FSc Part 1 Physics Chapter 7 Complete Guide Book — Waves and Vibrations

Throw a stone into a still pond and watch what happens. The stone itself sinks and stays put, but the rings it makes keep travelling outward, getting wider and wider. That travelling pattern is a wave. The book puts it neatly: a wave is a regular disturbance or variation that carries energy and spreads out from the source. Sunlight is the grandest example — energy from the Sun crosses millions of kilometres of empty space and reaches us as light waves, which are electromagnetic waves and need no medium at all.

Inside a medium the picture is a little different, and it is worth getting straight early. The particles of the medium do not travel with the wave. Each particle just wobbles up and down, or to and fro, about its own mean position, and hands the disturbance on to its neighbour. The energy moves; the material stays more or less where it was. That is why a cork on a pond bobs up and down as a ripple passes but does not get carried to the bank.

Six quantities describe any wave, and examiners love all six, so learn them as a set. Amplitude (A) is the biggest displacement of a particle from its rest position — how tall the wave is. Frequency (f) counts how many oscillations pass a point every second, in hertz. Wavelength (λ) is the distance between two neighbouring points that are doing exactly the same thing at the same time — two crests, for instance. Period (T) is the time for one complete oscillation, and it is simply the upside-down of frequency, T = 1/f. Speed (v) is how fast the pattern travels, and phase (φ) tells you where a point is in its cycle at a given moment.

Speed, frequency and wavelength lock together in the one equation you will use again and again in this chapter. A crest moves one full wavelength λ in one period T, so v = λ/T, and since 1/T = f, v = fλ (equation 7.1). If frequency goes up while speed stays fixed, wavelength must shrink — the two share the same product, so one cannot grow without the other shrinking.

Worked example 7.2 (book): a radio signal has λ₁ = 1500 m at f₁ = 200 kHz. What wavelength belongs to a transmitter at f₂ = 1000 kHz, and how fast do the radio waves travel? Both signals are radio waves, so both travel at the same speed: v = f₁λ₁ = (2.0 × 10⁵ Hz)(1500 m) = 3.0 × 10⁸ m s⁻¹ — the speed of light, as expected. Then λ₂ = v/f₂ = (3.0 × 10⁸)/(1.0 × 10⁶) = 300 m. Five times the frequency, one-fifth the wavelength, same speed.

Not all waves are the same kind of thing, and the book sorts them into four families. Mechanical waves need a material medium — solid, liquid or gas — because they are disturbances of the material itself: water waves, sound waves in air, water or solids, and earthquake (seismic) waves. Electromagnetic waves need no medium and happily cross a vacuum: radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays, in that order of rising frequency. Quantum waves belong to particles — matter waves and de Broglie waves of electrons in atoms, and photon waves of light quanta. Surface waves run along a boundary between two media, like wind-driven waves on the sea surface or seismic surface waves on the ground.

Then comes the split that matters most for this chapter: which way do the particles vibrate compared with the direction the energy travels? In a transverse wave the particles vibrate at right angles to the direction of travel — think of a rope flicked up and down, or the crests and troughs of a water wave. In a longitudinal wave the particles vibrate along the direction of travel, bunching up into compressions and spreading out into rarefactions — exactly what sound does in air. A stretched spring (slinky) shows both beautifully: flick it sideways for transverse, push and pull its end for longitudinal.

📌 Book note (Do you know?): the book points out that these wave types are not just textbook labels — they are the ideas behind whole branches of physics, engineering and everyday life, from medical imaging with X-rays to the sound of a plucked guitar string. Whenever a later section says "transverse" or "longitudinal", come back to this picture of particle motion and the rest will follow.

What happens when two waves arrive at the same particle at the same time? The particle cannot follow two instructions at once, so it does the sensible thing: it adds them. That is the principle of superposition — if a particle is acted on by two waves at once, its resultant displacement is the algebraic sum of the individual displacements. Two crests meeting make a taller crest; a crest meeting a trough cancels out.

Work through the two clean cases. If the waves are in phase (crest on crest), y = y₁ + y₂, and if the two amplitudes are equal, y = 2y₁ — the wave doubles. If they are exactly out of phase (crest on trough), y = y₁ + (−y₂) = y₁ − y₂, and with equal amplitudes the result is zero — the particle does not move at all at that instant. The book generalises at once: for n waves, y = y₁ + y₂ + … + yₙ, or in function form, y(x, t) = y₁(x, t) + y₂(x, t) + … + yₙ(x, t). Here y(x, t) is the wave function: it tells you the displacement at position x at time t, i.e. the shape of the wave and how it changes.

