Noise, acoustics and sound reproduction
Why can a concert of a few hours damage hearing permanently, even though the sound does not feel painful?
Prior knowledge
- The sound intensity level $L$ is given in decibels, and it always rises by the same amount when the intensity is multiplied by the same factorchapter Loudness of sound
- A wave is reflected from an obstacle according to the law of reflection and bends behind the edge of an obstaclechapter A wave meets an obstacle or a gap
- Waves that meet add together, and they can strengthen or weaken each otherchapter When waves combine: superposition and interference
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 1.
Think about it
Saturday's festival gig was great. In the front row, in front of the speaker towers, the bass could be felt in the chest, but the ears did not hurt. On the way home a friend's speech sounded muffled, and at night there was ringing in the ears. By morning the ringing was almost gone. The friend's father says that he wears hearing protectors all day in the noisy hall at his workplace, even though the hall is nowhere near as loud as the gig.
What decides how much noise the ear can take, and how can hearing be protected?
Recap: two sources
Predict
Make your predictions before investigating 2 questions
Prediction 1: level and time
Prediction 2: how can sound silence sound?
Investigate
Investigate 1: the same dose at different levels
Values calculated by a model
The table shows noise levels $L$ and the daily times that bring the same sound energy to the ear as $8\,\text{h}$ at a level of $85\,\text{dB}$. The last column shows how many times greater the intensity is than at a level of $85\,\text{dB}$.
| $L$ (dB) | time | relative intensity |
|---|---|---|
| $85$ | $8\,\text{h}$ | $1$ |
| $88$ | $4\,\text{h}$ | $2$ |
| $91$ | $2\,\text{h}$ | $4$ |
| $94$ | $1\,\text{h}$ | $8$ |
| $97$ | $30\,\text{min}$ | $16$ |
Look at how the time changes when the level rises by $3\,\text{dB}$, and what happens to the product of the intensity and the time.
Continue the rule of the table: how many minutes may one stay at a level of $103\,\text{dB}$? Answer in minutes.
Hint 1: a guiding question
How many steps of $3\,\text{dB}$ are there from a level of $97\,\text{dB}$ to a level of $103\,\text{dB}$?
Hint 2: what is the connection
Each step halves the time: $30\,\text{min} \to 15\,\text{min} \to \ldots$
Hint 3: an intermediate step
A level of $103\,\text{dB}$ is six steps above $85\,\text{dB}$: $480\,\text{min} : 2^6$.
Show the solution
Given
the rule of the table: $t_0 = 8\,\text{h}$ at a level of $L_0 = 85\,\text{dB}$, and every rise of $3\,\text{dB}$ halves the time; level $L = 103\,\text{dB}$.
Find
the permitted time $t_\text{perm}$ in minutes.
Principle
In the table the intensity doubles and the time halves with every step of $3\,\text{dB}$, so the product of intensity and time, that is, the energy reaching the ear, stays the same. The same rule continues for higher levels.
Formula
$t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L - L_0}{3\,\text{dB}}$
Conversions
The answer is asked in minutes:
Substitution
Answer
$7.5\,\text{min}$
The rule is a rule of thumb, so two significant figures are enough.
Reasonableness
From the last row of the table ($97\,\text{dB}$, $30\,\text{min}$) two halvings lead to the result: $30\,\text{min}$, then $15\,\text{min}$ and finally $7.5\,\text{min}$. The intensity is $64$ times as great, and
so the dose is the same.
Investigate 2: a level measured with a phone
Made-up example data
Invented sample data: students measured, with a sound level app on a phone, the level of a small loudspeaker playing steady noise at different distances in two places: on an empty sports field and in an empty classroom whose walls, ceiling and floor are hard. The settings of the loudspeaker were the same in both places.
| distance | field | classroom |
|---|---|---|
| $1\,\text{m}$ | $82\,\text{dB}$ | $82\,\text{dB}$ |
| $2\,\text{m}$ | $76\,\text{dB}$ | $78\,\text{dB}$ |
| $4\,\text{m}$ | $70\,\text{dB}$ | $76\,\text{dB}$ |
| $8\,\text{m}$ | $64\,\text{dB}$ | $75\,\text{dB}$ |
The reading of an uncalibrated phone app can differ from the true level by several decibels, but readings from the same phone with the same settings can be compared with each other.
Conclusion: what do the tables show?
