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The physics of the ear

The faintest sound you can hear moves your eardrum by less than the width of an atom, and the loudest you can bear carries a million million times the power. No linear scale survives that, so the ear does not use one, and neither does the instrument that measures it.

Year 13AQA 3.10.2

Builds on Progressive waves and Longitudinal and transverse waves.

IN THIS TOPIC

  • Describe the transmission of sound through the outer, middle and inner ear.
  • Explain why the middle ear raises the pressure before the sound reaches the cochlea.
  • Use I = P/A, and use the intensity level equation in both directions.
  • Explain why a logarithmic scale is the right one, and say what the dBA scale adds.
  • Read an equal loudness curve, say how one is produced, and say what damage does to it.

WHAT YOU PROBABLY THINK

A sound at 80 dB carries twice the power of one at 40 dB.

Air, bone, fluid

The ear is a chain of three sections, and each one hands the vibration to the next in a different medium. The outer ear is the visible pinna, which collects sound and funnels it into the ear canal. The canal is a tube closed at the far end by the eardrum, and like any closed pipe it resonates. Its first resonance falls near 3 kHz, which is one reason the ear is at its most sensitive there.

At the end of the canal the eardrum vibrates with the arriving pressure variation. That is the boundary of the middle ear, an air-filled cavity spanned by three small linked bones, the ossicles, which carry the motion across to a much smaller membrane called the oval window.

The ear as a transducer chain: pinna and canal in air, eardrum and ossicles as a lever, oval window and cochlea in fluidpinna and canaleardrumossicles: a leveroval windowcochlea: fluid and nerve endingsouter (air)middle (bone)inner (fluid)pressure raised by area ratio and lever
FIG. 1Sound in air is collected by the pinna, moves the eardrum, is carried across the middle ear by the three ossicles, and enters the fluid of the cochlea through the much smaller oval window.

Two things happen on the way across, and both raise the pressure. The ossicles are linked as a lever, so the force at the far end is larger than the force at the near end. More importantly the eardrum has perhaps twenty times the area of the oval window, and the same force spread over a twentieth of the area is twenty times the pressure. Together they lift the pressure by a factor of the order of twenty or thirty.

That is not decoration. Beyond the oval window lies the inner ear, and the cochlea inside it is full of fluid. Sound passing straight from air to fluid would meet a huge mismatch and almost all of it would reflect, exactly as an ultrasound pulse reflects at an air gap. The middle ear is the matching stage that gets the energy across. Inside the cochlea the vibration travels along a membrane whose stiffness changes with distance, so each frequency disturbs its own region most, and hair cells there fire nerve impulses to the brain. Position along the cochlea is how the ear codes pitch.

Intensity, and why the scale is logarithmic

Sound carries energy, and the quantity that matters at a detector is how much of it arrives per second on each square metre facing the wave. That is the intensity,

I=PAI = \frac{P}{A}NOT ON THE DATA SHEET: LEARN IT

measured in W m−2, with A the area at right angles to the direction the sound travels. An eardrum of area 60 mm2 in a sound of intensity I therefore absorbs a power IA, and that is the calculation an exam builds towards.

Now look at the range the ear covers. The quietest sound a good ear detects has an intensity of 1.0 × 10−12 W m−2, and a sound a million million times more intense is painful rather than deafening. A linear scale cannot show both ends of that on one axis. Worse, it would misrepresent what the listener reports, because the ear judges by ratio: multiply the intensity by ten and the sound seems one step louder, multiply it by ten again and it seems one more step louder. Equal ratios feel like equal steps, and that is precisely what a logarithm turns into equal intervals.

So sounds are quoted as an intensity level in decibels, measured against the threshold of hearing:

intensity level=10logII0\text{intensity level} = 10\,\text{log}\frac{I}{I_{0}}ON YOUR DATA SHEET

with I0 = 1.0 × 10−12 W m−2, the booklet value. Read it once and the numbers stop being arbitrary. Ten times the intensity is +10 dB, a hundred times is +20 dB, and doubling the intensity is close to +3 dB.

