Waves — Class XI Physics Notes | eduPhysics
eduPhysics / Notes / Class XI / Waves
Class XI · Chapter 14

Waves

The final chapter of Class XI: how disturbances travel through a medium without carrying the medium itself along — and why a passing train's horn changes pitch as it goes by.

● Hard ⏱ 32 min read 🎯 42 practice questions 📊 Not yet revised
0% complete — mark sections read as you go

1Transverse and Longitudinal Waves

A wave transports energy and disturbance through a medium, without the medium's particles themselves travelling along with it — each particle simply oscillates around its own fixed position (Chapter 13's SHM, in fact) as the disturbance passes through.

  • Transverse waves: particles oscillate perpendicular to the direction the wave travels — like waves on a plucked string, or light itself (Class XII, Chapter 10).
  • Longitudinal waves: particles oscillate parallel to the direction of travel, as compressions and rarefactions — sound waves in air are the classic example.

2Displacement Relation in a Travelling Wave

Progressive (Travelling) Wave
y(x, t) = A sin(kx − ωt + φ)
A = amplitude, k = wave number (= 2π/λ), ω = angular frequency (= 2π/T), φ = initial phase. This single expression describes the displacement of every particle in the medium, at every position x and every time t.

3Speed of a Travelling Wave

Wave Speed
v = ω/k = λ/T = f λ
The wave speed depends entirely on the properties of the medium it travels through — not on the wave's amplitude or frequency, which is exactly why every colour of light travels at the same speed in vacuum, and every musical note travels through air at the same speed.

4Speed of a Wave on a Stretched String

Wave Speed on a String
v = √(T/μ)
T = tension in the string, μ = mass per unit length. Tighter strings (higher T) or lighter strings (lower μ) both carry waves faster — exactly why tuning a guitar string tighter raises its pitch.

5Speed of Sound in a Medium

Speed of Sound (Solids/Liquids)
v = √(B/ρ)
B = bulk modulus (Class XI, Chapter 8), ρ = density. Stiffer, less dense materials carry sound faster — which is exactly why sound travels faster through steel than through air.
Speed of Sound in a Gas (Laplace's Correction)
v = √(γP/ρ)
γ = ratio of specific heats (Cp/Cv, Chapter 12), P = pressure. Newton originally proposed v = √(P/ρ) assuming isothermal compression, but this under-predicted the measured speed of sound; Laplace corrected it by recognising that sound compressions happen too fast for heat to escape — making the process adiabatic instead, which is where the γ comes from.

6Principle of Superposition

Superposition of Waves
y = y₁ + y₂
When two or more waves overlap at the same point, the net displacement is simply the algebraic sum of what each wave would produce alone. This one simple rule underlies interference, beats, and standing waves — everything in the rest of this chapter builds on it.

7Reflection of Waves and Standing Waves

A wave reflecting off a fixed boundary comes back inverted; off a free boundary, it reflects without inversion. When a wave and its own reflection overlap and superpose, they can combine into a standing wave — a pattern that oscillates in place rather than travelling, with fixed points of zero displacement (nodes) and maximum displacement (antinodes).

Standing Wave (String Fixed at Both Ends)
y = 2A sin(kx) cos(ωt)
Notice this factors into a purely spatial part and a purely time-dependent part — every point oscillates in place with an amplitude that depends on its position, unlike a travelling wave where the whole pattern moves.

8Normal Modes of a Stretched String

A string fixed at both ends can only sustain standing waves where both ends are nodes — this constraint allows only specific, discrete frequencies.

Allowed Frequencies (String, Both Ends Fixed)
fn = n v / (2L), n = 1, 2, 3, ...
L = string length. n = 1 gives the fundamental (lowest) frequency; higher n gives successive harmonics (overtones) — exactly the physics behind every stringed instrument's tuning and timbre.

9Normal Modes of Air Columns

Pipe Open at Both Ends
fn = n v / (2L), n = 1, 2, 3, ...
All harmonics allowed — same form as the string case.
Pipe Closed at One End
fn = n v / (4L), n = 1, 3, 5, ... (odd only)
The closed end must be a displacement node, and the open end an antinode — a constraint that only permits odd harmonics, which is exactly why instruments like a clarinet (effectively closed at the reed end) sound noticeably different in tone from an open flute of similar length.

10Beats

Superpose two sound waves of slightly different frequencies, and the resulting sound periodically swells and fades in loudness — a slow, audible throbbing called beats, caused by the two waves drifting in and out of phase with each other over time.

Beat Frequency
fbeat = |f₁ − f₂|
Musicians use this directly: tune an instrument against a reference pitch, and listen for the beats to slow down and vanish as the two frequencies converge — no electronic tuner required.

11The Doppler Effect

The frequency you actually hear changes when there's relative motion between a sound source and an observer — higher pitch when they're approaching, lower when receding. This is the Doppler effect, instantly familiar from a passing ambulance's siren shifting pitch as it passes.

Doppler Effect (Source and/or Observer Moving)
f′ = f × (v ± vo) / (v ∓ vs)
v = speed of sound, vo = observer's speed, vs = source's speed. Signs are chosen based on whether the motion is toward or away from the other party — approach always raises the observed frequency, receding always lowers it.

Formula Summary

Travelling Wave
y = A sin(kx−ωt)
Wave Speed
v = fλ
Speed on a String
v = √(T/μ)
Speed of Sound (Gas)
v = √(γP/ρ)
String, Both Ends Fixed
f_n = nv/2L
Pipe, One End Closed
f_n = nv/4L (odd n)
Beat Frequency
f_beat = |f₁−f₂|
Doppler Effect
f′ = f(v±vₒ)/(v∓v_s)

Solved Examples

Example 1 · Wave Speed on a String

A string of mass 5 g and length 2 m is under a tension of 60 N. Find the speed of a transverse wave on the string.

Solution: μ = mass/length = 0.005 kg / 2 m = 0.0025 kg/m.

v = √(T/μ) = √(60/0.0025) = √24000 ≈ 154.9 m/s.

Example 2 · Doppler Effect

A train sounds its horn at 400 Hz while moving toward a stationary observer at 30 m/s. Find the apparent frequency heard. (Speed of sound = 340 m/s)

Solution: f′ = f × v/(v − vs) = 400 × 340/(340 − 30) = 400 × 340/310 ≈ 438.7 Hz.

The observer hears a noticeably higher pitch than the horn's actual 400 Hz, exactly as expected for an approaching source.

Quick Check

1. The speed of a transverse wave on a stretched string depends on:
Frequency and wavelength only
✓ Tension and mass per unit length (correct)
Amplitude only
The wave's phase constant
2. The beat frequency produced by two sound waves equals:
The sum of the two frequencies
✓ The difference between the two frequencies (correct)
The average of the two frequencies
The product of the two frequencies
← Previous · Chapter 13

Oscillations

Back
🎓 Syllabus Complete

All 14 Class XI chapters done

View All Chapters

Share this:

Like this:

Like Loading…