Write, on the basis of the tables: how does the permitted time change when the noise level rises by $3\,\text{dB}$, and why? What does the measurement in the classroom tell us about why a room with hard surfaces is noisy?
Model answer
When the level rises by $3\,\text{dB}$, the intensity doubles and the time halves, so the product of intensity and time, that is, the energy reaching the ear, stays the same. That is why the same dose builds up in an hour already at $94\,\text{dB}$. In the classroom the level dropped only $7\,\text{dB}$ over a distance of $8\,\text{m}$, whereas on the field it dropped $18\,\text{dB}$. Hard surfaces reflect sound, and the reflected sound fills the room, so it is not quiet even far from the source. The numbers are values calculated with the rule of thumb and invented sample data, not real measurements.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Noise and hearing
Noise is sound that is not wanted or that is harmful. The same sound can be music to one person and noise to another, but whether it harms hearing does not depend on taste: what matters is the level of the sound and how long one listens to it. Even moderate noise weakens concentration and sleep and increases stress, so noise affects well-being even when hearing is not damaged.
Hearing damage arises in the cochlea of the inner ear. There the hair cells change the vibration of sound into nerve signals. A loud sound strains and damages the hair cells. The dulling of hearing and the ringing in the ears felt after a concert are signs of strain. They often pass, but repeated strain leads to permanent hearing loss, because human hair cells do not regenerate. Permanent ringing in the ears, or tinnitus, is a common result of noise. Sound causes pain only at a level of about $120\,\text{dB}$, so pain does not warn in time.
Dose: level and time
The dose that harms hearing builds up in the same way as energy: what matters is how much energy the sound brings to the ear during the whole exposure. In the chapter Loudness of sound we learned that a rise of $3\,\text{dB}$ in the intensity level corresponds to roughly a doubling of the intensity. When the intensity doubles, twice as much energy reaches the ear every second, so the same dose is reached in half the time. That is why a noise level only a few decibels higher can shorten the safe time many times over.
Noise at work in law
In Finland, noise at work is regulated by Government Decree 85/2006 on the protection of workers from the risks arising from noise. The decree compares a worker's daily noise exposure, that is, the average noise level over an eight-hour working day, with action values. When the lower action value $80\,\text{dB}$ is exceeded, the employer must make hearing protectors available to the worker and inform the worker about the dangers of noise. When the upper action value $85\,\text{dB}$ is exceeded, protectors must be used and noise must be reduced according to a plan. The daily exposure at the ear, taking the attenuation of the hearing protectors into account, must not exceed the limit value $87\,\text{dB}$. The values are A-weighted: the meter weights frequencies roughly as the ear hears them. The decree also has its own C-weighted limits for peak values of sound, which apply to a gunshot or an impact, for example. The law does not set the same limits for the noise of concerts and headphones, but the ear cannot tell leisure from work.
Permitted exposure time
The upper action value corresponds to eight hours of exposure to a level of $85\,\text{dB}$. The same dose can be received in a shorter time at a higher level, and every rise of $3\,\text{dB}$ halves the time. This rule of thumb follows from the fact that noise exposure is calculated as an average of energy. The permitted daily exposure time is
$$t_\text{perm} = \frac{t_0}{2^{\,n}}$$
$$n = \frac{L - L_0}{3\,\text{dB}}$$
where $L$ is the noise level at the ear (dB), $L_0 = 85\,\text{dB}$ is the upper action value, $t_0 = 8\,\text{h}$ is the corresponding time, and $n$ tells how many steps of $3\,\text{dB}$ the level $L$ is above the action value. For example, at a level of $94\,\text{dB}$ we have $n = 3$, so the time is halved three times: $8\,\text{h} : 8 = 1\,\text{h}$. When $n$ is not a whole number, the power is calculated with a calculator.
The formula is an estimate, not a guarantee: people's hearing withstands noise in different ways, and exposure should be kept clearly below the limit. The noise level at the ear is obtained with the rules of the chapter Loudness of sound: two equally strong sources raise the level by about $3\,\text{dB}$, and doubling the distance from a point source lowers the level in an open space by about $6\,\text{dB}$. A hearing protector lowers the level reaching the ear by its attenuation, when it is well fitted.