The decibel scale compresses twelve powers of ten of intensity into 0 to 120 dB, so every factor of ten of intensity adds ten decibelsintensity / W m⁻²intensity level10⁻¹²0 dBthreshold of hearing10⁻⁹30 dBa quiet room10⁻⁶60 dBconversation10⁻³90 dBa busy road10⁰120 dBthe threshold of paineach step here is 1000 times the intensity and 30 dB more
FIG. 2Five intensities a thousandfold apart, from the threshold of hearing to the threshold of pain, and the intensity levels they correspond to: 0, 30, 60, 90 and 120 dB.

The claim at the top of the page dies here. From 40 dB to 80 dB is 40 decibels, so four factors of ten, and the intensity ratio is 104. Ten thousand times, not twice.

WORKED EXAMPLE

The power arriving on an eardrum

A tone measured at 78 dB falls on an eardrum of area 60 mm2. Find the power the eardrum receives.

First the intensity. 78 = 10 log(I/I0), so I/I0 = 107.8 = 6.3 × 107, and I = 6.3 × 107 × 1.0 × 10−12 = 6.3 × 10−5 W m−2.

Then the power. A = 60 mm2 = 60 × 10−6 m2, so P = IA = 6.3 × 10−5 × 60 × 10−6 = 3.8 × 10−9 W.

Four nanowatts, and the ear reports it comfortably. Converting square millimetres to square metres is where this question is usually lost: the factor is 10−6, not 10−3.

Two sounds can also be compared with each other, without either one being referred to the threshold. Their relative intensity level is

relative intensity level=10logI2I1\text{relative intensity level} = 10\,\text{log}\frac{I_{2}}{I_{1}}NOT ON THE DATA SHEET: LEARN IT

which is just the difference of their two levels, and it is what a question means by asking how many decibels louder one sound is than another.

Equal loudness curves

Intensity level is a measurement. Loudness is a judgement, and the two part company as soon as the frequency changes, because the ear is not equally sensitive across its range of about 20 Hz to 20 kHz. An equal loudness curve maps the difference.

Producing one takes a listener and two tones. A reference tone at 1 kHz is set to a chosen intensity level. A test tone at some other frequency is then adjusted until the listener judges the two equally loud, and the level it needed is plotted at that frequency. Repeat across the range, join the points, and the curve shows every combination of frequency and intensity level that sounds equally loud to that listener. Each curve is named by its own level at 1 kHz, and the results are averaged over many listeners.

Equal loudness curves: every point on one curve sounds equally loud, each curve is named by its intensity level at 1 kHz, and all of them dip near 3 kHz201001 k3 k20 kfrequency / Hz (logarithmic)intensitylevel / dB020406080most sensitive near 3 kHzthe lowest curve is the threshold of hearing: 0 dB at 1 kHz
FIG. 3A family of equal loudness curves. Each is labelled by its intensity level at the 1 kHz reference, each dips to its lowest near 3 kHz, and the higher curves are flatter than the lowest one.

Three readings come off that family, and questions ask for all three. The ear is most sensitive between about 2 kHz and 5 kHz, where the curves dip lowest, so least intensity is needed there. Sensitivity falls away at both ends, and the bass end is the worse of the two, which is why a very low note needs a far higher intensity level to match a mid tone. And the curves flatten as they rise, so a loud passage is heard with a more even frequency balance than a quiet one.

The lowest curve of all is the threshold of hearing, the minimum intensity a normal ear can detect, quoted at 1 kHz as 1.0 × 10−12 W m−2. That is where the 0 dB of the decibel scale comes from, so 0 dB does not mean silence; it means the faintest audible sound at the reference frequency.

A meter reading in plain dB weights every frequency equally, so it will call a rumbling low-frequency noise louder than a listener does. The dBA scale fixes that by filtering the signal with a frequency weighting shaped like the ear's own response before the level is computed, discounting the frequencies the ear is poor at. A figure in dBA therefore tracks perceived loudness, and it is the one written into noise-at-work limits.