One condition matters and is easy to forget in exams: superposition in this simple form applies to linear waves, or waves of small amplitude. And notice what the principle buys you — three whole phenomena fall out of it, and the rest of the chapter is really about them: two waves of the same frequency travelling the same way give interference; two of slightly different frequencies travelling the same way give beats; two of equal frequency travelling in opposite directions give stationary waves. Keep that three-way split in mind and the chapter stops feeling like a list.

📌 Book note (Ponder upon!): beside this section the book shows an interference pattern made by white light — bright and dark bands side by side. It is the same adding-up idea, only with light instead of water or sound. If two waves can cancel, darkness can be made by adding light to light — a strange thought the first time you meet it, and exactly what the next section explains.

Interference is superposition with a strict condition: the two waves must have the same frequency and travel in the same direction. The book's experiment makes it concrete. Two loudspeakers S₁ and S₂ are driven by one audio generator, so they vibrate in step — sources like this are called coherent sources. A microphone wired to a CRO (cathode ray oscilloscope) is moved around in front of them, and the CRO screen shows how big the sound is at each point.

At some points, labelled P₁, P₃ and P₅, a compression from one speaker arrives together with a compression from the other, and rarefaction meets rarefaction. The displacements add, the CRO shows a large wave, and your ear hears loud sound. That is constructive interference. The book works out the path difference ΔS = S₂P − S₁P at P₁ as 4½λ − 3½λ = λ, and at P₃ and P₅ it finds 0 and λ. The pattern is obvious: whenever the path difference is a whole number of wavelengths, the waves add. So the condition is ΔS = nλ, where n = 0, ±1, ±2, ±3, …

At other points, P₂ and P₄, a compression meets a rarefaction and the two cancel — the CRO line goes flat and the sound dies away. That is destructive interference. There the path difference works out to 4λ − 3½λ = ½λ, and at P₄ also ½λ in the book's figure. So the condition is a half-wavelength, three halves, five halves, and so on: ΔS = (2n + 1)λ/2. If you remember only one pair of formulas from this chapter, make it these two — and remember which one has the "+1".

Worked example 7.1 (book): two speakers 3.0 m apart emit a 344 Hz tone. A microphone moves on a line 4.0 m from the speaker line and hears maximum loudness at the centre point O (path difference zero, constructive) and at the next maximum P. For that next maximum the path difference must be one wavelength: λ = S₂P₁ − S₁P₁. From the right triangle, S₂P₁ = √(3² + 4²) = √25 = 5 m, while S₁P₁ = 4 m, so λ = 5 − 4 = 1 m. Then v = fλ = 344 × 1 = 344 m s⁻¹, the speed of sound. Notice the trick: the geometry gives you the wavelength, and the wave equation turns it into a speed.

📌 Book note (Interesting fact!): the same interference paints the coloured bands on a thin soap film lit by white light — different wavelengths interfere constructively at different thicknesses, so you see colour where the film is thicker or thinner. Sound interference on a CRO and colours on a soap bubble are one idea wearing two costumes.

Here is superposition doing a job in your pocket. A noise-cancelling headphone does not block sound with padding alone — it destroys it with sound. Step by step, as the book lists it: first, a microphone on the headphone picks up the ambient noise around you, the engine rumble or background chatter. Second, that signal goes to an amplifier and a processing unit inside the headphone. Third, the processor builds an "anti-noise" signal — the exact opposite of the incoming noise, same amplitude, opposite phase. Fourth, this anti-noise is played through the earpiece together with your music. Fifth, when anti-noise meets the real noise at your ear, the two cancel by destructive interference, and you hear mostly the music.

The book is honest about the limits, and so should you be in an answer: real headphones are an oversimplification of that five-step story. They use several microphones and complex algorithms to chase changing noise, and they work best on steady, low-frequency droning sounds rather than sudden sharp ones. But the core physics never changes — it is the out-of-phase case from section 7.2, y₁ + (−y₂) = 0, engineered into a product. If an exam asks for an application of superposition, this is the one to write first.

Now flip the third case of superposition: two waves of the same frequency and amplitude travelling in opposite directions. Their sum does not travel at all. The pattern stays in one place, swelling up and down, and the book calls it a stationary wave (also called a standing wave) — "standing" because the shape stands in space, oscillating between positive and negative values without marching forward or back. Guitar strings and organ pipes both live on this idea.