Noise control
Noise can be reduced in three places. At the source, a quieter machine is chosen, it is serviced, its vibration is isolated, or it is enclosed in a sound-absorbing casing. Along the path, the distance is increased or a barrier such as a noise wall is built. The barrier stops the direct sound, but sound bends over its edge because of diffraction (chapter A wave meets an obstacle or a gap), and the lower the sound, the more it bends. That is why a barrier dampens low sounds less well than high ones. Soft, porous surfaces dampen reflections along the path. At the ear, hearing protectors are used: ear defenders or earplugs, which typically attenuate by $20$–$30\,\text{dB}$.
Control at the source is the most effective, because it protects everyone. A hearing protector protects only when it is on: at a level of $100\,\text{dB}$ the dose for the whole day is reached in just $15\,\text{min}$, so taking the protector off for a moment cancels a large part of its benefit. At a concert, earplugs, distance from the loudspeakers and breaks in a quiet place are simple means, and the volume of headphones should be kept moderate.
Room acoustics
Reflection and absorption
In a room, sound reaches the listener directly and reflected from the walls, ceiling and floor. Reflected sound travels a longer distance, so it arrives slightly later. Hard, smooth surfaces such as concrete, glass and tiles reflect most of the sound. Soft, porous materials such as mineral wool panels, textiles and people's clothes absorb sound: in them the vibration of the air changes into internal energy of the material. In a reverberant room the reflected sound fills the whole room, so the level falls less with distance from the source than outdoors.
Reverberation
When a sound source falls silent, reflections can still be heard for a moment. This dying sound is called reverberation. The reverberation time $T$ is the time in which the sound level falls by $60\,\text{dB}$ after the source has fallen silent; the intensity has then fallen to a millionth. Long reverberation muddles the syllables of speech together, and the room feels noisy, because people raise their voices over each other. In a classroom a reverberation time of about half a second is often aimed for, while in a concert hall music benefits from a time of about two seconds. Absorbing panels in the ceiling and on the walls shorten the reverberation time.
Active noise cancelling
Noise-cancelling headphones have a microphone that hears the noise of the environment. The electronics of the headphones form from it a signal that is in the opposite phase, and the loudspeaker of the headphones plays it. In the ear canal the pressure variations of the noise and of the anti-noise add together and largely cancel each other: this is destructive interference (chapter When waves combine: superposition and interference). The method works best for steady, low hum, such as the sound of an engine or ventilation. The period and the wavelength of a low sound are long, so the anti-noise easily stays in the opposite phase. At high frequencies even a small timing error spoils the cancellation, and they are dampened by the cups of the headphones. Noise-cancelling headphones allow music to be listened to at a lower volume, but they do not replace hearing protectors in loud noise.
Loudspeaker and microphone
A loudspeaker changes an electrical signal into sound. The alternating current from the amplifier flows in a coil that is attached to a cone and sits around the pole of a permanent magnet. The magnet exerts on the current-carrying coil a force whose direction changes with the direction of the current, so the coil and the cone vibrate at the same frequency as the current. The cone compresses and rarefies the air in front of it, and a sound wave is produced. Energy changes from electrical energy into the kinetic energy of the cone and further into sound. Most of the electrical energy, however, changes into internal energy of the coil: the efficiency of a loudspeaker is often only about $1\,\%$, so an electrical power of $50\,\text{W}$ produces a sound power of only about $0.5\,\text{W}$. The interplay of magnetism and electricity is studied in more detail on later courses.
A microphone works the other way round. The pressure variations of the sound move a thin diaphragm. In a dynamic microphone the diaphragm moves a coil around the pole of a magnet, and an alternating voltage that varies in step with the sound is produced in the coil. Energy changes from sound into the kinetic energy of the diaphragm and from that into an electrical signal. A loudspeaker and a microphone are thus similar in construction, but energy changes in opposite directions in them.
Common misconception
“The ear gets used to noise.” Noise may stop feeling disturbing, but that is because the brain learns to filter out steady sound. The strain on the hair cells does not decrease with getting used to it; the damage accumulates unnoticed over the years. Hearing loss is often noticed only when it becomes difficult to distinguish speech in a hubbub, and by then the damage can no longer be repaired.
Summary
- Noise is unwanted or harmful sound. The dose that harms hearing depends on both level and time.
- The upper action value $85\,\text{dB}$ corresponds to $8\,\text{h}$, and every rise of $3\,\text{dB}$ halves the permitted time: $t_\text{perm} = t_0 / 2^{\,n}$.
