Questions on the ear and the transmission of sound

Question 1

State what is meant by the intensity of a sound wave, give the equation defining it, and state its unit.

Accepted Answer

Intensity is the power carried by the wave through unit area at right angles to its direction of travel (1). I = P/A, measured in W m⁻² (1).

Question 2

State what is meant by the threshold of hearing, and give its value.

Accepted Answer

The minimum intensity that a normal ear can detect, quoted at a frequency of 1 kHz (1). Its value is I₀ = 1.0 × 10⁻¹² W m⁻² (1).

Question 3

Describe the path taken by a sound from the air outside the ear to the nerve impulses that leave it, naming the outer, middle and inner sections.

Accepted Answer

Outer ear: the pinna collects the sound and funnels it along the ear canal to the eardrum, which vibrates (1). Middle ear: the three ossicles carry the vibration from the eardrum across to the smaller oval window. Inner ear: the oval window drives the fluid in the cochlea, where hair cells convert the vibration into nerve impulses (1).

Question 4

State what an equal loudness curve shows, and state what quantity is plotted on each axis.

Accepted Answer

It joins the combinations of frequency and intensity level that a listener judges to be equally loud (1). Intensity level in dB is plotted against frequency, on a logarithmic frequency axis (1).

Question 5

Calculate the intensity level of a sound of intensity 1.0 × 10⁻⁶ W m⁻².

Accepted Answer

Intensity level = 10 log(I/I₀) = 10 log[(1.0 × 10⁻⁶)/(1.0 × 10⁻¹²)] (1) = 10 log(10⁶) = 60 dB (1).

Question 6

State the approximate range of frequencies that a healthy young person can hear.

Accepted Answer

About 20 Hz to 20 kHz (1). The upper limit falls with age long before the lower one does.

Question 7

A sound has an intensity of 3.0 × 10⁻⁵ W m⁻². Calculate its intensity level.

Accepted Answer

I/I₀ = (3.0 × 10⁻⁵)/(1.0 × 10⁻¹²) = 3.0 × 10⁷ (1). Intensity level = 10 log(3.0 × 10⁷) (1) = 75 dB (1).

Question 8

A sound is measured at 85 dB. Calculate its intensity.

Accepted Answer

85 = 10 log(I/I₀), so I/I₀ = 10^8.5 (1) = 3.16 × 10⁸ (1). I = 3.16 × 10⁸ × 1.0 × 10⁻¹² = 3.2 × 10⁻⁴ W m⁻² (1).

Question 9

A tone of intensity level 66 dB falls on an eardrum of cross-sectional area 55 mm². Calculate the power incident on the eardrum.

Accepted Answer

I = I₀ × 10^6.6 = 4.0 × 10⁻⁶ W m⁻² (1). A = 55 mm² = 55 × 10⁻⁶ m² (1). P = IA = 4.0 × 10⁻⁶ × 55 × 10⁻⁶ = 2.2 × 10⁻¹⁰ W (1). The area conversion carries a factor of 10⁻⁶; a factor of 10⁻³ converts a length, not an area.

Question 10

Explain how the middle ear increases the pressure of the vibration reaching the inner ear, and state why this increase is necessary.

Accepted Answer

The three ossicles are linked as a lever, so the force delivered to the oval window is larger than the force the eardrum receives (1). The eardrum also has roughly twenty times the area of the oval window, so the force acts over a much smaller area and the pressure is correspondingly greater (1). It is necessary because the cochlea beyond the oval window is filled with fluid: without the gain almost all the sound would reflect at the air-to-fluid boundary rather than entering (1).

Question 11

Sound A has an intensity of 4.0 × 10⁻⁶ W m⁻² and sound B an intensity of 8.0 × 10⁻⁵ W m⁻². Calculate how many decibels louder B is than A.

Accepted Answer

Relative intensity level = 10 log(I_B/I_A) (1). I_B/I_A = (8.0 × 10⁻⁵)/(4.0 × 10⁻⁶) = 20 (1). 10 log 20 = 13 dB (1). Neither sound has to be referred to the threshold: the difference of two levels is the log of their ratio.