The book builds the picture point by point along a string, marking seven points 1 to 7, spaced λ/4 apart, and watching them at t = 0, T/4, T/2, 3T/4 and T. Points 1, 3, 5 and 7 always end up with zero resultant displacement — whatever one wave does, the other undoes. These are nodes, points of zero amplitude, and neighbouring nodes sit λ/2 apart. Points 2, 4 and 6, by contrast, swing with the sum of both amplitudes — these are antinodes, points of maximum amplitude, sitting midway between nodes, so a node and the next antinode are λ/4 apart. The whole pattern of nodes and antinodes is the stationary wave.

Why does the energy not escape? Because a node never moves, energy cannot flow past it. It stays trapped between nodes, changing form as the wave oscillates: when the antinodes are at their extreme positions the energy is all potential, and when they whip through their equilibrium positions it is all kinetic. A stationary wave is therefore a store of energy standing in place, not a delivery of energy from one place to another — the exact opposite of the progressive waves you will meet in section 7.10.

Take a string of length ℓ, clamp both ends, and stretch it with tension F. Pluck it and two waves run out from the pluck point, reflect off the clamped ends, and settle into a stationary wave. The ends cannot move, so they must be nodes. What happens in between depends on where you plucked.

(a) Plucked at the middle. The string vibrates in a single loop, with nodes at the two ends and one antinode in the middle. The distance between the two nodes is half a wavelength, so ℓ = λ₁/2, i.e. λ₁ = 2ℓ (7.2). The speed of waves on the string depends only on tension and mass per unit length m, not on where you plucked: v = √(F/m) (7.3). Frequency follows from v = fλ, so f₁ = v/λ₁ = v/(2ℓ), and substituting v gives f₁ = (1/2ℓ)√(F/m) (7.4). Only waves of this frequency are set up in this mode.

(b) Plucked at a quarter of its length. Now the string vibrates in two loops, with an extra node in the middle. Two node-gaps make one wavelength, so ℓ = λ₂ and λ₂ = ℓ (7.5) — half the wavelength of the first mode. Speed is unchanged (same tension, same string), so f₂ = v/λ₂ = v/ℓ = 2f₁. Two loops, double the frequency. The book's point is worth underlining: plucking harder does not change the pitch; plucking at a different point selects a different mode.

(c) Plucked at an arbitrary point. In general the string can vibrate in n loops with (n + 1) nodes and n antinodes, and the frequency is fₙ = n(v/2ℓ) = nf₁ (7.6) with wavelength λₙ = 2ℓ/n (7.7), for n = 1, 2, 3, … More loops, higher frequency, shorter wavelength — but fλ always stays equal to v. Three conclusions follow, and the book states them plainly. First, the allowed frequencies f₁, 2f₁, 3f₁, …, nf₁ form the harmonic series; f₁ is the fundamental (first harmonic), 2f₁ the second harmonic, and so on. Second, frequencies are quantised: only integral multiples are allowed, and waves at other frequencies are quickly damped out. Third, a musician tunes a string by changing tension (tightening the pegs) or changing the vibrating length (fingers on the neck) — both change f₁ through the formula above.

Worked example 7.3 (book): a stationary wave is set up in a string 120 cm long, fixed at both ends, vibrating in four segments at 120 Hz. Find its wavelength and fundamental frequency. (i) ℓ = 1.2 m, n = 4, and ℓ = nλₙ/2, so λ₄ = 2ℓ/n = (2 × 1.2)/4 = 0.6 m. (ii) fₙ = nf₁, so 120 = 4f₁ and f₁ = 30 Hz. Four segments means the fourth harmonic is sounding; the string's own fundamental is a quarter of that.

📌 Book note (Brain teaser): the book asks — a guitar string plucked at its centre, what harmonic is produced? From (a) above: one loop, nodes only at the ends, so it is the first harmonic (fundamental). And a second teaser on the next page asks how many nodes and antinodes form on a 2 m string sounding at 100 Hz — ⚠️ flagged honestly: as printed, that teaser cannot be answered uniquely, because the number of loops (the mode n) is not given; the frequency and length alone do not fix it. With n loops the answer would be (n + 1) nodes and n antinodes — but n itself must be known first, so no single number is invented here.