- Noise is controlled at the source, along the path (distance, barrier, absorbing surfaces) and at the ear (hearing protectors for the whole exposure).
- Hard surfaces reflect and soft ones absorb sound. The reverberation time is the time in which the level falls by $60\,\text{dB}$.
- Anti-noise cancels low noise by destructive interference.
- A loudspeaker changes an electrical signal into the motion of a cone and into sound, a microphone changes sound into the motion of a diaphragm and into an electrical signal.
Optional Return to your predictions
Compare your predictions with the table and the Theory summary. Then return to the opening question: why can a gig of a few hours in the front row strain hearing more than a working day in a noisy hall with protectors on? What means could be used at a gig to protect hearing?
Would you change your prediction?
Apply
Worked example
Problem
A drummer practises with a band for $2.0\,\text{h}$ a day, and the noise level at the drummer's ear is $100\,\text{dB}$. By at least how many decibels must the hearing protectors attenuate, so that the noise exposure of the practice does not exceed the dose that corresponds to the upper action value $85\,\text{dB}$ for eight hours?
Step 1/7
Given and find. Noise level $L = 100\,\text{dB}$, exposure time $t = 2.0\,\text{h}$, reference dose $L_0 = 85\,\text{dB}$ for a time $t_0 = 8\,\text{h}$. We are asked for the minimum attenuation $\Delta L$ of the hearing protectors.
Step 2/7
Principle. The dose depends on both the level and the time, and every rise of $3\,\text{dB}$ halves the permitted time. We look for the level $L_\text{perm}$ at which the permitted time is exactly $2.0\,\text{h}$. The protectors must lower the level reaching the ear to this value, so the attenuation is the difference of the levels.
Step 3/7
Formula solved. From the formula $t = \tfrac{t_0}{2^{\,n}}$ we get
$$\begin{aligned} 2^{\,n} &= \frac{t_0}{t} \\ L_\text{perm} &= L_0 + n \cdot 3\,\text{dB} \\ \Delta L &= L - L_\text{perm} \end{aligned}$$
Step 4/7
Conversions. Not needed: both times are in hours, so the unit cancels, and the levels are in decibels.
Step 5/7
Substitution.
so $n = 2$, because $2^2 = 4$. The permitted level is
and the attenuation is
Step 6/7
Answer. The hearing protectors must attenuate by at least $9\,\text{dB}$. The levels are given as whole decibels, so the result is given as whole decibels.
Step 7/7
Reasonableness. Without protectors the permitted time would be $8\,\text{h} : 2^5 = 15\,\text{min}$, so a two-hour practice would give eight times the dose. Eight times the intensity corresponds to three steps of $3\,\text{dB}$, that is, $9\,\text{dB}$, as we obtained. Ordinary protectors attenuate by $20$–$30\,\text{dB}$, so the requirement is easy to meet; musicians often use plugs that attenuate evenly so that playing sounds natural.
Find the error
Apply to a new situation
A summer worker helps while another person saws boards with a circular saw outdoors. The noise level of the saw is $100\,\text{dB}$ at a distance of $1.0\,\text{m}$, and the helper stands $4.0\,\text{m}$ from the saw. Assume the saw is a point source in an open space.
For how many hours a day may the helper be in this noise without hearing protectors, when the reference is the upper action value? Answer in hours.
Hint 1: a guiding question
First calculate the level at the helper's position. How many times does the distance double when $1.0\,\text{m}$ becomes $4.0\,\text{m}$?
Hint 2: what is the connection
Every doubling lowers the level by $6\,\text{dB}$: $L = 100\,\text{dB} - 2 \cdot 6\,\text{dB}$. Then use the formula $t_\text{perm} = t_0 / 2^{\,n}$.
Hint 3: an intermediate step
At the helper's position $L = 88\,\text{dB}$, so $n = 1$ and $t_\text{perm} = 8\,\text{h} : 2$.
Show the solution
Given
$L_1 = 100\,\text{dB}$ at a distance of $r_1 = 1.0\,\text{m}$, the helper's distance $r_2 = 4.0\,\text{m}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$; point source in an open space.
Find
the helper's permitted daily time $t_\text{perm}$ in hours.
Principle
In an open space the intensity of a point source falls to a quarter every time the distance doubles, so the level falls by $6\,\text{dB}$ (chapter Loudness of sound). From the helper's level the permitted time follows from the dose rule, in which every $3\,\text{dB}$ halves the time.
