Sound, Doppler & Ultrasound

Sound, Doppler & Ultrasound

Part 1 of 2 — Sound: Intensity, Decibels & Hearing

Where this fits. The previous lesson, Waves & Interference, covered periodic motion, wave basics, superposition, beats, standing waves and resonance. This lesson closes the fluids-and-waves sequence with sound itself: intensity, decibels and hearing, then the Doppler effect, ultrasound and shock waves. The next lesson, Optics, moves on to light.

How Sound Is Made

Every sound starts with something vibrating: vocal folds, a speaker cone, a tuning fork's prongs. As the surface pushes outward it crowds the air ahead of it into a compression (pressure slightly above normal); as it pulls back it leaves a rarefaction (slightly below). These zones travel outward as a longitudinal pressure wave, while each air molecule only jiggles back and forth about its resting position.

• Sound needs a medium. In a vacuum there is nothing to compress, so no sound travels, although light still crosses it.
• Its speed is set by the medium (the Waves & Interference lesson): about $343\;\text{m/s}$ in room-temperature air, faster in warmer air, faster still in liquids, and fastest in solids.
• The frequency is the vibration rate of the source.

Passage Strategy: Decide What Is Fixed Before You Compute

Most wave errors come from changing the wrong variable. Before writing a formula, ask what is FIXED:

The last row is a passage favourite: whenever a length fixes the wavelength, the frequency must follow the speed of sound.

• Helium voice: the vocal folds vibrate at about the same rate, but sound travels faster in helium, so the vocal tract's resonances rise and the voice's timbre (its blend of harmonics, covered later in this part) shifts toward the higher harmonics.

Intensity and the Inverse-Square Law

Intensity is the power a wave carries through each square metre of area perpendicular to its travel:

$I = \frac{P}{A} \qquad (\text{W/m}^2)$

A small source radiating equally in all directions (a point source) spreads its power over a growing sphere of area $4\pi r^2$:

$I = \frac{P}{4\pi r^2} \quad\Rightarrow\quad I \propto \frac{1}{r^2}$

Double the distance and the intensity drops to $1/4$; triple it, $1/9$; ten times farther, $1/100$. Intensity also grows with the square of the wave's amplitude, as all wave energy does.

The Decibel Scale

Human ears respond to intensities from about $10^{-12}\;\text{W/m}^2$ (barely audible) to about $1\;\text{W/m}^2$ (painful), a range of a trillion. To tame that range, the sound level $\beta$ uses a logarithm:

$\beta = 10\log\frac{I}{I_0}, \qquad I_0 = 10^{-12}\;\text{W/m}^2$

$\beta$ is in decibels (dB), and $I_0$ is the reference intensity, the threshold of human hearing, defined as $0\;\text{dB}$. Keep the two quantities apart: intensity $I$ is physical power per area; level $\beta$ is a logarithmic rating of it.

Rules that follow:

• Never add decibels directly. Two identical $70\;\text{dB}$ sources double the intensity, giving about $73\;\text{dB}$, not $140$. Ten identical sources give ten times the intensity: $+10\;\text{dB}$.
• Unequal sources: the louder one dominates. A $60\;\text{dB}$ and a $50\;\text{dB}$ source together make only about $60.4\;\text{dB}$, because the quieter one adds just a tenth more intensity.
• A level difference is a ratio, never a fraction of the levels. From $50$ to $70\;\text{dB}$ is $\times 100$, not $\times 70/50$.
• Sounds fainter than $I_0$ have negative levels: $10^{-13}\;\text{W/m}^2$ is $-10\;\text{dB}$.

Rough landmarks (approximate; no need to memorize): a whisper about $30\;\text{dB}$, conversation about $60\;\text{dB}$, heavy traffic about $80\;\text{dB}$, a siren up close $110$ to $120\;\text{dB}$, the threshold of pain about $120$ to $130\;\text{dB}$.

• Occupational guidelines treat long exposure above about $85\;\text{dB}$ as a hearing-damage risk.

Practice: Intensity & Decibels 🎯

Worked Example — Decibel Arithmetic Without a Calculator

Problem 1 — Levels and intensities. Use $\beta = 10\log(I/I_0)$ with $I_0 = 10^{-12}\;\text{W/m}^2$. Every factor of 10 in intensity is $10\;\text{dB}$.

• (a) A monitor alarm measures $40\;\text{dB}$ at a nurse's station and $70\;\text{dB}$ beside the patient. How much more intense is it beside the patient?
• (b) A blender delivers $10^{-5}\;\text{W/m}^2$ to your ear. What is its sound level?
• (c) A siren is rated at $90\;\text{dB}$. What intensity is that?

