157.Physics of the Ear

Physics Factsheet www.curriculum-press.co.uk

Number 157

The Physics of the Ear To understand how our hearing works the first thing is to understand the ear. Fundamentally the ear’s job is to convert the vibrations of sound through the air into electrical impulses that the brain can interpret. Each part of the ear has a role to play in this process.

Balance canals

Bones of the middle ear

Stirrup

To brain

Anvil Auditory nerve

Hammer

Ear canal Ear drum Middle ear Oval window Round window

The Outer ear and ear Canal: This section of the ear collects the air-vibrations and channels them inwards to the eardrum (sometimes called the tympanic membrane) in the middle ear.

Cochlea

The cochlea and auditory nerve: The cochlea is a spiral-shaped tube with lines of tiny hairs along its inner-surface. It contains a fluid which transmits the vibrations from the bones of the middleear to the lines of hairs along the inside of the cochlea, making them sway back and forth. This swaying stimulates nerves at the base of the hairs, producing electrical impulses which are sent to the brain down the auditory nerve. The vibrations are transmitted into the fluid through ‘the oval window’ which is noticeably smaller than the eardrum. As a result, the pressure created in the fluid of the cochlea is greatly increased (about 30 times), amplifying the sound further.

The ear canal itself is a confined column of air and has its own resonant frequency of around 3000Hz (3kHz). The upshot of this is that human hearing is usually most sensitive or responsive to sounds at around this frequency. We will look at how our hearing responds to different frequencies later on. The eardrum and middle-ear: The sound-waves striking the eardrum cause it to vibrate, and these vibrations are passed on through the three bones in the middle ear – the ‘hammer’, the ‘anvil’ and the ‘stirrup’. They are named simply due to their shape.

The balance canals (or semi-circular canals) detect movement and help the body balance. They play no part in hearing. Example 1: Can you suggest how a ‘perforated eardrum’ (where the eardrum is torn slightly) might affect a person’s hearing?

These are the smallest bones in the human body and are incredibly lightweight. This means that the vibration from the eardrum can be carried easily, as they have very little mass so they’re easy to move back and forth.

Answer: The pressure-waves of the sound travelling into the ear-canal would hit the eardrum but the air would be able to pass through the perforations to some extent. As a result, the sound would not make the eardrum vibrate as much and the person’s hearing would seem quieter than normal.

They act as a lever and multiply the force of vibration from the eardrum by about 1.5 times.

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157. The Physics of the Ear Sensitivity

Example 3: The largest speaker in the unit shown (the woofer) is rated as 100W. It has a diameter of 50cm or 0.5m.

As we’ve already mentioned, human hearing is better at detecting some frequencies than others. Frequencies around 3000Hz or 3kHz are the easiest for us to detect due to the resonance effects in the ear canal at around that frequency.

Since the speaker is circular, it has a surface area of:

Example 2: Use your understanding of resonance in a closed air column to suggest why air in the ear-canal, which is about 2.5cm long, resonates at around 3000Hz. The speed of sound in air is approximately 300m/s.

Area = π × r2 = π × 0.252 = 0.196m2 This gives a sound intensity of: 100W ÷ 0.196m2 = 509.3W/m2 Question: The smallest speaker in the unit (the tweeter) is rated as 20W, and has a diameter of 7cm or 0.07m. What sound intensity would it produce?

Answer: Fundamental resonance in an air column happens when the air-column is ¼ of the wavelength long. If 2.5cm = ¼ λ then λ = 10cm = 0.1m Using f = v/λ f = 300/0.1 = 3000Hz.

Answer: Area = π × r2 = π × 0.0352 = 3.85×10-3m2 This gives a sound intensity of: 20W ÷ 3.85×10-3m2 = 5200W/m2

Differences in sensitivity can be shown on a Frequency Response Curve.

The threshold of human hearing – the lowest intensity of sound we can detect – is about 1×10-12 W/m2 and occurs at our most sensitive range of frequencies (around 1-3kHz). At this point we hit a problem. The lowest intensity we can hear is 1×10-12 W/m2. The greatest intensity we can bear without serious damage is 100W/m2. This gives an enormous range of audible intensities.