Formula
$L_2 = L_1 - k \cdot 6\,\text{dB}$, where $2^{\,k} = \dfrac{r_2}{r_1}$, and $t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L_2 - L_0}{3\,\text{dB}}$
Conversions
Not needed: the distances are in metres, so their ratio has no unit, and the time comes out directly in hours.
Substitution
so
Then
Answer
$4.0\,\text{h}$
Two significant figures as in the distances $1.0\,\text{m}$ and $4.0\,\text{m}$.
Reasonableness
Next to the saw, at a level of $100\,\text{dB}$, the time would be only $15\,\text{min}$. At the helper's position the intensity is $16$ times smaller, so the time is
A working day is longer than that, however, so even the helper should wear protectors.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The noise level rises from $85\,\text{dB}$ to $95\,\text{dB}$. How many times greater is the intensity, and in how long does the same dose build up as in $8\,\text{h}$ at a level of $85\,\text{dB}$?
Answer
A rise of $10\,\text{dB}$ corresponds to ten times the intensity (chapter Loudness of sound). The same energy builds up in a tenth of the time: $480\,\text{min} : 10 = 48\,\text{min}$. The rule of thumb $t_\text{perm} = t_0 / 2^{\,n}$ gives the same: $n = 10/3$ and $t_\text{perm} \approx 48\,\text{min}$.
- After a concert the ears ring, but by morning the ringing has gone. Does that mean that hearing was not damaged?
Answer
Not necessarily. Ringing and temporary dulling of hearing show that the hair cells of the inner ear were strained. Hearing may recover almost fully, but repeated strain leads to permanent hearing loss and may cause permanent tinnitus, because hair cells do not regenerate.
1. Reverberation
2. Earplugs at a concert
In the front row of a concert the noise level is $106\,\text{dB}$. A listener wears earplugs whose attenuation is $15\,\text{dB}$.
For how many hours can the listener stay in the front row before the ear receives the daily dose that corresponds to the upper action value? Answer in hours.
Hint 1: a guiding question
What level reaches the ear through the plugs?
Hint 2: what is the connection
At the ear $L = 106\,\text{dB} - 15\,\text{dB}$. How many steps of $3\,\text{dB}$ is that above $85\,\text{dB}$?
Hint 3: an intermediate step
$L = 91\,\text{dB}$, so $n = 2$ and $t_\text{perm} = 8\,\text{h} : 2^2$.
Show the solution
Given
noise level $L_\text{noise} = 106\,\text{dB}$, attenuation of the plugs $\Delta L = 15\,\text{dB}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$.
Find
the permitted time $t_\text{perm}$ in hours.
Principle
The plugs reduce the level reaching the ear by their attenuation. The dose depends on the level at the ear, and every $3\,\text{dB}$ above the level that corresponds to the upper action value halves the permitted time.
Formula
$L = L_\text{noise} - \Delta L$ and $t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L - L_0}{3\,\text{dB}}$
Conversions
Not needed: the levels are in decibels and the time comes out directly in hours.
Substitution
Answer
$2.0\,\text{h}$
; the levels are whole decibels and the rule is an estimate, so two significant figures are enough.
Reasonableness
Without plugs the time would be $8\,\text{h} : 2^7 \approx 0.063\,\text{h}$, that is, about $4\,\text{min}$. An attenuation of $15\,\text{dB}$ is five steps, so the time grows to $2^5 = 32$ times as long:
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Loudspeaker and microphone
Explain how a loudspeaker changes the electrical signal of an amplifier into sound and how a microphone works the other way round. Say which forms of energy appear in each and where most of the electrical energy given to a loudspeaker ends up.
Model answer
In a loudspeaker the alternating current of the amplifier flows in a coil that is around the pole of a permanent magnet and attached to a cone. The magnet exerts on the coil a force whose direction changes with the current, so the cone vibrates at the frequency of the current and compresses and rarefies the air: sound is produced. Energy changes from electrical energy into kinetic energy and sound, but most of it, often about $99\,\%$, changes into internal energy of the coil. In a microphone the pressure variations of the sound move a diaphragm, and in a dynamic microphone the coil attached to the diaphragm moves around the pole of the magnet, so an alternating voltage that follows the sound is produced in the coil. Energy changes from sound into kinetic energy and into an electrical signal.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Sound wave
- A longitudinal wave of compressions and rarefactions travelling through a medium.