Air columns do the same thing as strings, but with longitudinal waves and one new rule about the ends. An organ pipe is a long hollow tube, open at both ends or closed at one end, in which sound sets up stationary waves. At an open end, air molecules are free to move, so the open end behaves as an antinode. At a closed end, the molecules are blocked, so it behaves as a node. Everything else is counting node-gaps.

Case 1 — open at both ends. (a) Fundamental: one node in the middle, antinodes at both ends. Two node-to-antinode gaps of λ₁/4 make the length, so ℓ = λ₁/2, λ₁ = 2ℓ (7.8), and f₁ = v/λ₁ = v/(2ℓ) (7.9). (b) Second mode: two nodes, three antinodes, ℓ = λ₂/4 + λ₂/2 + λ₂/4 = λ₂, so f₂ = v/ℓ = 2f₁. (c) In general, fₙ = n(v/2ℓ) = nf₁ and λₙ = 2ℓ/n (7.10) for n = 1, 2, 3, … — all harmonics, odd and even, are present, exactly like the string.

Case 2 — closed at one end. (a) Fundamental: one node at the closed end, one antinode at the open end — a single node-to-antinode gap, so ℓ = λ₁/4, λ₁ = 4ℓ (7.11), and f₁ = v/(4ℓ). A closed pipe of the same length sounds an octave lower than an open one — half the frequency, because its fundamental wavelength is twice as long. (b) Second mode: two nodes and two antinodes, ℓ = λ₃/4 + λ₃/2 = 3λ₃/4, so λ₃ = 4ℓ/3 and f₃ = 3v/(4ℓ) = 3f₁. Note the label: this is the third harmonic, even though it is only the second mode — the second harmonic is missing. (c) In general, fₙ = n(v/4ℓ) = nf₁ and λₙ = 4ℓ/n (7.13), but only for n = 1, 3, 5, … — only odd harmonics. The book's conclusion is a favourite exam line: a pipe open at both ends is richer in harmonics than one closed at one end.

Worked example 7.4 (book): an organ pipe 50 cm (0.5 m) long, speed of sound 350 m s⁻¹. (a) Open at both ends: f₁ = 350/(2 × 0.5) = 350 Hz, next harmonic f₂ = 2 × 350 = 700 Hz. (b) Closed at one end: f₁ = 350/(4 × 0.5) = 175 Hz, next harmonic (n = 3, not 2!) f₃ = 3 × 350/(4 × 0.5) = 525 Hz. Watch that trap: in a closed pipe the "next" harmonic after the fundamental is the third, not the second.

Worked example 7.5 (book): a church organ has pipes open at one end only, from 30 mm to 4 m long, speed of sound 340 m s⁻¹. Shortest pipe gives the highest fundamental: f_max = 340/(4 × 30 × 10⁻³) = 2833.33 Hz. Longest pipe gives the lowest: f_min = 340/(4 × 4) = 21.25 Hz. So the fundamental range is roughly 21 Hz to 2833 Hz — nearly the whole range of human hearing from one instrument, just by changing pipe length.

📌 Book notes: the book adds that in a real organ pipe the sound is driven by a wavering sheet of air, like a flute's jet, interacting with the pipe's upper lip to keep the oscillation going — and it teases you with a flute player whose flute sounds slightly sharp. The honest physics of that teaser: a warmer air column (or harder blowing) raises the speed of sound in the pipe, and since f = v/λ with λ fixed by the pipe length, the pitch rises — the flute goes sharp. That reasoning is derived here, not printed as a book answer, so treat it as explanation, not quotation.

Stationary waves are not only a string-and-pipe trick — they happen with microwaves too, and the book turns that into a measurable experiment. Microwaves are electromagnetic waves, called "micro" because their wavelengths are typically millimetres or centimetres, much shorter than radio waves. Confine them in a cavity or waveguide and they bounce back and forth, and when the microwave frequency matches the resonant frequency of the cavity, the incident and reflected waves superpose into a standing pattern of nodes and antinodes. Your kitchen microwave oven is the same physics: a magnetron makes the microwaves, a metal waveguide feeds them into a metal box, and the food sits in the standing pattern — which is also why the turntable rotates, so no part of the food stays stuck at a node.

The bench experiment is simple and elegant. A microwave source (transmitter) faces a metal reflector plate, with a probe detector on a stand between them, wired to a CRO. The waves travel to the plate, reflect back, and the outgoing and returning waves superpose into a stationary wave. Slide the probe (or the plate) along the line and the CRO shows the intensity rising and falling: strong at antinodes, weak at nodes. Measure the distance from one antinode to the next, and you have half a wavelength — double it for λ. One moving probe, one standing pattern, one wavelength: a complete measurement from a single idea.

