Sound Intensity, Decibels and Applications of Sound – Sound Waves (Senior 6 Physics) | Ganzaa

Sound Waves

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Senior 6 · Physics

Sound Waves

5 lessons

Lesson 5 of 5

Sound Intensity, Decibels and Applications of Sound

25 min

How loud is too loud?

A wedding band with big speakers, a motorcycle engine, a workshop grinder: every day many people in Kigali and elsewhere spend hours near sounds loud enough to damage their hearing slowly and permanently. To judge the danger we need to measure sound, and the ear itself gives a clue about how: it responds to an enormous range of powers, from the faintest whisper to a jet engine, a range of more than a million million. This lesson turns that range into the decibel scale and then shows how sound is put to work.

Intensity and the inverse-square law

The intensity of a wave is the power it carries per unit area perpendicular to its direction of travel. A small source radiating sound power $P$ equally in all directions spreads it over a sphere of area $4\pi r^2$ , so

I = \frac{P}{A} = \frac{P}{4\pi r^2} \qquad \Rightarrow \qquad \frac{I_2}{I_1} = \left(\frac{r_1}{r_2}\right)^{2}

Figure 5.1: The same power spreads over 4 times the area at twice the distance.

Intensity also depends on the wave itself. The energy of each oscillating layer of air is proportional to $\omega^2 s_m^2$ , and a detailed calculation gives

I = \tfrac{1}{2}\rho v \omega^2 s_m^2 = \frac{(\Delta p_m)^2}{2\rho v}

So intensity is proportional to the square of the amplitude and, for a fixed displacement amplitude, to the square of the frequency.

Definition

Sound intensity $I$ is the sound power passing through unit area at right angles to the direction of travel: $I = \frac{P}{A}$ , unit $\mathrm{W\,m^{-2}}$ . The threshold of hearing $I_0 = 1.0 \times 10^{-12}\,\mathrm{W\,m^{-2}}$ is the faintest sound a healthy ear detects at 1000 Hz; at about $1\,\mathrm{W\,m^{-2}}$ sound becomes painful.

The decibel scale

Because the ear responds to ratios of intensity, roughly equally to each tenfold step, sound is measured on a logarithmic scale. The sound intensity level is

\beta = 10\log_{10}\left(\frac{I}{I_0}\right) \ \mathrm{dB}, \qquad I_0 = 1.0 \times 10^{-12}\,\mathrm{W\,m^{-2}}

Useful consequences: multiplying $I$ by 10 adds 10 dB; doubling $I$ adds $10\log_{10}2 = 3\,\mathrm{dB}$ ; doubling the distance from a point source divides $I$ by 4 and so subtracts 6 dB. The threshold of hearing is 0 dB and the threshold of pain about 120 dB.

Sound level	Example	Safe listening time per week
30 dB	soft whispering	unlimited
60 dB	normal conversation	unlimited
80 dB	doorbell	40 hours
85 dB	heavy traffic (inside a car)	12 h 30 min
90 dB	shouted conversation	4 hours
95 dB	motorcycle	1 h 15 min
100 dB	hair dryer	20 minutes
110 dB	shouting in the ear	2.5 minutes
120 dB	standing near a siren	12 seconds

Levels, examples and safe listening times from the World Health Organization, "Deafness and hearing loss: safe listening" (questions and answers).

Notice the pattern: every 3 dB rise doubles the intensity and roughly halves the safe time (80 dB for 40 hours is the same sound energy as 83 dB for 20 hours).

Definition

The sound intensity level $\beta$ of a sound of intensity $I$ is $\beta = 10\log_{10}\frac{I}{I_0}$ , measured in decibels (dB), where $I_0 = 1.0 \times 10^{-12}\,\mathrm{W\,m^{-2}}$ . It compares a sound with the threshold of hearing; it is a ratio, not an amount of energy.

Worked example 1: a loudspeaker at a party

A loudspeaker radiates $0.20\,\mathrm{W}$ of sound equally in all directions. Find the intensity and the intensity level 5.0 m away.

I = \frac{P}{4\pi r^2} = \frac{0.20}{4\pi \times 5.0^2} = 6.4 \times 10^{-4}\,\mathrm{W\,m^{-2}}

\beta = 10\log_{10}\frac{6.4 \times 10^{-4}}{1.0 \times 10^{-12}} = 10\log_{10}(6.4 \times 10^{8}) = 88\,\mathrm{dB}

By the WHO table (safe time halves for every 3 dB above 85 dB), roughly 6 hours a week at this level is the safe limit.

Worked example 2: adding sources, moving away

(a) One grinder in a workshop gives 80 dB at a worker's position. What level do two identical grinders give? (b) A generator gives 90 dB at 2.0 m. How far away is the level 70 dB?