- Noise
- Unwanted or disturbing sound.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 16-1Core
On a building site the noise level of a hammer drill at the user's ear is $91\,\text{dB}$.
For how many hours a day may the user use the tool without hearing protectors, when the reference is the upper action value, $8\,\text{h}$ at a level of $85\,\text{dB}$? Answer in hours.
Answer
$2.0\,\text{h}$
Show the solution
Given
noise level $L = 91\,\text{dB}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$
Find
the permitted daily time $t_\text{perm}$ in hours
Principle
Noise exposure is calculated as an average of energy: every rise of $3\,\text{dB}$ doubles the intensity, so the same dose is reached in half the time.
Formula
$t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L - L_0}{3\,\text{dB}}$
Conversions
Not needed: the levels are in decibels and the time comes out in hours.
Substitution
$n$$= \dfrac{91\,\text{dB} - 85\,\text{dB}}{3\,\text{dB}}$$= 2$$t_\text{perm}$$= \dfrac{8\,\text{h}}{2^2}$$= 2.0\,\text{h}$Answer
$2.0\,\text{h}$
The levels are whole decibels and the rule is an estimate, so two significant figures are enough.
Reasonableness
Two halvings: $8\,\text{h} \to 4\,\text{h} \to 2\,\text{h}$. The intensity is four times as great, and $4 \cdot 2\,\text{h} = 8\,\text{h}$, so the dose is the same as in the reference.
- Exercise 16-2Core
A ventilation unit has two identical fans. When only one is running, the noise level at the door of the machine room is $70\,\text{dB}$.
What is the level at the same place when both fans are running?
Answer
$73\,\text{dB}$
Show the solution
Given
level of one fan $L_1 = 70\,\text{dB}$; two identical fans
Find
the combined level $L$ in decibels
Principle
The intensities of two sources add, not the levels. Two equally strong sources double the intensity (chapter Loudness of sound).
Formula
$L$$= L_1 + 10\,\text{dB} \cdot \lg \dfrac{2 I_1}{I_1}$$= L_1 + 10\,\text{dB} \cdot \lg 2$Conversions
Not needed.
Substitution
$L$$= 70\,\text{dB} + 10\,\text{dB} \cdot \lg 2$$= 70\,\text{dB} + 3.01\,\text{dB}$$= 73.01\,\text{dB}$Answer
$73\,\text{dB}$
In whole decibels like the given value.
Reasonableness
The level rises only a little even though the number of noisy devices doubles. A result of $140\,\text{dB}$ would be above the threshold of pain, which does not suit two fans.
- Exercise 16-3
The level of a loudspeaker at a garden party is $88\,\text{dB}$ at a distance of $4.0\,\text{m}$.
What is the level at a distance of $8.0\,\text{m}$? Treat the loudspeaker as a point source in an open space with no reflections.
Answer
$82\,\text{dB}$
Show the solution
Given
$L_1 = 88\,\text{dB}$ at a distance $r_1 = 4.0\,\text{m}$, new distance $r_2 = 8.0\,\text{m}$; point source in an open space
Find
the level $L_2$ at the distance $r_2$
Principle
The intensity of a point source is inversely proportional to the square of the distance, so double the distance gives a quarter of the intensity. That lowers the level by $10\,\text{dB} \cdot \lg 4 \approx 6\,\text{dB}$ (chapter Loudness of sound).
Formula
$L_2 = L_1 - 6\,\text{dB}$, when $r_2 = 2 r_1$
Conversions
Not needed:
$r_2 : r_1$$= 8.0\,\text{m} : 4.0\,\text{m}$$= 2$Substitution
$L_2$$= 88\,\text{dB} - 6\,\text{dB}$$= 82\,\text{dB}$Answer
$82\,\text{dB}$
In whole decibels like the given value.
Reasonableness
The level decreases but stays of the same order of magnitude: a quarter of the intensity is not a quarter of the decibel number. In a reflective space the drop would be smaller.
- Exercise 16-4
The noise level of a pneumatic hammer at the user's ear is $106\,\text{dB}$. The user wears ear defenders whose attenuation is $23\,\text{dB}$.
What is the noise level reaching the ear?