(a) A ratio from a difference. $\Delta\beta = 30\;\text{dB} \Rightarrow \dfrac{I_2}{I_1} = 10^{30/10} = 10^3 = 1000$

Thirty decibels is three factors of ten. Never treat the levels as a ratio ($70/40$): the scale is logarithmic.

(b) From intensity to level. $\beta = 10\log\dfrac{10^{-5}}{10^{-12}} = 10\log(10^{7}) = 10(7) = 70\;\text{dB}$

Shortcut: count the powers of ten above $10^{-12}$ and multiply by 10.

(c) From level to intensity. Run the shortcut from (b) backward: $90\;\text{dB}$ is nine powers of ten above $I_0$, so $I = 10^{9} \times 10^{-12} = 10^{-3}\;\text{W/m}^2$.

Problem 2 — Ratios that are not powers of ten. A factor of 10 in intensity is $10\;\text{dB}$, and a factor of 2 is about $3\;\text{dB}$ (because $\log 2 \approx 0.3$).

• (a) A sound becomes five times as intense. About how much does its level rise?
• (b) About how much does it rise if the sound becomes four times as intense?
• (c) What happens to the level when the wave's amplitude doubles?

(a) Five times. Write $5 = 10/2$: that is $+10\;\text{dB}$ and then $-3\;\text{dB}$, so about $+7\;\text{dB}$.

(b) Four times. $4 = 2 \times 2$: about $+3 + 3 = +6\;\text{dB}$.

(c) Double the amplitude. Intensity grows as amplitude squared, so doubling the amplitude gives $\times 4$ the intensity: about $+6\;\text{dB}$, as in (b).

Problem 3 — Several sources, and distance. One infusion pump hums at $40\;\text{dB}$. A loudspeaker, treated as a point source, gives $90\;\text{dB}$ at $1\;\text{m}$.

• (a) What level do two identical pumps make together?
• (b) What level do ten identical pumps make?
• (c) What level does the loudspeaker give at $10\;\text{m}$?
• (d) What level does it give at $2\;\text{m}$?

(a) Two pumps. Intensity $\times 2$, about $43\;\text{dB}$ (never $80$).

(b) Ten pumps. Intensity $\times 10$, $50\;\text{dB}$.

(c) Ten times as far. The intensity is $1/10^2 = 1/100$ as great, which is $-20\;\text{dB}$, so $70\;\text{dB}$.

(d) Twice as far. The intensity is $1/4$, about $-6\;\text{dB}$, so about $84\;\text{dB}$. The strategy in every case: convert the question into an INTENSITY RATIO, then turn each factor of 10 into $10\;\text{dB}$ and each factor of 2 into $3\;\text{dB}$.

Attenuation: Sound Fades for Two Reasons

The inverse-square law describes only geometric spreading: the same power covering a larger area. A real medium also absorbs sound, converting its energy into heat, so intensity falls even faster than $1/r^2$. This extra loss is attenuation.

• Higher frequencies are absorbed more strongly. That is why distant thunder is a low rumble while nearby thunder cracks, and why music through a wall keeps its bass and loses its treble.
• Soft tissue absorbs ultrasound heavily, more so at higher frequencies; that trade-off decides which probe frequency a clinician chooses (Part 2).
• Reflection and scattering at boundaries also remove energy from the forward beam.

• In soft tissue, ultrasound attenuation is roughly $0.5\;\text{dB}$ per centimetre for each megahertz of frequency.

Pitch, Loudness and Hearing

The physics maps onto perception:

A violin and a flute playing the same note share the same fundamental frequency, so they have the same pitch. They sound different because each adds a different blend of overtones, the harmonics above the fundamental (the Waves & Interference lesson).

• Human hearing spans about $20\;\text{Hz}$ to $20\;\text{kHz}$. The upper limit falls with age and noise exposure, so high frequencies are lost first. Sensitivity peaks near $2$ to $5\;\text{kHz}$.
• Infrasound lies below $20\;\text{Hz}$ (elephant calls, earthquakes). Ultrasound lies above $20\;\text{kHz}$: bats and dolphins echolocate with it, and medical imaging uses 1 to 20 MHz.
• Where pitch is decoded: each place along the cochlea's basilar membrane vibrates best at one frequency, high near the base and low near the apex (the Sensory Systems lesson).
• Amplitude changes loudness, not pitch: strike a tuning fork harder and it gets louder at the same frequency.

