Gr 11 · Waves & Sound · Deep Study

Understanding Waves

Waves carry energy without carrying matter. This guide builds from basic wave properties through the pendulum, Doppler effect, interference, and sound intensity — with the intuition behind every formula.

5 sections Worked examples with steps Checkpoint questions Intuition-first explanations
1
Wave Properties
All waves obey the same fundamental relationships between speed, frequency and wavelength

All waves — light, sound, water — obey the same fundamental relationships between speed, frequency and wavelength. Master these and you can analyse any wave.

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Why does wave speed stay constant in a medium?
The speed of a wave is determined by the medium it travels through — the stiffness and density of the material. The source only determines how often it vibrates (frequency). Once the wave is in the medium, the medium controls how fast that disturbance propagates. Change the source and you change the wavelength, not the speed.
A λ T (one period) distance / time displacement equilibrium
Anatomy of a Wave — amplitude A (orange), wavelength λ (green), period T (purple), dashed equilibrium line
The wave equation
v = f × λ
v = wave speed (m/s)  ·  f = frequency (Hz)  ·  λ = wavelength (m)
Period & Frequency
T = 1/f    f = 1/T
Wave speed
v = fλ
Sound in air (20°C)
≈ 340 m/s
★ Easy
Finding wavelength from frequency
A sound wave has frequency 440 Hz in air (v = 340 m/s). What is its wavelength?
Show solution
1
Identify known values
f = 440 Hz    v = 340 m/s    find: λ
2
Rearrange v = fλ
λ = v / f = 340 / 440 = 0.773 m
Answer: λ = 0.773 m (≈ 77 cm)
★★ Medium
Wave entering a new medium
A wave has λ = 0.25 m and T = 0.002 s. It then enters a new medium where v = 250 m/s. What is the new wavelength?
Show solution
1
Find frequency (stays constant across media)
f = 1/T = 1/0.002 = 500 Hz
2
Verify original speed
v = fλ = 500 × 0.25 = 125 m/s
3
New wavelength in new medium (f unchanged)
λ = v/f = 250/500 = 0.5 m
Answer: New λ = 0.5 m (frequency stays 500 Hz; wavelength doubles as speed doubles)
✓ Checkpoint 1
a) A sound wave has f = 200 Hz in air (v = 340 m/s). Find λ.
b) A wave has λ = 0.4 m and v = 320 m/s. Find f and T.
c) When a wave enters a denser medium and slows down, does the frequency change? Why?

a) λ = v/f = 340/200 = 1.7 m

b) f = v/λ = 320/0.4 = 800 Hz    T = 1/f = 1/800 = 0.00125 s

c) No. Frequency is set by the source and is unchanged as the wave crosses into a new medium. Only wavelength changes (it shrinks as v decreases, since f stays constant and v = fλ).

2
Pendulum & SHM
Why a pendulum clock keeps perfect time regardless of how far it swings

The pendulum is the classic example of Simple Harmonic Motion (SHM). Its period depends only on length and gravity — not on mass or how far it swings (for small angles). This property is called isochronism and is why pendulum clocks work.