Answer
$83\,\text{dB}$
Show the solution
Given
noise level $L_\text{noise} = 106\,\text{dB}$, attenuation of the protectors $\Delta L = 23\,\text{dB}$
Find
the level $L$ reaching the ear
Principle
The attenuation of a hearing protector tells by how many decibels it lowers the level reaching the ear, so the attenuation is subtracted from the noise level.
Formula
$L = L_\text{noise} - \Delta L$
Conversions
Not needed.
Substitution
$L$$= 106\,\text{dB} - 23\,\text{dB}$$= 83\,\text{dB}$Answer
$83\,\text{dB}$
In whole decibels like the given values.
Reasonableness
The level stays below the upper action value $85\,\text{dB}$, so the protectors are enough for the whole working day if they are well fitted and worn all the time. Without protectors the permitted time would be about $4\,\text{min}$.
- Exercise 16-5
To which point of noise control, the source, the path or the ear, does each measure belong? a) a noise barrier beside a motorway, b) enclosing a compressor in a sound-absorbing casing, c) earplugs, d) sound-absorbing panels in the ceiling of a factory hall.
Answer
a) path, b) source, c) ear, d) path
Show the solution
Find
whether each measure belongs to the source, the path or the ear
Principle
At the source the sound produced is reduced, along the path the travel of sound from the source to the listener is blocked or dampened, and at the ear the sound reaching the ear is reduced.
Model answer
a) A noise barrier stops the direct sound between the road and the homes, so it is a measure along the path. b) A casing encloses the sound right around the machine, so it controls noise at the source. c) Earplugs attenuate sound only at the ear. d) Panels fitted in the ceiling do not make the machine quieter, but they absorb the reflections through which the sound travels to the worker, so they are a measure along the path.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 16-6Core
A factory hall has two identical machines. Each alone produces a level of $85\,\text{dB}$ at the worker's position.
For how many hours a day may the worker stay in the hall without hearing protectors when both machines are running? The reference is the upper action value. Answer in hours.
Answer
$4.0\,\text{h}$
Show the solution
Given
two identical machines, each $L_1 = 85\,\text{dB}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$
Find
the permitted daily time $t_\text{perm}$ in hours
Principle
The intensities of two equally strong sources add, so the intensity doubles and the level rises by about $3\,\text{dB}$. Then every $3\,\text{dB}$ above the action value halves the permitted time.
Formula
$L = L_1 + 3\,\text{dB}$ and $t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L - L_0}{3\,\text{dB}}$
Conversions
Not needed.
Substitution
$L$$= 85\,\text{dB} + 3\,\text{dB}$$= 88\,\text{dB}$$n$$= \dfrac{88\,\text{dB} - 85\,\text{dB}}{3\,\text{dB}}$$= 1$$t_\text{perm}$$= \dfrac{8\,\text{h}}{2^1}$$= 4.0\,\text{h}$Answer
$4.0\,\text{h}$
The rule is an estimate, so two significant figures are enough.
Reasonableness
The second machine brings as much energy again, so the total energy builds up twice as fast and the dose is reached in half the time. The result is consistent: $2 \cdot 4\,\text{h} = 8\,\text{h}$.
- Exercise 16-7Core
A student listens to music with headphones on bus journeys for a total of $1.5\,\text{h}$ a day, and the level at the student's ear is about $94\,\text{dB}$.
What percentage is the student's daily dose of the dose that corresponds to the upper action value ($8\,\text{h}$ at a level of $85\,\text{dB}$)?
Answer
$150\,\%$
Show the solution
Given
level at the ear $L = 94\,\text{dB}$, listening time $t = 1.5\,\text{h}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$
Find
the dose as a percentage of the reference dose
Principle
First calculate the permitted time at a level of $94\,\text{dB}$: in that time exactly the reference dose builds up. The share of the dose is the ratio of the listening time to the permitted time, because the level is the same all the time.
Formula
$t_\text{perm} = \dfrac{t_0}{2^{\,n}}$, where $n = \dfrac{L - L_0}{3\,\text{dB}}$, and share $= \dfrac{t}{t_\text{perm}} \cdot 100\,\%$
Conversions
Not needed: the times are in hours, and the unit cancels.