Practice: Distance, Attenuation & Hearing 🎯

Key Takeaways

• Inverse-square law. A point source gives $I = P/(4\pi r^2)$, so doubling the distance cuts the intensity to $1/4$.
• The decibel scale. $\beta = 10\log(I/I_0)$ with $I_0 = 10^{-12}\;\text{W/m}^2$; $+10\;\text{dB}$ is $\times 10$ intensity, $+3\;\text{dB}$ about $\times 2$.
• Attenuation. Absorption adds to spreading and is stronger at higher frequencies, so distant sounds lose their treble.
• Pitch, loudness, timbre. Pitch follows frequency, loudness intensity, timbre the harmonic mix; hearing spans $20\;\text{Hz}$ to $20\;\text{kHz}$.

Common traps

• Adding decibels. Two $70\;\text{dB}$ sources double the intensity: about $73\;\text{dB}$, never $140$.
• Taking a ratio of levels. From $40$ to $70\;\text{dB}$ is $\times 1000$ in intensity, not $\times 70/40$.
• Dropping the square. Intensity falls as $1/r^2$: ten times farther is $1/100$ the intensity, or $-20\;\text{dB}$.
• Louder means higher. A bigger amplitude makes a sound louder at the same pitch; pitch follows frequency alone.

• Low-yield extras: the $85\;\text{dB}$ occupational exposure guideline; the attenuation coefficient of soft tissue.

Sound, Doppler & Ultrasound

Part 2 of 2 — Doppler, Ultrasound & Shock Waves

Doppler: Relative Approach Raises Pitch

When a sound source and a listener move toward each other, the listener hears a higher frequency than the source emits; when they move apart, a lower one. That is the Doppler effect:

$f' = f\,\frac{v \pm v_o}{v \mp v_s}$

$f$ is the emitted frequency, $f'$ the frequency heard and $v$ the speed of sound; $v_o$ and $v_s$ are the observer's and source's speeds relative to the air, entered as positive numbers. The two signs are independent, and the ritual is approach raises: pick each so that motion TOWARD the other party raises $f'$.

• Observer moving toward the source: $+v_o$ on top (bigger top); moving away: $-v_o$.
• Source moving toward the observer: $-v_s$ below (smaller bottom); moving away: $+v_s$.
• Anyone standing still contributes zero.

Why it happens. A moving SOURCE chases its own wavefronts, bunching them ahead (shorter $\lambda$) and spreading them behind; the air still carries them at $v$, so $f = v/\lambda$ rises ahead. A moving OBSERVER leaves $\lambda$ untouched but meets wavefronts more often. The speed of sound never changes.

• Only motion along the line joining source and listener counts; motion straight across that line gives no shift at that instant.
• Slow-motion shortcut: for a relative approach speed $u \ll v$, the fractional shift is $\Delta f/f \approx u/v$.

Echoes Off a Moving Target: The Double Shift

Bounce sound off a moving target and two Doppler shifts happen in sequence: the target first RECEIVES the sound as a moving observer, then RE-EMITS it as a moving source. For a target moving at speed $u$ (much slower than sound), the echo's shift is about twice the one-way shift:

$\Delta f \approx \frac{2fu}{v}$

The shift is upward for an approaching target and downward for a receding one. Bats, police radar guns (with radio waves) and Doppler ultrasound all use this.

Worked Example — Doppler Direction Check

Problem. An ambulance siren at $600\;\text{Hz}$ approaches a stationary pedestrian at $34\;\text{m/s}$. Take $v = 340\;\text{m/s}$ for clean arithmetic.

• (a) Before calculating, roughly what should the pedestrian hear as the ambulance approaches, and after it passes?
• (b) Exactly what frequency does she hear while it approaches?
• (c) What does she hear after it passes?
• (d) If instead she runs at $34\;\text{m/s}$ toward a parked ambulance, what does she hear?

(a) Estimate first. $u/v = 34/340 = 10\%$, so expect a shift of about $60\;\text{Hz}$: roughly $660\;\text{Hz}$ approaching and $540\;\text{Hz}$ receding. Answer choices are usually far enough apart for the estimate to pick the right one.

(b) Approaching source. The source approaches, so subtract in the denominator:

$f' = 600\,\dfrac{340}{340 - 34} = 600\,\dfrac{340}{306} \approx 667\;\text{Hz}$

Approach, so the answer must exceed $600$: the minus sign was chosen to make that happen. (The room-temperature value, $343\;\text{m/s}$, gives about $666\;\text{Hz}$, the same to the precision tested.)

(c) After it passes. Now the source recedes, the denominator becomes $340 + 34 = 374$, and $f' = 600(340/374) \approx 545\;\text{Hz}$: the classic drive-by pitch drop.