(a) Intensities add, not decibels: $\beta = 80 + 10\log_{10}2 = 83\,\mathrm{dB}$ (not 160 dB).

(b) A fall of 20 dB means $I$ is divided by $10^{2} = 100$ . Since $I \propto \frac{1}{r^2}$ , $r$ must be multiplied by $\sqrt{100} = 10$ : $r = 20\,\mathrm{m}$ .

Worked example 3: pressure amplitude of a loud sound

Find the pressure amplitude of a 100 dB sound in air ( $\rho = 1.20\,\mathrm{kg\,m^{-3}}$ , $v = 343\,\mathrm{m\,s^{-1}}$ ).

I = I_0 \times 10^{\frac{100}{10}} = 1.0 \times 10^{-2}\,\mathrm{W\,m^{-2}}

\Delta p_m = \sqrt{2\rho v I} = \sqrt{2 \times 1.20 \times 343 \times 1.0 \times 10^{-2}} = 2.9\,\mathrm{Pa}

Less than three thousandths of one percent of atmospheric pressure, yet harmful after 20 minutes a week.

Experiment 5.1: testing the inverse-square law

Apparatus: a loudspeaker and signal generator (a steady 1 kHz tone), a sound level meter or a phone with a sound level app, a 10 m tape measure, open ground away from walls (Figure 5.2).

Figure 5.2: Measuring the sound level at doubling distances.

Set a steady tone. Measure the background level with the tone off.
With the tone on, record the level at 1, 2, 4 and 8 m, holding the meter at the height of the speaker.
If the inverse-square law holds, each doubling of distance lowers the level by 6 dB. One class found 86, 80, 75 and 69 dB (illustrative results).

Reflections from the ground and nearby walls, and background noise at large distances, are the main reasons real results differ from exactly 6 dB per doubling.

Warning

Safety. Keep the level at 1 m below about 85 dB and never put your ear close to the loudspeaker. Keep the tests short. Choose an area without traffic, and do not leave the tone running while people are nearby.

Applications of sound and ultrasound

Ultrasound imaging

An ultrasound probe contains a piezoelectric crystal that sends short pulses of ultrasound, typically a few megahertz, into the body and detects the echoes. The time of each echo gives the depth of a boundary ( $d = \frac{vt}{2}$ , with $v \approx 1540\,\mathrm{m\,s^{-1}}$ in soft tissue), and the strength of each echo depends on how different the two materials are. That difference is measured by the acoustic impedance $Z = \rho v$ . For a wave meeting a boundary head-on, the fraction of the intensity reflected is

R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^{2}

Figure 5.3: Echoes from tissue boundaries. The percentages are the reflected fractions of intensity calculated in Worked example 4.

Medium	$Z$ ( $\mathrm{kg\,m^{-2}\,s^{-1}}$ )
air	$429$
fat	$1.34 \times 10^{6}$
muscle	$1.70 \times 10^{6}$
bone	$7.8 \times 10^{6}$ (upper value)

Impedances from OpenStax College Physics, Table 17.7.1.

Ultrasound is used because its short wavelength shows small detail, and it is not ionising radiation, unlike X-rays; at the low intensities used for imaging it is considered safe, which is why it is used to check on a baby before birth.

Doppler ultrasound

Ultrasound reflected from blood cells moving at speed $u$ towards the probe is Doppler shifted twice: once as the cells receive it (moving observer) and once as they re-emit it (moving source):

f_{\text{echo}} = f\,\frac{v + u}{v - u} \quad \Rightarrow \quad \Delta f = f\,\frac{2u}{v - u} \approx \frac{2fu}{v} \quad (u \ll v)

The machine measures $\Delta f$ as a beat frequency and so finds the speed of blood flow.

Other uses

SONAR and echo sounding: ships find depth, shoals of fish and underwater objects from echo times.
Echolocation: bats and dolphins locate prey from ultrasonic echoes.
Cleaning and industry: ultrasonic baths shake dirt off jewellery and instruments; pulses find cracks inside metal castings.
Architecture: halls are designed so reverberation is neither too long nor too short.
Noise control: ear defenders, sound-absorbing panels and, in headphones, anti-noise signals that interfere destructively with the noise.

Worked example 4: why the gel?

Use the table to find the fraction of ultrasound intensity reflected (a) at an air-to-muscle boundary, (b) at a fat-muscle boundary, (c) at a muscle-bone boundary.

\text{(a)}\; R = \left(\frac{1.70 \times 10^{6} - 429}{1.70 \times 10^{6} + 429}\right)^{2} = 0.999

\text{(b)}\; R = \left(\frac{1.70 - 1.34}{1.70 + 1.34}\right)^{2} = 0.014

\text{(c)}\; R = \left(\frac{7.8 - 1.70}{7.8 + 1.70}\right)^{2} = 0.41

(a) An air gap between probe and skin would reflect 99.9% of the ultrasound, so almost none would enter the body: the gel removes the air. (b) Soft-tissue boundaries reflect about 1.4%, enough to see and weak enough that most of the pulse carries on deeper. (c) Bone reflects about 41%, which is why it shows bright and casts a "shadow" behind it.