Sensitivity relative to maximum

100%

Intensity level and the decibel scale As a result the Decibel Scale is used for intensity level, and unsurprisingly it is a logarithmic scale, to try and make the massive range of intensities more manageable.

10%

The intensity level is obtained from the sound intensity (I) compared to the minimum that can be heard (Io = 1×10-12W/m2), in other words from I/Io.

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Freq (Hz)

The logarithm (log10) is calculated for this ratio and the answer is then multiplied by 10. So: Intensity Level = 10 × log (I/Io)

This is just a graph that shows how the ear responds to sounds of different frequencies.

Example 4: Consider our ‘woofer’ speaker from earlier. It was producing sound of intensity 509.3 W/m2 .

As you can see from the graph given, our response is best at around 1-3kHz, and becomes less at frequencies either side of this. There are points (at around 20Hz and 20kHz) where our ears cannot detect sound at all, as you can see from where the line intercepts the x axis.

Intensity Level = 10 × log (I/Io)

I/Io = 509.3 ÷ (1×10-12) = 5.093× 1014

So sound intensity = 10 × log (2.546×1014) = 10× 14.71 = 147.1 dB

These frequencies are beyond our hearing range. Frequencies lower than the left intercept are described as InfraSound; those higher than the right intercept are described as UltraSound.

Question: Use this technique to confirm that the lowest and highest Intensity Levels on the decibel scale, corresponding to 1×10-12W/m2 and 100 W/m2, are 0dB and 140dB. Answer: Lowest Intensity Level. I = 1×10-12W/m2 I/Io = 1×10-12W/m2 / 1×10-12W/m2 = 1 10×Log (I/Io) = 10×Log (1) = 10×0 = 0 dB

At this point it is worth identifying how sound intensity is handled.

Intensity Like light-intensity, sound intensity is measured in terms of the power (in Watts) of sound being delivered per square metre area.

Answer: Highest Intensity Level. I = 100W/m2 I/Io = 100W/m2 / 1×10-12W/m2 = 1014 10 × Log (I/Io) = 10 × Log (1014) = 10×14 = 140 dB

Exam Hint: Values for ‘I/Io’ and Log (I/Io) have NO units. Since the decibel scale is a logarithmic scale, an increase in Intensity Level of 10dB represents a ten-fold increase in intensity (i.e the sound intensity got 10 times greater.)

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157. The Physics of the Ear Although the Decibel scale is the most commonly used system for describing sound intensity levels, there is also the dBA scale. This is a modified version which takes into account the fact that we have a better response to some frequencies than others, and sets a common standard across all frequencies. When sound level meters are used, they incorporate the dBA scale as it is a better reflection of the sound level we would hear rather than the absolute sound intensity level.

Hearing loss 0

Frequency /kHz 5

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Loudness and loudness graphs

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Loudness is a way of describing a person’s interpretation of a sound. It is a combined effect of the intensity of the sound and the brain/ear’s processing of it. As a result there is no direct link between loudness and intensity.

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Equal loudness in phons Loss of hearing/dB

120

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Intensity in decibels

90 80

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Example 6: My gran, who’s 96 years old, cannot hear her modern, electronic doorbell very easily. Would a more old-fashioned metal-chime doorbell be better?

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This sort of hearing loss is most significant at the higher frequencies, and as a result the threshold of hearing changes in a distinctive way.

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100

Answer: Yes. Modern doorbells use speakers which are often very tinny and produce mainly high-frequency sound, which (due to her age) is now beyond her hearing range. A metal chime would doubtless produce a deeper sound with more low-frequencies, which would be well within her hearing range despite its limitations.

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1000 Frequency (Hz)

10,000

To test a person’s loudness profile (their interpretation or recognition of the loudness of sounds at various frequencies) the person will be asked to judge when the test-sound (at different frequencies) sounds to them to be as loud as a reference sound (usually 1kHz) often called a standard source. The intensity of the test-sound will then be adjusted until the person judges the loudness of the two sounds to be the same.