(d) Switch who moves. A moving observer changes the numerator: $f' = 600(340 + 34)/340 = 600(1.1) = 660\;\text{Hz}$. Close to the $667\;\text{Hz}$ of (b), but not identical, because only a moving source changes the wavelength.

Practice: The Doppler Effect 🎯

Ultrasound Imaging

Medical ultrasound uses frequencies of about 1 to 20 MHz, far above the $20\;\text{kHz}$ limit of hearing. A probe held on the skin contains a crystal that turns a brief electrical pulse into a burst of sound, then listens, turning returning echoes back into electrical signals.

The crystal is piezoelectric: it deforms when a voltage is applied and generates a voltage when squeezed.

Depth from echo time. Each tissue boundary reflects part of the pulse. The machine times each echo and assumes sound moves at $1540\;\text{m/s}$ in soft tissue. The pulse travels to the boundary AND back, so:

$d = \frac{vt}{2}$

Forgetting the factor of 2 doubles every depth.

Resolution vs. penetration. An ultrasound beam cannot resolve details much smaller than its wavelength, and $\lambda = v/f$: about $1\;\text{mm}$ at $1.5\;\text{MHz}$, about $0.1\;\text{mm}$ at $15\;\text{MHz}$. So higher frequency gives FINER detail. But tissue absorbs higher frequencies more strongly (attenuation, Part 1 of this lesson), so higher frequency penetrates LESS deeply.

Typical choices (approximate; no need to memorize the numbers):

Why echoes happen. How strongly a boundary reflects depends on the mismatch in acoustic impedance, $Z = \rho v$, the product of a tissue's density and its speed of sound.

• Neighbouring soft tissues differ only slightly, so they return the weak echoes that build up the image.
• Tissue and bone differ more: bone reflects strongly and casts an acoustic shadow behind it, as do gallstones.
• Tissue and air differ enormously, so an air gap reflects almost all the sound: gas-filled lungs and bowel block the view, and even a thin film of air under the probe would.
• Ultrasound uses no ionizing radiation, which is why it is the imaging method of choice in pregnancy.

• The fraction of intensity reflected at a boundary is $\left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$.

Doppler Ultrasound and Blood Flow

Doppler ultrasound measures blood velocity from the double shift of echoes off moving red blood cells.

• Blood flowing toward the probe shifts the echo up, flowing away shifts it down, and the size of the shift is proportional to the speed.
• Only the velocity component along the beam counts: at beam-to-flow angle $\theta$, $\Delta f = 2fu\cos\theta/v$. A perpendicular beam gives zero shift, so the probe is tilted along the vessel; in a ratio of two shifts taken at the same angle, $\theta$ cancels.
• The shifts are typically a few kilohertz, which is audible: Doppler machines play them as the familiar whooshing pulse.

Grading a stenosis. A stenosis is a narrowing of a vessel's lumen, the open channel the blood flows through. Continuity gives $A_1u_1 = A_2u_2$ ($u$ = blood speed; the Fluids in Motion lesson writes it as $v$), so blood speeds up where the lumen narrows, and the Doppler shift is proportional to that speed. A stenosis therefore shows up as a local jump in the shift, and the shift ratio gives the area ratio.

• Colour Doppler displays flow toward the probe in red and away in blue (remembered as "BART": blue away, red toward).

Shock Waves and Sonic Booms

As a source speeds up toward the speed of sound, the wavefronts ahead of it crowd closer and closer. At $v_s = v$ they pile into a single intense front, and the Doppler formula's denominator reaches zero: the formula no longer applies.

When the source moves FASTER than sound ($v_s > v$), it outruns its own wavefronts. They overlap along a cone trailing behind the source, the Mach cone, and the cone's surface is a shock wave: a thin zone of abrupt, large pressure change. The Mach number is $M = v_s/v$; the faster the source, the narrower the cone.

• A sonic boom is the shock cone sweeping past a listener. The aircraft drags the cone along for as long as it flies supersonically, so the boom is not a one-time event at the moment of "breaking the sound barrier". Each listener under the flight path hears it once, AFTER the plane has passed overhead.
• Medicine uses shock waves too: lithotripsy focuses shock waves generated outside the body onto a kidney stone and shatters it.

• The cone's half-angle $\theta$ satisfies $\sin\theta = 1/M$. A cracking whip is a small sonic boom: the tip briefly moves faster than sound.

Worked Example — Depth, Frequency and a Stenosis

Problem 1 — How deep? A pulse leaves the probe, and an echo from the far edge of the liver returns $130\;\mu\text{s}$ later ($1\;\mu\text{s} = 10^{-6}\;\text{s}$; $v = 1540\;\text{m/s}$).