Worked example 5: blood flow

A 4.0 MHz Doppler probe points straight along an artery in which blood moves towards it at $0.30\,\mathrm{m\,s^{-1}}$ . Find the frequency shift ( $v = 1540\,\mathrm{m\,s^{-1}}$ ), and the smallest detail the 4.0 MHz wave can show.

\Delta f \approx \frac{2fu}{v} = \frac{2 \times 4.0 \times 10^{6} \times 0.30}{1540} = 1.6 \times 10^{3}\,\mathrm{Hz}

\lambda = \frac{v}{f} = \frac{1540}{4.0 \times 10^{6}} = 3.9 \times 10^{-4}\,\mathrm{m} \approx 0.4\,\mathrm{mm}

The shift of about 1.6 kHz is in the audible range, so many machines play it through a loudspeaker as a "whooshing" sound.

Warning

Common mistakes. (1) Adding decibels: two 80 dB sources give 83 dB, not 160 dB. (2) Forgetting to square in $I \propto \frac{1}{r^2}$ or $R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^{2}$ . (3) Using $\log$ to base $e$ instead of base 10. (4) Using $v$ in air for ultrasound in the body: use about 1540 m s⁻¹.

Tip

Decibel shortcuts to learn: ×2 intensity = +3 dB; ×10 = +10 dB; ×100 = +20 dB; double the distance = −6 dB.

How does ultrasound work? (NIBIB, the US National Institute of Biomedical Imaging and Bioengineering). Watch how ultrasound uses echoes to “see” inside the body, and compare it with Figure 5.3.

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Practice

Find the intensity level of a sound of intensity $3.2 \times 10^{-5}\,\mathrm{W\,m^{-2}}$ .
A sound has an intensity level of 95 dB. Find its intensity.
At 3.0 m from a small generator the intensity level is 84 dB. Find the level at 12 m, assuming the inverse-square law.
How many identical machines, each giving 75 dB at a point, together give 85 dB there?
An ultrasound echo from a boundary returns $65\,\mu\mathrm{s}$ after the pulse is sent. How deep is the boundary ( $v = 1540\,\mathrm{m\,s^{-1}}$ )?
Using the impedance table, explain why an ultrasound scan cannot see through the lungs well.
A factory worker is exposed to 90 dB for 8 hours on each of 5 days a week. Use the WHO table to evaluate the risk and suggest two ways to reduce it.

Show answersHide answers

$\beta = 10\log_{10}\frac{3.2 \times 10^{-5}}{1.0 \times 10^{-12}} = 10\log_{10}(3.2 \times 10^{7}) = 75\,\mathrm{dB}$
$I = 1.0 \times 10^{-12} \times 10^{9.5} = 3.2 \times 10^{-3}\,\mathrm{W\,m^{-2}}$
The distance is multiplied by 4, so $I$ is divided by 16: $\Delta\beta = 10\log_{10}16 = 12\,\mathrm{dB}$ , giving 72 dB.
+10 dB needs 10 times the intensity: 10 machines.
$d = \frac{vt}{2} = \frac{1540 \times 65 \times 10^{-6}}{2} = 0.050\,\mathrm{m} = 5.0\,\mathrm{cm}$
The lungs are full of air. The impedance of air (429) is tiny compared with that of tissue (about $1.7 \times 10^{6}$ ), so almost all the ultrasound (99.9%) is reflected at the first tissue-air boundary and almost none passes through.
The weekly safe time at 90 dB is 4 hours; the worker receives 40 hours, ten times the safe dose, so permanent hearing loss is likely. Reduce it by wearing ear defenders or earplugs, enclosing or silencing the machines, moving the worker further away (−6 dB per doubling of distance) and shortening the time spent near them.

Key points

$I = \frac{P}{4\pi r^2}$ for a point source; $I \propto s_m^2$ and $I = \frac{(\Delta p_m)^2}{2\rho v}$ .
$\beta = 10\log_{10}\frac{I}{I_0}$ dB with $I_0 = 1.0 \times 10^{-12}\,\mathrm{W\,m^{-2}}$ ; +3 dB doubles the intensity.
Ultrasound imaging uses echoes at impedance boundaries, $R = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^{2}$ ; Doppler ultrasound measures blood speed from $\Delta f \approx \frac{2fu}{v}$ .

1How loud is too loud?

2Intensity and the inverse-square law

3The decibel scale

4Experiment 5.1: testing the inverse-square law

5Applications of sound and ultrasound