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Why is period independent of amplitude?
Swing wider and the bob travels a longer arc. But it also moves faster (gravity pulls harder from further out). These two effects exactly cancel for small angles: more distance, more speed — same time. This is a deep property of any system where restoring force is proportional to displacement.
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Why is period independent of mass?
A heavier bob experiences more gravitational force, but it also has more inertia — harder to accelerate. The two effects cancel exactly. This is the same reason all objects fall at the same rate in a vacuum: gravity scales with mass, and so does inertia.
L max Eₚ zero Eₖ max Eₚ zero Eₖ max Eₖ zero Eₚ T = 2π√(L/g)
Simple Pendulum — Energy and Period. Purple bob at center has maximum kinetic energy; grey bobs at extremes have maximum potential energy.
Pendulum period
T = 2π √(L / g)
L = string length (m)  ·  g = 9.8 m/s²  ·  T in seconds
Pendulum period
T = 2π√(L/g)
Spring-mass period
T = 2π√(m/k)
Restoring force
F = −kx
★ Easy
Pendulum period on Earth
A pendulum has L = 1.0 m. Find its period and frequency on Earth (g = 9.8 m/s²).
Show solution
1
Apply T = 2π√(L/g)
T = 2π × √(1.0/9.8) = 2π × √(0.1020) = 2π × 0.3194 = 2.007 s
2
Find frequency
f = 1/T = 1/2.007 = 0.498 Hz
Answer: T ≈ 2.01 s    f ≈ 0.498 Hz
★★ Medium
Finding length from period; spring-mass system
A pendulum has T = 3.0 s on Earth. Find L. Also: a spring (k = 200 N/m) holds a 0.5 kg mass. Find its period.
Show solution
1
Rearrange T = 2π√(L/g) for L
T/(2π) = √(L/g)  →  (T/2π)² = L/g  →  L = g(T/2π)²
2
Substitute values
L = 9.8 × (3.0 / 6.283)² = 9.8 × (0.4775)² = 9.8 × 0.2280 = 2.23 m
3
Spring-mass period
T = 2π√(m/k) = 2π√(0.5/200) = 2π√(0.0025) = 2π × 0.05 = 0.314 s
Answer: L ≈ 2.23 m    Spring-mass T ≈ 0.314 s
✓ Checkpoint 2
a) Find T for a pendulum with L = 0.5 m on Earth (g = 9.8 m/s²).
b) Same pendulum on the Moon (g = 1.62 m/s²). Find T.
c) Why does doubling the mass of a pendulum bob not change the period?

a) T = 2π√(0.5/9.8) = 2π√(0.051) = 2π × 0.226 = 1.42 s

b) T = 2π√(0.5/1.62) = 2π√(0.309) = 2π × 0.556 = 3.49 s  (much slower on the Moon)

c) Mass cancels out. Gravity pulls harder on a heavier bob, but that same mass has more inertia and resists acceleration equally. The T = 2π√(L/g) formula contains no mass term — they cancel algebraically.

3
Doppler Effect
Why a siren sounds higher as an ambulance approaches and lower as it recedes

When a source of waves and an observer are moving relative to each other, the observed frequency differs from the emitted frequency. The actual wave speed through the medium does not change.

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Why does an approaching source sound higher-pitched?
As the source moves toward you, each successive wavefront is emitted from a position closer to you. The wavefronts pile up — they arrive more frequently. More wavefronts per second means higher observed frequency. When the source recedes, wavefronts spread apart and arrive less frequently — lower pitch.
Doppler formula
fobs = fs × (v ± vobs) / (v ∓ vs)
Top sign: + when observer moves toward source, − when moving away. Bottom sign: − when source moves toward observer, + when moving away.
⚠️
Doppler changes frequency, NOT wave speed. The wave still travels at v = 340 m/s through air regardless of how fast the ambulance drives. The observer hears more (or fewer) waves per second, not faster waves.
▶ src vₛ 👁 higher frequency wavefronts closer 👁 lower frequency wavefronts farther
Doppler Effect — Moving Source Compresses Wavefronts. Source moves right: wavefronts bunch up ahead (higher pitch for right observer), spread out behind (lower pitch for left observer).
★ Easy
Approaching ambulance
An ambulance siren emits f = 800 Hz. It drives toward a stationary observer at vs = 30 m/s. vsound = 340 m/s. What frequency does the observer hear?
Show solution
1
Identify: source moving toward stationary observer
Use − on bottom (source approaching): fobs = fs × v / (v − vs)
2
Substitute
fobs = 800 × 340 / (340 − 30) = 800 × 340 / 310 = 800 × 1.097 = 876 Hz
Answer: fobs ≈ 876 Hz (higher than 800 Hz because source approaches)
★★ Medium
Moving observer; receding source
Source emits f = 600 Hz (stationary). Observer moves toward source at 20 m/s. Then: source moves away at 25 m/s from a stationary observer. vsound = 340 m/s.
Show solution
1
Case A: observer approaching, source stationary
fobs = 600 × (340 + 20) / 340 = 600 × 360/340 = 600 × 1.059 = 635 Hz
2
Case B: source receding, observer stationary
fobs = 600 × 340 / (340 + 25) = 600 × 340/365 = 600 × 0.932 = 559 Hz
Answer: Case A: 635 Hz (higher)    Case B: 559 Hz (lower)
✓ Checkpoint 3
a) A train horn emits 500 Hz and approaches at 40 m/s. vsound = 340 m/s. What does a stationary observer hear?
b) Same train receding at 40 m/s. What frequency does the observer hear?
c) Does the Doppler effect change the speed of sound reaching the observer?