• (a) How far does the pulse travel, there and back?
• (b) How deep is the far edge of the liver?
• (c) How soon would an echo return from a carotid artery about $2\;\text{cm}$ below the skin?

(a) Round trip. $vt = (1540)(1.30 \times 10^{-4}) \approx 0.20\;\text{m}$

(b) Halve it. The $0.20\;\text{m}$ from (a) covers the depth twice: $d \approx 0.10\;\text{m} = 10\;\text{cm}$.

(c) Run it backward. The round trip is $2(0.02) = 0.04\;\text{m}$, so $t = 0.04/1540 \approx 26\;\mu\text{s}$.

Problem 2 — Which probe? A sonographer must image a carotid artery about $2\;\text{cm}$ deep and a liver $10\;\text{cm}$ deep, with a $12\;\text{MHz}$ or a $3\;\text{MHz}$ probe ($v = 1540\;\text{m/s}$).

• (a) What wavelength does the $12\;\text{MHz}$ probe give, and is it fine enough for the carotid wall?
• (b) What wavelength does the $3\;\text{MHz}$ probe give?
• (c) Which probe should image the liver, and why?

(a) The 12 MHz probe. $\lambda = 1540/(1.2 \times 10^{7}) \approx 0.13\;\text{mm}$, fine enough to see the vessel wall of the shallow carotid.

(b) The 3 MHz probe. $\lambda = 1540/(3 \times 10^{6}) \approx 0.51\;\text{mm}$: coarser.

(c) The liver. The $3\;\text{MHz}$ probe. Its $0.51\;\text{mm}$ wavelength from (b) is coarser, but at $10\;\text{cm}$ deep a $12\;\text{MHz}$ beam would be absorbed before its echo returned.

Problem 3 — Grading a stenosis with Doppler. A $4.0\;\text{MHz}$ beam is aligned with carotid flow ($v = 1540\;\text{m/s}$). The echo shift is $2.0\;\text{kHz}$ upstream and $8.0\;\text{kHz}$ inside a narrowing.

• (a) How fast is the blood moving upstream, and inside the narrowing?
• (b) How does the lumen's area inside the narrowing compare with the upstream area?
• (c) How does its radius compare?
• (d) What does the Doppler signal show just beyond the narrowing?

(a) Speeds from the double shift. $u = \dfrac{v\,\Delta f}{2f}$. Upstream: $u = \dfrac{(1540)(2000)}{2(4.0 \times 10^{6})} \approx 0.39\;\text{m/s}$. Inside: four times the shift, so four times the speed, about $1.5\;\text{m/s}$.

(b) Area by continuity. At (nearly) constant flow, $A \propto 1/u$. With four times the speed from (a), the lumen's area is one-quarter of the upstream area.

(c) Radius. $A \propto r^2$, so one-quarter the area means the radius is halved, about a 50% diameter narrowing. Note that a ratio question never needs the factor of 2 or the probe frequency; they cancel.

(d) Interpret. Beyond the stenosis, turbulence widens the spread of shifts. The pressure and resistance consequences of the narrowing (Bernoulli and Poiseuille) are in the Fluids in Motion lesson.

Practice: Ultrasound, Stenosis & Shock Waves 🎯

Key Takeaways

• Approach raises pitch. In $f' = f(v \pm v_o)/(v \mp v_s)$ pick each sign so approach raises $f'$; estimate with $u/v$.
• Ultrasound imaging. Depth is $d = vt/2$; higher frequency resolves finer detail but penetrates less deeply.
• Doppler blood flow. Echoes off moving blood shift by $2fu/v$; continuity turns a shift ratio into an area ratio.
• Shock waves. A source faster than sound trails a Mach cone, whose boom sweeps each listener after the plane passes.

Common traps

• A faster-moving sound. A moving source changes the wavelength, never the speed of sound, which the medium sets.
• Forgetting the factors of 2. Echoes shift twice ($2fu/v$), and echo depth halves the round trip ($vt/2$).
• Higher frequency, deeper view. Higher-frequency ultrasound is absorbed sooner, so it penetrates less, not more.
• Squaring the speed ratio. Continuity makes area inverse to speed: three times the shift means one-third the area.

• Low-yield extras: the reflection-fraction formula; the colour Doppler red/blue convention; the Mach-cone angle formula; the cracking whip.

Next up: the Optics lesson, which turns from sound to light: reflection, refraction, mirrors, lenses and the wave behaviour of light.