution

Linear mass density μ = mass/length = (4.0 × 10⁻³ kg)/(0.80 m) = 5.0 × 10⁻³ kg/m.

Wave speed v = √(T/μ) = √(100 / 5.0 × 10⁻³) = √(2.0 × 10⁴) = 141 m/s (approx.).

Example 2 – Fundamental frequency of a closed pipe

A glass tube 34 cm long is closed at one end. Taking the speed of sound in air as 340 m/s, find its fundamental frequency.

Solution

For a pipe closed at one end, the fundamental (n = 1) has wavelength λ₁ = 4L.

Here L = 0.34 m, so λ₁ = 4 × 0.34 = 1.36 m.

Fundamental frequency f₁ = v/λ₁ = 340 / 1.36 = 250 Hz.

Example 3 – Beat frequency

Two tuning forks are sounded together. One has frequency 256 Hz; the other is slightly de-tuned. A listener hears 4 beats per second. What are the possible frequencies of the second fork?

Solution

Beat frequency = |f₁ − f₂| = 4 Hz.

So f₂ = 256 + 4 = 260 Hz or f₂ = 256 − 4 = 252 Hz.

Without additional information (such as loading one fork with wax and noting the change in beat frequency), both values are possible.

Common mistakes

• Confusing wavelength with amplitude → wavelength is the spatial period (distance for one complete cycle), amplitude is the maximum displacement from equilibrium.
• Using f = n v/(2L) for a closed pipe → closed pipes support only odd harmonics; the correct formula is f = n v/(4L) with n = 1, 3, 5, …
• Forgetting that wave speed depends on the medium, not on frequency or amplitude → changing frequency changes wavelength, not speed in a given medium.
• Sign errors in the Doppler formula → decide a positive direction first and apply signs consistently; motion towards increases frequency, motion away decreases it.
• Treating nodes and antinodes in standing waves as travelling points → they are fixed in space.

Quick revision

1. A wave transfers energy, not matter, through a medium.
2. v = fλ; for a string v = √(T/μ).
3. Standing waves have fixed nodes and antinodes; frequencies depend on boundary conditions.
4. Beat frequency equals the absolute difference of the two interfering frequencies.
5. Doppler effect: approach raises apparent frequency; recession lowers it.
6. Open pipe has all harmonics; closed pipe has only odd harmonics.