Put a wall in a wave's way with a gap in it, and the wave does not simply march straight through the gap like a marching column. It bends around the edges and spreads out beyond the barrier. That bending and spreading is diffraction. It happens when a wave meets an obstacle or an opening (a slit) whose size is comparable to the wavelength of the wave. The rule of thumb is beautifully simple: the longer the wavelength, the greater the spreading, and vice versa. Sound bends around corners (which is why you hear someone in the next room before you see them), X-rays diffract off the regular rows of atoms in a crystal (the atomic spacing is of the order of the X-ray wavelength), and water waves do it in a tank where you can watch it happen.

The ripple tank is the apparatus that lets you watch it. It holds water, a vibrator (often a motorised needle) that makes straight or circular waves, obstacles such as small rectangular blocks or semicircular barriers, and gaps of different widths. Make the gap, run the vibrator, and repeat with different gap widths. When the gap is small compared with the wavelength, diffraction is strong: the waves bend well around the obstacle and spread into a semicircular pattern. As the gap widens relative to the wavelength, diffraction weakens and the waves pass through with less bending. The experiment shows the qualitative effect of gap width cleanly — narrow gap, big spread; wide gap, small spread — and that relationship, gap width versus wavelength, is exactly what the exam wants you to state.

Worked example 7.6 (book): in a ripple tank a wave generator makes 500 pulses in 10 s. Find the frequency and time period. Frequency is pulses per second: f = 500/10 = 50 Hz. Period is its reciprocal: T = 1/f = 1/50 = 0.02 s. A short calculation, but notice the method: count, divide by time, invert. The same two steps answer any "pulses in so many seconds" question.

📌 Book notes (For your information / Interesting information): a fine diffraction grating shows the colours of white light spread out — and the book points out that the fine rulings on a compact disc, each about 0.5 µm wide, act as a diffraction grating too: shine a small white source on a disc and the reflected light forms coloured lanes, each a diffraction pattern from the rulings. Your old CDs are diffraction gratings you already own.

Sound two tuning forks of slightly different frequencies together and you do not hear two notes — you hear one note whose loudness slowly swells and fades, swells and fades. That periodic rise and fall of loudness is a beat. The book's definition is precise: when two waves of slightly different frequencies travelling in the same direction overlap, the resulting periodic variation of sound between maximum and minimum loudness is called beats.

Follow the book's numbers and the mechanism clicks into place. Two forks A and B both at 32 Hz sounded together give a single pure note — you can barely tell them apart. Load fork B with a little wax and its frequency drops slightly, say to 30 Hz. Sound them together now and the note alternately grows loud and fades. At some instant X the two waves displace the air in the same direction, so the resultant is large and the sound is loud. A quarter of a second later the displacement from one fork is opposite to the other, the resultant is minimum at Y, and the sound is faint or gone. Another quarter second on, they agree again at Z and the sound is loud again. In one second you hear two loud sounds and two faint ones — and the book's worked line gives the count directly: number of beats = f_A − f_B = 32 − 30 = 2 beats per second. In general, beat frequency f_beat = |f₁ − f₂|, the difference of the two frequencies. If the difference grows past about 10 Hz, the beats come too fast to pick out and the effect is lost.

Musicians turn this nuisance into a tool. To tune a string instrument, play a note against a note of known frequency and adjust the string — tightening or loosening it — until no beats are heard at all. No beats means the frequencies match. The book lists where beats show up in instruments: a guitar's two strings tuned a few cents apart, a piano's adjacent white and black keys, a violin's two strings under slightly different bow pressure or speed, two drums tuned slightly differently, a flute's two notes with slightly different embouchure, an organ's two pipes tuned slightly apart, and a synthesiser's two oscillators at slightly different frequencies. In every case the small frequency difference makes the amplitude breathe in and out, and that breathing is the beat. Sometimes players use beats deliberately for rhythmic colour; sometimes they are an unwanted wobble to be tuned away.