a) fobs = 500 × 340/(340−40) = 500 × 340/300 = 567 Hz

b) fobs = 500 × 340/(340+40) = 500 × 340/380 = 447 Hz

c) No. The speed of sound in air is determined by the medium (~340 m/s at 20°C) and is unaffected by source or observer motion. Only the observed frequency changes.

4
Interference & Standing Waves
What happens when two waves meet, and how guitars and pipes produce musical notes

When two waves overlap in the same medium, their displacements add (superposition). If crests meet crests, the result is louder (constructive). If a crest meets a trough, they cancel (destructive). Standing waves form when a wave reflects back on itself, creating fixed nodes and antinodes.

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Why do nodes stay perfectly still?
At a node, the forward-travelling wave and the reflected wave always arrive with exactly opposite displacements — they cancel at every instant. It's permanent destructive interference at that point. Between nodes, the wave oscillates up and down (antinode region) but the nodal points themselves never move.
In phase → constructive Out of phase → destructive resultant: double amplitude wave 1 wave 2 result resultant: zero (flat line) wave 1 wave 2 (inverted) result
Constructive interference (left): in-phase waves add to double amplitude. Destructive interference (right): opposite-phase waves cancel completely.
n=1 N A N fundamental (λ = 2L) n=2 N A N A N 2nd harmonic (λ = L) n=3 N A N A N A N 3rd harmonic (λ = 2L/3)
Standing Waves — Harmonics on a String. N = node (no movement), A = antinode (maximum displacement). Solid and dashed lines show the two extreme positions.
Interference conditions
Constructive: path diff = nλ     Destructive: path diff = (n + ½)λ
n = 0, 1, 2, 3, …
String / open pipe
fn = nv/(2L), n=1,2,3…
Closed pipe
fn = nv/(4L), n=1,3,5…
★ Easy
Guitar string harmonics
A guitar string has L = 0.65 m and v = 400 m/s. Find the fundamental frequency and the second harmonic.
Show solution
1
Fundamental (n = 1)
f1 = 1 × v/(2L) = 400/(2 × 0.65) = 400/1.30 = 307.7 Hz
2
Second harmonic (n = 2)
f2 = 2 × 307.7 = 615.4 Hz
Answer: f1 ≈ 308 Hz    f2 ≈ 615 Hz
★★ Medium
Two-speaker interference
Two speakers emit f = 340 Hz (v = 340 m/s, so λ = 1 m). Point P is 4.0 m from speaker 1 and 4.5 m from speaker 2. Constructive or destructive?
Show solution
1
Find wavelength
λ = v/f = 340/340 = 1.0 m
2
Calculate path difference
path difference = 4.5 − 4.0 = 0.5 m = ½λ
3
Compare to conditions

path diff = (0 + ½)λ → this matches the destructive condition with n = 0.

