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Chapter 6 · 6 hours

Noise pollution

Practice questions

Practice questions and answers

5 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Explain the nature of sound (wave properties, frequency, wavelength and speed). With a neat sketch, describe the structure and working of the human ear and state the audible range.

Answer

Nature of sound

Sound is a longitudinal pressure wave that travels through an elastic medium (air, liquid or solid) by alternate compression and rarefaction of particles. It cannot travel in vacuum.

  • Frequency ff (Hz): number of cycles per second. Time period T=1/fT = 1/f.
  • Wavelength λ\lambda (m): distance between consecutive compressions.
  • Speed cc: c=fλc = f\lambda. In air c=331.4+0.6 Tc = 331.4 + 0.6\,T (T in ∘^\circC) ≈343\approx 343 m/s at 20 ∘^\circC; about 1480 m/s in water and 5000 m/s in steel.
  • Amplitude (sound pressure) decides loudness; frequency decides pitch.
  • Audible range for a healthy young person: 20 Hz to 20 000 Hz; below is infrasound, above is ultrasound.

Human ear

  Outer ear      Middle ear           Inner ear
 +--------+   +-----------------+   +--------------+
 | pinna  |   |  ossicles       |   |  cochlea     |
 | canal  |-->|  hammer-anvil-  |-->|  (fluid +    |--> auditory
 |   |    |   |  stirrup        |   |  hair cells) |    nerve
 |eardrum |   |  (Eustachian    |   |  semicircular|    to brain
 +--------+   |   tube)         |   |  canals      |
              +-----------------+   +--------------+
  1. Outer ear: pinna collects sound and the ear canal guides it to the eardrum (tympanic membrane), which vibrates.
  2. Middle ear: three small bones (malleus, incus, stapes) amplify and transmit the vibration to the oval window. The Eustachian tube equalises pressure.
  3. Inner ear: the fluid-filled cochlea contains the basilar membrane with thousands of hair cells. Vibration makes the fluid move, bending the hair cells that convert it into nerve signals, carried by the auditory nerve to the brain. High frequencies are sensed near the base and low frequencies near the apex of the cochlea. The semicircular canals do not hear but control balance.

Damage to hair cells by loud noise is permanent.

  • Practice · 4 marks

(a) Define sound pressure level (SPL) and sound power level (PWL). (b) A sound has an rms pressure of 0.2 Pa. Find its SPL. (c) A machine radiates 0.01 W of sound power. Find its PWL, and its SPL at 10 m on open flat ground (hemispherical radiation). Reference values: p_ref = 20 micro-Pa and W_ref = 10^-12 W.

Answer

(a) Definitions

  • SPL: Lp=10log⁡10(p2pref2)=20log⁡10(ppref)L_p = 10\log_{10}\left(\dfrac{p^2}{p_{ref}^2}\right) = 20\log_{10}\left(\dfrac{p}{p_{ref}}\right) dB, with pref=20 μp_{ref} = 20\ \muPa (threshold of hearing at 1 kHz). It depends on the distance from the source.
  • PWL: Lw=10log⁡10(WWref)L_w = 10\log_{10}\left(\dfrac{W}{W_{ref}}\right) dB, with Wref=10−12W_{ref} = 10^{-12} W. It is the property of the source and is independent of distance and surroundings.

(b) SPL

Lp=20log⁡10(0.220×10−6)=20log⁡10(104)=80 dBL_p = 20\log_{10}\left(\frac{0.2}{20\times10^{-6}}\right) = 20\log_{10}(10^4) = 80\ \text{dB}

(c) PWL and SPL at 10 m

Lw=10log⁡10(0.0110−12)=10log⁡10(1010)=100 dBL_w = 10\log_{10}\left(\frac{0.01}{10^{-12}}\right) = 10\log_{10}(10^{10}) = 100\ \text{dB}

For hemispherical spreading, intensity I=W/(2πr2)I = W/(2\pi r^2), so

Lp=Lw−20log⁡10r−8=100−20log⁡10(10)−8=100−20−8=72 dBL_p = L_w - 20\log_{10} r - 8 = 100 - 20\log_{10}(10) - 8 = 100 - 20 - 8 = 72\ \text{dB}

(The constant is 10log⁡102π=810\log_{10}2\pi = 8.)

Answer: (b) SPL = 80 dB; (c) PWL = 100 dB; SPL at 10 m = 72 dB.

  • Practice · 4 marks

Three machines in a workshop produce sound pressure levels of 85 dB, 88 dB and 80 dB at a worker's position when operating individually. Find the combined SPL when all work together. What is the level if the 80 dB machine is switched off? Also, why can two equal sources never raise the level by more than 3 dB?