Worked example 7.7 (book) — the classic trap, read it twice: two forks give 3 beats per second; one fork is 256 Hz; loading it with wax lowers its frequency slightly, and the beat frequency becomes 1 Hz. Find the second fork's frequency. Before loading, f₂ = 256 ± 3, so 259 Hz or 253 Hz. Try 259 Hz first: if the first fork is loaded, its frequency falls below 256 Hz, moving away from 259 Hz, so beats would increase past 3 — but the observed beats decreased to 1. So 259 Hz is wrong. With 253 Hz the story fits: loading the first fork moves it towards 253 Hz, beats shrink, and indeed 256 − 253 = 3 before, and after a small fall the difference becomes 1. Answer: f₂ = 253 Hz. The lesson: when a fork is loaded, always test both candidates against what happens to the beat count — the direction of change tells you which candidate is real.

📌 Book note (Do you know?): the tuning fork itself was invented in 1711 by John Shore, a royal trumpeter and lutenist — a musician's tool invented by a musician, which is fitting for a section that ends with musicians tuning by ear.

So far the chapter has mostly dealt with waves that stand still. A progressive wave (travelling wave) is the opposite kind: it moves steadily in one direction, carrying energy from one point to another. Water waves, sound waves and light waves are all progressive waves. And once energy is on the move, you can ask how much of it passes through each square metre every second. That is intensity (I): the amount of energy transmitted per unit area per unit time in the direction of propagation, measured in watts per square metre (W m⁻²).

From the definition, I = E/(A × t), and since E/t is power P, I = P/A. For a source spreading evenly in all directions, the area is the surface of a sphere, A = 4πr², so intensity falls as the square of the distance — move twice as far away and the same power is spread over four times the area. The second relation is the one that surprises students: in a mechanical wave the energy is stored as kinetic and potential energy of the medium's particles, and the more a particle is displaced from its mean position, the more energy it carries. So intensity is proportional to the square of the amplitude: I ∝ A², or I = kA² (7.14), where k depends on the wave and the medium.

That square matters enormously. Double the amplitude and the intensity does not double — it quadruples. Triple it and intensity goes up nine times (this is exercise MCQ 7.8, and now you know why the answer is 9, not 3). It also explains a small everyday fact the book tucks in as a tidbit: the frequency and amplitude of a travelling wave are independent of each other, which is why you can turn up a song's volume (amplitude) without changing its pitch (frequency).

Worked example 7.8 (book): (a) A wave has intensity 0.5 W m⁻² at 3.0 m from the source. Its power? I = P/(4πr²), so P = I × 4πr² = 0.5 × 4 × 3.14 × (3.0)² = 0.5 × 113.04 = 56.52 W. (b) Two progressive waves have intensities 0.5 and 0.25 W m⁻²; their total intensity is simply the sum, I = 0.5 + 0.25 = 0.75 W m⁻². Worked example 7.9 (book): a speaker gives 50 W, spread evenly, and the sphere at 5 m has area about 314 m² (4π × 25 ≈ 314). Intensity I = P/A = 50/314 ≈ 0.159 W m⁻². Same formula, three different disguises — learn to spot P, A and r in the question and the rest is arithmetic.

📌 Book notes (Tidbits / Brain teaser): sound cannot travel through a vacuum, because there are no particles to carry it — a tidbit that also explains why the Sun's energy must arrive as electromagnetic waves, not sound. And the book's teaser asks for the decibel level of a wave of intensity 10 W m⁻² — ⚠️ flagged honestly: the chapter as printed gives no reference intensity and no decibel formula, so that teaser cannot be completed from this chapter alone; no number is invented here. (With the usual reference 10⁻¹² W m⁻² the level would be 130 dB, but that reference is not in the book's text, so it is not presented as the book's answer.)

Stand on a railway platform as a train approaches hooting, and the whistle sounds higher than it does after the train has passed and is moving away. The train's whistle has not changed — what changed is the spacing of the waves reaching your ear. That apparent change in frequency (or pitch) caused by relative motion between source and observer is the Doppler effect. It was first noticed by Johann Doppler in the light of stars: light from a star moving relative to Earth arrived with a slightly different frequency than the same light made on Earth.

Start with a source at rest emitting wavelength λ, with sound speed v. The observer receives f = v/λ waves per second. Now let the source move towards a stationary observer A with speed uₛ. Each new wave is emitted from a position a little closer to A, so the waves get squeezed together: in one second the source advances uₛ, and that advance is shared among f waves, so each wavelength shrinks by Δλ = uₛ/f. The wavelength A actually receives is λ_A = λ − Δλ = v/f − uₛ/f = (v − uₛ)/f. The apparent frequency is f_A = v/λ_A = [v/(v − uₛ)]f. Since v/(v − uₛ) is greater than 1, f_A is greater than f — higher pitch, exactly as the platform hears it. The change Δλ is called the Doppler shift.