Question 12

Explain why sound is measured on a logarithmic scale rather than a linear one.

Accepted Answer

The audible range runs from about 1.0 × 10⁻¹² W m⁻² up to about 1 W m⁻², twelve powers of ten, and no linear axis can show both ends usefully (1). The ear also judges by ratio rather than by difference, so multiplying the intensity by a fixed factor is heard as a fixed step of loudness (1). A logarithmic scale turns those equal ratios into equal intervals, so equal steps in decibels correspond to equal steps in perceived loudness (1).

Question 13

Describe how an equal loudness curve is produced.

Accepted Answer

A reference tone of 1 kHz is set to a chosen intensity level (1). A test tone at some other frequency is played and its intensity is adjusted until the listener judges the two tones to be equally loud; the intensity level it needed is recorded at that frequency (1). This is repeated across the audible range and the points joined, and the results are averaged over many listeners; the curve is labelled by its own intensity level at 1 kHz (1).

Question 14

On an equal loudness curve, a tone of frequency 100 Hz has to be at an intensity level of 60 dB to sound as loud as a 1 kHz tone at 30 dB. Calculate the ratio of the intensity of the 100 Hz tone to that of the 1 kHz tone, and explain what this shows about the ear. State where on such a curve the ear is most sensitive.

Accepted Answer

The two levels differ by 60 − 30 = 30 dB (1). 30 = 10 log(I₁₀₀/I₁₀₀₀), so the ratio is 10³ = 1000 (1). The 100 Hz tone must carry a thousand times the intensity to be heard as equally loud, so the ear is far less sensitive at 100 Hz than at 1 kHz (2). The curves dip lowest between about 2 kHz and 5 kHz, near 3 kHz, and that is where the ear is most sensitive (1).

Question 15

A machine in a workshop produces an intensity level of 92 dB at a bench. A second, identical machine is switched on beside it. Calculate the new intensity level at the bench, and determine how many such machines would be needed to raise the level to 102 dB.

Accepted Answer

Intensities add, not levels: two identical machines give twice the intensity (1). 10 log 2 = 3.0 dB, so the new level is 92 + 3.0 = 95 dB (2). For 102 dB the level has risen by 10 dB above one machine (1), which is a factor of 10¹ in intensity, so 10 machines are needed (1). Adding the decibel figures themselves would predict 184 dB, which is above the threshold of pain and plainly wrong.

Question 16

A worker is exposed to a noise of 96 dBA. Ear defenders reduce the intensity reaching the ear by a factor of 250. Calculate the intensity level inside the defenders, and explain why a workplace noise limit is written in dBA rather than in dB.

Accepted Answer

The reduction in intensity level is 10 log 250 = 24 dB (2). The level inside is therefore 96 − 24 = 72 dBA (1). A reading in plain dB weights every frequency equally, so it overstates the effect of low-frequency rumble that the ear barely registers (1). The dBA scale applies a standard weighting curve, drawn to approximate the ear's response at moderate levels, before the level is computed, so its single number counts the frequencies the ear is sensitive to most heavily; that makes it an exposure metric rather than a measurement of perceived loudness, and an exposure metric is the fair basis for a limit (1).

Question 17

Compare the equal loudness curves of a person with normal hearing with those of a person who has spent years working in a noisy factory, and explain the practical difficulty the second person is likely to report.

Accepted Answer

The threshold curve of the damaged ear lies above that of the normal ear everywhere, so a greater intensity is needed before anything is heard at all (1). The lift is uneven: the high-frequency end rises far more than the middle of the range, and noise damage typically leaves the worst loss around 4 kHz (1). The higher curves change much less, so loud sounds still seem loud and the loss is easy to overlook (1). Consonants such as s, f and t carry their information at high frequency and low intensity, so those are lost first and following speech in a noisy room becomes difficult (1).