Answer

Sound levels are logarithmic and cannot be added directly. Add the intensities (or mean-square pressures):

Ltotal=10log⁡10(∑i10Li/10)L_{total} = 10\log_{10}\left(\sum_i 10^{L_i/10}\right)

All three machines

108.5=3.162×108108.8=6.310×108108.0=1.000×108Sum=10.472×108\begin{aligned} 10^{8.5} &= 3.162\times10^{8} \\ 10^{8.8} &= 6.310\times10^{8} \\ 10^{8.0} &= 1.000\times10^{8} \\ \text{Sum} &= 10.472\times10^{8} \end{aligned} Ltotal=10log⁡10(10.472×108)=10×9.0200=90.2 dBL_{total} = 10\log_{10}(10.472\times10^{8}) = 10\times9.0200 = 90.2\ \text{dB}

Without the 80 dB machine

L=10log⁡10[(3.162+6.310)×108]=10log⁡10(9.472×108)=89.8 dBL = 10\log_{10}\left[(3.162+6.310)\times10^{8}\right] = 10\log_{10}(9.472\times10^{8}) = 89.8\ \text{dB}

Switching off the quietest source reduces the total only by 0.4 dB.

Two equal sources

Ltotal=10log⁡10(2×10L/10)=L+10log⁡102=L+3 dBL_{total} = 10\log_{10}(2\times10^{L/10}) = L + 10\log_{10}2 = L + 3\ \text{dB}

Doubling the energy adds 3 dB. If the two levels differ, the increase is smaller (for a difference of 10 dB it is only about 0.4 dB), so 3 dB is the maximum. Thus nn equal sources give L+10log⁡10nL + 10\log_{10}n.

Answer: Combined level = 90.2 dB; without the 80 dB machine = 89.8 dB.

  • Practice · 5 marks

Give typical sound levels (in dB) at different places and activities. Explain the harmful effects of noise on human beings.

Answer

Typical noise levels

Source / placeLevel (dB)
Threshold of hearing0
Rustling leaves, quiet bedroom20-30
Library, quiet residential area at night30-40
Normal conversation (1 m), office55-65
Busy street traffic, restaurant70-80
Heavy truck, factory, loud radio85-95
Motorcycle, power tools, disco95-105
Aircraft take-off at 100 m, thunder110-120
Threshold of pain120-130
Jet engine at close range, firecrackers140

Guideline values (WHO): about 55 dB(A) outdoors in the daytime in residential areas, 30 dB(A) bedroom at night. Industrial limit for an 8-hour exposure is typically 85-90 dB(A).

Effects of noise

  1. Hearing damage: temporary threshold shift and permanent hearing loss from long exposure above 85 dB(A), or sudden very loud sound (acoustic trauma). Tinnitus (ringing in the ears).
  2. Physiological: raised blood pressure and heart rate, stress hormone release, heart disease, headache, digestive disturbance.
  3. Psychological: annoyance, irritability, anxiety, reduced concentration, fatigue.
  4. Sleep disturbance: waking, poor sleep quality.
  5. Communication and performance: speech interference, accidents, reduced work efficiency and learning in children.
  6. Other: damage to buildings from vibration (e.g. sonic boom), and stress in animals.
  • Practice · 8 marks

Explain the methods of controlling noise at the source, along the transmission path and at the receiver. Include the use of barriers, enclosures and absorbing materials.

Answer

Every noise problem has three parts: source, path, receiver. Control is best applied at the source, then along the path, and last at the receiver.

  SOURCE  ---------- PATH ---------->  RECEIVER
  (reduce)      (barrier, enclosure,     (ear plugs,
                absorber, distance)       muffs)

1. Control at the source

  • Choose quieter machines and processes (belt drives rather than gears, electric rather than pneumatic tools).
  • Proper design, balancing of rotating parts, lubrication and maintenance to cut vibration and friction.
  • Vibration isolation: spring mounts, rubber pads and flexible connections.
  • Mufflers (silencers) for engine and fan exhaust: reactive (expansion chamber) and dissipative (absorptive lining) types.
  • Reduce impact forces and speed, use damping materials on panels, and keep pipe velocity low.
  • Legislation: limit noise from vehicles, horn restrictions, and no loudspeakers at night.

2. Control along the path

  • Distance: noise falls 6 dB for each doubling of distance (point source).
  • Barriers: a solid wall or earth bund between source and receiver gives 5-15 dB reduction for a wall that is high enough to break the line of sight; it should be dense (about 10 kg/m2^2), without gaps.
  • Enclosures: the machine is placed in a sealed box lined inside with absorbing material, with vibration breaks and silenced ventilation openings; gives 15-30 dB.
  • Absorbing materials: porous materials (mineral wool, glass wool, foam, acoustic tiles, curtains, carpets) turn sound energy into heat and reduce the reverberant level inside rooms. Absorption coefficient α\alpha ranges from 0 to 1.
  • Transmission loss (TL): heavy partitions and double walls with an air gap give high TL (mass law: 6 dB for each doubling of mass or frequency).
  • Planning: zoning, buffer zones, trees (small effect, mostly psychological) and layout of buildings away from highways.

3. Protection at the receiver

  • Ear plugs (reduce about 20-30 dB) and ear muffs (about 25-35 dB), limiting exposure time by job rotation, regular audiometric testing, and sound-proof control rooms for operators.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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