What damage does to the curves

Hearing is lost in two ways the specification asks about: injury from exposure to excessive noise, and gradual deterioration with age. Both show up as a change in the curves rather than as silence.

Hearing loss lifts the threshold curve, and lifts it most at the high-frequency end, so the faintest audible sound has to be louder201001 k4 k20 kfrequency / Hz (logarithmic)thresholdlevel / dBafter noise or agehealthy thresholdnotch near 4 kHza higher threshold means a fainter sound is no longer heard
FIG. 4A healthy threshold curve against a damaged one. The damaged ear needs a greater intensity at every frequency, by far the most at the high-frequency end, and noise injury leaves an extra dip in performance near 4 kHz.

The threshold curve lifts, meaning a greater intensity is now needed before anything is heard at all, and it lifts unevenly. The high-frequency end goes first, so the top of the range shrinks well before the middle is affected. Loud noise typically leaves a dip in performance around 4 kHz; ageing raises the whole treble end steadily. The higher equal loudness curves change less, so loud sounds still seem loud while quiet ones vanish.

The practical consequence is worth stating in an answer, because it is what a patient reports. Consonants such as s, f and t carry their information at high frequency and at low intensity, so they are the first things to go, and speech becomes hard to follow in a noisy room long before anyone would say they were deaf.

THE EXAM BIT

  • Transmission questions want the chain in order and the medium at each stage. Pinna and canal in air, eardrum, three ossicles across the middle ear, oval window, fluid in the cochlea, hair cells, nerve impulses.
  • Give both reasons for the pressure gain. The ossicles act as a lever, and the eardrum's area is far larger than the oval window's, so the same force acts over a much smaller area.
  • Say what the pressure gain is for. The cochlea is fluid-filled, and without the middle ear almost all the sound would reflect at the air-to-fluid boundary instead of entering.
  • Working back from a level to an intensity is one line: I = I0 × 10L/10. Then multiply by the area, in square metres, if a power is wanted.
  • Quote the threshold of hearing properly. It is the minimum intensity a normal ear can detect at 1 kHz, and dropping the frequency loses a mark.
  • Intensity and intensity level are different quantities with different units. W m−2 against dB, and a question asking for one will not accept the other.
  • For hearing loss, describe the shift in the curves rather than the biology. The specification excludes the physiological changes, so marks are for the threshold rising and for the high-frequency end rising most.

CHECK YOURSELF

A machine produces a sound of intensity 2.0 × 10−4 W m−2 at a worker's ear. Find the intensity level in dB. A second machine is 6.0 dB louder: find its intensity. Explain why the site's noise limit is written in dBA rather than dB.

Show a hint

One equation, used forwards and then backwards. The last part is about which frequencies the ear actually notices.

Show the answer

I/I0 = (2.0 × 10−4)/(1.0 × 10−12) = 2.0 × 108, and 10 log(2.0 × 108) = 83 dB.

6.0 dB more means 10 log(I2/I1) = 6.0, so I2/I1 = 100.6 = 4.0 and I2 = 8.0 × 10−4 W m−2.

A plain dB reading treats every frequency alike, so it overstates the effect of low-frequency rumble that the ear barely registers. The dBA scale weights the spectrum the way the ear's own response does, so it measures what the worker is exposed to rather than what a microphone sees.

Intensity is power per unit area; intensity level is ten log of it against 1.0 × 10⁻¹² W m⁻².

The scale is logarithmic because the ear judges by ratio, so +10 dB is always ten times the intensity.

The middle ear raises the pressure by a lever and an area ratio, so the sound can get into the fluid of the cochlea.

Equal loudness curves dip near 3 kHz, flatten as they rise, and lift at the treble end when hearing is damaged.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic
15 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

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  • Describe the transmission of sound through the outer, middle and inner ear.
  • Explain why the middle ear raises the pressure before the sound reaches the cochlea.
  • Use I = P/A, and use the intensity level equation in both directions.
  • Explain why a logarithmic scale is the right one, and say what the dBA scale adds.
  • Read an equal loudness curve, say how one is produced, and say what damage does to it.

Open the full revision checklist to track your progress across the whole unit.