Now the source moves away from a stationary observer B with speed uₛ. The same reasoning runs in reverse: each wave is emitted from a little farther away, the waves stretch, λ_B = λ + Δλ = (v + uₛ)/f, and the apparent frequency is f_B = [v/(v + uₛ)]f. The factor is less than 1, so f_B is less than f — lower pitch as the train recedes. One pair of formulas, two signs, and the sign is the whole story: minus in the denominator for approaching (waves squeezed, frequency up), plus for receding (waves stretched, frequency down). The book's learning outcomes cover only the moving-source, stationary-observer case, and that is what these formulas describe; a moving observer is not required here.

Worked example 7.10 (book) — relative motion done carefully: trucks P and Q travel the same way; P ahead at 12 m s⁻¹, Q behind at 20 m s⁻¹ sounding its horn, and P's driver estimates the horn at 830 Hz. What frequency does Q's own horn actually emit? (Speed of sound 340 m s⁻¹.) Q is closing on P at uₛ = 20 − 12 = 8 m s⁻¹ relative to P, so P hears the approaching-source frequency: 830 = [340/(340 − 8)] × f_Q, giving f_Q = (830 × 332)/340 = 810.47 Hz. The horn's true note is lower than what P hears, because Q is gaining on P and squeezing the waves.

Worked example 7.11 (book): a train sounds its horn at rest and the platform observer estimates 1200 Hz; 50 s after departure the horn sounds again and the observer estimates 1140 Hz. (Speed of sound 340 m s⁻¹.) The train is now moving away, so f′ = [v/(v + uₛ)]f: 1140 = [340/(340 + uₛ)] × 1200, so 340 + uₛ = (340 × 1200)/1140 = 357.89, and uₛ = 17.89 m s⁻¹. Starting from rest with steady acceleration, the average speed over 50 s is (0 + 17.89)/2, so distance S = 8.945 × 50 ≈ 448 m from the station. Two Doppler steps and one kinematics step, chained together.

📌 Book notes (Info Corner / Do you know?): hearing ranges differ wildly across species — the book's table gives dolphin 150–150,000 Hz, bat 1000–120,000 Hz, cat 60–70,000 Hz, dog 15–50,000 Hz, human 20–20,000 Hz — which is why bats can use ultrasound we cannot hear at all. And when a plane accelerates past the speed of sound, the wavefronts pile up into a cone trailing behind it like a boat's wake: that piled-up front is the shock wave, heard as a sonic boom.

The Doppler effect also works for electromagnetic waves, and that opens a whole shelf of applications — the book walks through them one by one, and each is worth a line in revision. RADAR (Radio Detection And Ranging) sends out radio waves and catches their reflections: if an aeroplane approaches the radar, the reflected wavelength is shorter; if it recedes, it is longer, and the shift gives the plane's speed. The same principle tracks satellites around the Earth. SONAR (Sound Navigation And Ranging) does the underwater version with sound or ultrasound echoes: the Doppler detection relies on the relative speed of target and detector, and its military uses include finding submarines, mine hunting and measuring sea depth. Bats, the book notes elsewhere, do sonar biologically — navigating and finding food by echolocation.

In astronomy, the Doppler shift of starlight is measured by comparing a star's line spectrum with the same lines made in a laboratory. Stars moving away show a red shift — their light is stretched to longer wavelengths, towards the red end — and stars moving towards us show a blue shift, compressed towards the blue end. Astronomers found that essentially all distant galaxies are red-shifted, and from the size of the shift they estimate how fast the galaxies are receding. A radar speed trap uses the same idea on the road: microwaves are sent in short bursts, reflected off a car, and the Doppler shift of the reflection is turned by computer into the car's speed. Satellite navigation uses Doppler shift to fix a satellite's velocity and position, and satellite communication compensates for it to keep signals stable. Doppler radar measures wind speed and rainfall patterns, and the shift even helps track the Earth's surface movement and deformation.

Then comes medicine, which is where the chapter's learning outcomes point: Doppler echocardiography measures blood-flow velocity and detects heart problems such as valve stenosis (narrowing) or regurgitation (leaking), and helps set pacemakers; Doppler ultrasound measures blood flow, calculates cardiac output, and detects narrowing or blockage of blood vessels. The same frequency shift that tells an astronomer a galaxy is fleeing tells a doctor a heart valve is leaking — one equation, two universes.