If a person has spent a long time in an environment where they were subjected to loud noises in a small range of frequencies (like machinery noise or engine noise) then the hearing becomes ‘desensitised’ in that range of frequencies. It’s possible that this is due either to damage to the tiny hairs in the cochlea that detect those particular frequencies, or damage to the nerves that they stimulate. As a consequence the hearing response is relatively unharmed at other frequencies but there is a notable decrease in the person’s threshold for hearing within the range.

You will notice that the unit ‘phons’ is used on a loudness graph. ‘Phons’ simply represent how many decibels the standard (reference) source is set at, so lines will have been obtained for the standard source being set at 10dB, 20dB, 30dB etc up to 120dB.

Frequency /kHz 0 0

For example, when the standard/reference source was set at 30dB this person judged that a 20Hz sound (where the line starts) was at the same loudness when it was at 85dB, and judged a 10000Hz sound to be at the same loudness when it was, in fact, set at 40dB.

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Example 5: With the reference level set at 40dB (i.e 40 ‘phons’) how loud did the 20Hz sound have to be for the person to judge it to be just as loud?

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Answer: 90dB

50

Hearing defects Hearing changes with age anyway – as we age the highest frequency we can detect gets progressively lower. You will probably be able to hear a noise at 15kHz (aged 17 or 18 you’re still able to detect sounds at the higher end of the hearing range) but it’s doubtful whether your parents will.

Loss of hearing/dB

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157. The Physics of the Ear Practice Questions 1. Sound travels by waves of increasing and decreasing air-pressure. Explain how this ultimately produces electrical impulses in the auditory nerve. You should mention the means by which the ‘sound’ is amplified in the middle ear. 4 marks max. 2. The upper-limit of human hearing is taken as approximately 100W/m2. Why is this set as the upper-limit of human hearing?

1 mark

3. A patient was examined to obtain the equal-loudness graph below. Use the graph to answer the questions. 120 100 80 Intensity (dB) 60 40 20 0 10

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46

100

215

464

1000

2154

4642

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Frequency (Hz)

(a) What was the sound intensity level (in dB) of the standard source for this curve? Explain how you decided upon your answer. (b) Describe how the graph might have been obtained

2 marks 3 marks

4. The headphones for an iPod are rated at 20mW. They have a diameter of 1cm. Calculate: (a) The surface area of the headphone. (b) The sound intensity (in W/m2) from the headphone. (c) The sound intensity level (in dB). (d) Comment on whether prolonged use of these headphones at maximum volume would be harmful to hearing.

2 2 3 2

marks marks marks marks

Answers 1. The increasing and decreasing pressure causes the eardrum to 4. (a) Area = π × r2 = π × 0.0052 = 7.85 × 10-5 m2 vibrate.3 (b) Sound intensity = 20 × 10-3 ÷ 7.85 × 10-5 = 254 W/m2. This vibration is transferred through the bones of the middle ear (c) Intensity Level = 10 × Log(I/Io) = 10 × Log (254/1×10-12) (hammer, anvil, stirrup) .v3 = 10 × Log (2.54×1014) The lever-effect of these bones amplifies the sound (by 1.5 times) v = 10 × 14.4 The stirrup sets up ‘shock-waves’/pressure-waves in the fluid of = 144dB the cochlea.3 (d) Yes it probably would, as 144dB is above the recommended The oval-window is far smaller than the eardrum so the pressure is limit. multiplied.3 The pressure-waves stimulate the tiny hairs inside the cochlea and trigger electrical impulsesv (max. 4) 2. Sounds of greater intensity than this could cause damage to hearing. 3 3. 60dB. 3 (a) The intensity is 60dB at a frequency of 1000Hz, and 1000Hz is taken as the reference frequency of the standard source.v (b) The standard source would be set at 60dB (or the same value as given in (a)) 3 The frequency would then be changed and the intensity then adjusted until the patient judged it to be of the same loudness 3 This would be repeated for a range of frequencies from (10Hz to 15000Hz) 3

Acknowledgements: This Physics Factsheet was researched and written by Beverly Rickwood The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

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