Answer: Destructive interference — P is a quiet point
✓ Checkpoint 4
a) A string has L = 0.8 m and v = 320 m/s. Find f1, f2, and f3.
b) Two waves have a path difference of 2.5λ. Is the interference constructive or destructive?
c) Where are the nodes on a standing wave, and why don't they move?

a) fn = nv/(2L):   f1 = 320/(1.6) = 200 Hz    f2 = 400 Hz    f3 = 600 Hz

b) 2.5λ = (2 + ½)λ, so n=2 → this is the destructive pattern. Destructive interference (quiet point).

c) Nodes sit at positions where the forward and reflected waves always arrive with opposite phase — they permanently destructively interfere. No matter what instant you look, the displacements cancel exactly at those points, so they never move.

5
Sound Intensity
Why doubling your distance from a speaker makes it much quieter than you expect

Sound intensity (I) is the power per unit area (W/m²). As sound spreads outward spherically from a point source, the same power is spread over an ever-larger sphere. The surface area of a sphere grows as r², so intensity falls as 1/r².

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Why does intensity follow an inverse square law?
Imagine the energy spreading over a sphere of radius r. Area = 4πr². If you double r, the area quadruples, so the energy per unit area (intensity) drops to one quarter. Move three times as far away and intensity drops to 1/9. This geometric spreading is why sound fades so quickly with distance.
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Why use decibels?
Our ears respond logarithmically. A sound 10 times more intense feels roughly twice as loud (not 10 times). The decibel scale compresses this enormous range (10&sup-;¹² to 10² W/m²) into a manageable 0–140 dB scale that reflects how we actually perceive loudness.
source r 2r 3r I₀ I₀/4 (4× area) I₀/9 (9× area) I ∝ 1/r² double distance → 1/4 intensity
Inverse Square Law — I ∝ 1/r². Same energy spreads over 4× the area at 2r, and 9× the area at 3r. Intensity drops sharply with distance.
Sound intensity level
L = 10 log(I / I0)   dB
I0 = 10&sup-;¹² W/m² (threshold of hearing)  ·  +10 dB = ×10 intensity
Inverse square law
I ∝ 1/r²
Intensity ratio
I2/I1 = (r1/r2
Threshold
I0 = 10&sup-;¹² W/m²
★ Easy
Inverse square law
The intensity at 2 m from a speaker is 0.04 W/m². What is the intensity at 4 m?
Show solution
1
Apply the inverse square ratio
I2 = I1 × (r1/r2)² = 0.04 × (2/4)² = 0.04 × 0.25
2
Calculate
I2 = 0.04 × 0.25 = 0.01 W/m²
Answer: I = 0.01 W/m² (one quarter of original, as expected when distance doubles)
★★ Medium
Decibel calculation; inverse square at new distance
I = 10&sup-;&sup5; W/m². Find the dB level. Then: intensity at 6 m = 2×10&sup-;&sup4; W/m². Find intensity at 18 m.
Show solution
1
Decibel level
L = 10 log(10&sup-;&sup5; / 10&sup-;¹²) = 10 log(10&sup7;) = 10 × 7 = 70 dB
2
Intensity at 18 m (distance triples from 6 m)
I = 2×10&sup-;&sup4; × (6/18)² = 2×10&sup-;&sup4; × (1/3)² = 2×10&sup-;&sup4; × 1/9 = 2.22×10&sup-;&sup5; W/m²
Answer: 70 dB    I at 18 m ≈ 2.22 × 10&sup-;&sup5; W/m²
✓ Checkpoint 5
a) I = 10&sup-;&sup8; W/m². What is the dB level?
b) The intensity of a sound triples. By how many dB does the level increase? (Hint: 10 log 3 ≈ 4.77)
c) Why does moving twice as far from a speaker make it sound much quieter than expected?

a) L = 10 log(10&sup-;&sup8;/10&sup-;¹²) = 10 log(10&sup4;) = 10 × 4 = 40 dB

b) ΔL = 10 log(3I/I) = 10 log(3) ≈ 10 × 0.477 = 4.77 dB

c) Because of the inverse square law. Doubling the distance reduces intensity to 1/4 (not 1/2). On the dB scale that's a drop of 10 log(1/4) = −6 dB. Our ears perceive this as noticeably quieter — the drop is steeper than intuition suggests because intensity falls with the square of distance.