Worked example 7.12 (book): a faint galaxy's Ca-α absorption line is found at 478 nm, while the same line measures 397 nm in the laboratory. (a) Is the galaxy moving towards or away from Earth? The observed wavelength is longer (478 > 397), so the light is red-shifted: the galaxy is moving away. The book computes the frequencies to show it: f = c/λ = (3 × 10⁸)/(397 × 10⁻⁹) = 7.56 × 10¹⁴ Hz in the lab, and f′ = (3 × 10⁸)/(478 × 10⁻⁹) = 6.28 × 10¹⁴ Hz observed, so f′ < f, again away. (b) Its speed? f′ = [c/(c + uₛ)]f gives 6.28 × 10¹⁴ = [3 × 10⁸/(3 × 10⁸ + uₛ)] × 7.56 × 10¹⁴, so 3 × 10⁸ + uₛ = (3 × 10⁸ × 7.56)/6.28 = 3.61 × 10⁸, and uₛ = 6.12 × 10⁷ m s⁻¹ — about a fifth of the speed of light. (The book's side note gives a similar case: a galaxy receding at 0.2c shows a line normally at 500 nm shifted longer.) Note the formula here is the sound-style Doppler formula with c in place of v, as the book itself uses it; a fully relativistic treatment would differ slightly, but the book's answer follows its own formula exactly.

This chapter is one idea used four ways. A wave carries energy without carrying matter (v = fλ), and when waves meet they add — superposition. Add same-frequency waves travelling together and you get interference (ΔS = nλ loud, ΔS = (2n + 1)λ/2 quiet). Add equal waves travelling oppositely and you get a standing wave with nodes and antinodes, whose allowed frequencies are quantised into harmonics. Let a wave bend around an obstacle and you get diffraction; let two nearly equal frequencies overlap and you get beats (f_beat = |f₁ − f₂|); let a source move and you get the Doppler shift (minus in the denominator approaching, plus receding). Intensity ties the travelling wave back to energy: I = P/A and I ∝ A².

Topics covered, with links:

• 7.1 Waves — definition, six parameters, v = fλ (Ex. 7.2: 300 m at 1000 kHz)
• Types of waves — mechanical, electromagnetic, quantum, surface; transverse vs longitudinal
• 7.2 Superposition — y = y₁ + y₂, in phase doubles, out of phase cancels
• 7.3 Interference — coherent sources, ΔS = nλ and (2n + 1)λ/2 (Ex. 7.1: 344 m s⁻¹)
• Noise-cancelling headphones — anti-noise, destructive interference in a product
• 7.4 Stationary waves — nodes λ/2 apart, node to antinode λ/4, energy trapped
• 7.5 String modes — fₙ = nf₁, v = √(F/m) (Ex. 7.3: λ = 0.6 m, f₁ = 30 Hz)
• 7.6 Air columns — open pipe all harmonics, closed pipe odd only (Ex. 7.4, 7.5)
• 7.7 Microwave experiment — source, probe, reflector, antinode spacing gives λ
• 7.8 Diffraction — ripple tank, small gap strong spreading (Ex. 7.6: 50 Hz, 0.02 s)
• 7.9 Beats — f_beat = |f₁ − f₂|, tuning by zero beats (Ex. 7.7: 253 Hz)
• 7.10 Intensity — I = P/A, I = kA² (Ex. 7.8: 56.52 W; Ex. 7.9: 0.159 W m⁻²)
• 7.11 Doppler effect — approaching [v/(v − uₛ)]f, receding [v/(v + uₛ)]f (Ex. 7.10, 7.11)
• 7.12 Doppler applications — radar, sonar, red/blue shift, medicine (Ex. 7.12: 6.12 × 10⁷ m s⁻¹)

Revision guidance: learn the chapter as three formula pairs — interference (nλ vs (2n + 1)λ/2), pipes (open v/2ℓ all n vs closed v/4ℓ odd n only), Doppler (minus approaching vs plus receding) — and everything else hangs off them. Practise the two trap questions until they are automatic: the closed pipe's "next" harmonic is the third, and a loaded tuning fork's candidate frequency is settled by whether the beats grew or shrank. For numericals, always convert units first (km h⁻¹ to m s⁻¹, hours to seconds, mm to m) — most lost marks in this chapter are unit marks, not physics marks.