Substitution
$n$$= \dfrac{94\,\text{dB} - 85\,\text{dB}}{3\,\text{dB}}$$= 3$$t_\text{perm}$$= \dfrac{8\,\text{h}}{2^3}$$= 1.0\,\text{h}$share $= \dfrac{1.5\,\text{h}}{1.0\,\text{h}} \cdot 100\,\% = 150\,\%$
Answer
$150\,\%$
Two significant figures as in the time $1.5\,\text{h}$.
Reasonableness
The listening time is longer than the permitted time, so the dose exceeds the reference, that is, more than $100\,\%$ is expected. If the sound were lowered by $3\,\text{dB}$, the dose would be halved to $75\,\%$.
- Exercise 16-8
The noise level of a lawn mower is $78\,\text{dB}$ at a distance of $2.0\,\text{m}$. A neighbour wants the level on their terrace to be at most $60\,\text{dB}$.
How far from the mower should the terrace be? Treat the mower as a point source in an open space.
Answer
$16\,\text{m}$
Show the solution
Given
$L_1 = 78\,\text{dB}$ at a distance $r_1 = 2.0\,\text{m}$, desired level $L_2 = 60\,\text{dB}$; point source in an open space
Find
the distance $r_2$ in metres
Principle
In an open space every doubling of the distance lowers the level by $6\,\text{dB}$, because the intensity falls to a quarter (chapter Loudness of sound). We calculate how many doublings are needed.
Formula
$k = \dfrac{L_1 - L_2}{6\,\text{dB}}$ and $r_2 = 2^{\,k} \cdot r_1$
Conversions
Not needed.
Substitution
$k$$= \dfrac{78\,\text{dB} - 60\,\text{dB}}{6\,\text{dB}}$$= 3$$r_2$$= 2^3 \cdot 2.0\,\text{m}$$= 16\,\text{m}$Answer
$16\,\text{m}$
Two significant figures as in the distance $2.0\,\text{m}$.
Reasonableness
The distances $2\,\text{m}$, $4\,\text{m}$, $8\,\text{m}$ and $16\,\text{m}$ give the levels $78\,\text{dB}$, $72\,\text{dB}$, $66\,\text{dB}$ and $60\,\text{dB}$. In a yard the walls of houses reflect sound, so in reality a slightly longer distance would be needed.
- Exercise 16-9
Hands are clapped once in an empty classroom. A sound level app on a phone shows that after the clap the level falls from $92\,\text{dB}$ to $62\,\text{dB}$ in $0.45\,\text{s}$. Assume the level falls steadily.
The reverberation time is the time in which the level falls by $60\,\text{dB}$, but it can be deduced even if the level cannot be followed through the whole drop of $60\,\text{dB}$.
What is the reverberation time of the classroom?
Answer
$0.90\,\text{s}$
Show the solution
Given
the level falls from $L_1 = 92\,\text{dB}$ to $L_2 = 62\,\text{dB}$ in a time $\Delta t = 0.45\,\text{s}$; the fall is steady
Find
the reverberation time $T$, that is, the time in which the level falls by $60\,\text{dB}$
Principle
When the level falls steadily, it falls by the same number of decibels every second. The reverberation time follows from the rate of fall: a fall of $60\,\text{dB}$ takes as many times longer as the measured fall is smaller. The measurement is often done this way, because the background noise of the room hides the full drop of $60\,\text{dB}$.
Formula
$T$$= \dfrac{60\,\text{dB}}{L_1 - L_2} \cdot \Delta t$Conversions
Not needed.
Substitution
$T$$= \dfrac{60\,\text{dB}}{92\,\text{dB} - 62\,\text{dB}} \cdot 0.45\,\text{s}$$= 2 \cdot 0.45\,\text{s}$$= 0.90\,\text{s}$Answer
$0.90\,\text{s}$
Two significant figures as in the time $0.45\,\text{s}$.
Reasonableness
The measured fall of $30\,\text{dB}$ is half of $60\,\text{dB}$, so the time doubles. The result is longer than the roughly half a second aimed for in classrooms, which suits an empty classroom with hard surfaces.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 16-10
A warehouse worker spends $4.0\,\text{h}$ of the working day at the loading dock, where the noise level is $88\,\text{dB}$, and the remaining $4.0\,\text{h}$ in the packing room, where the level is $82\,\text{dB}$.
What percentage is the worker's daily dose of the dose that corresponds to the upper action value ($8\,\text{h}$ at a level of $85\,\text{dB}$)?
Answer
$125\,\%$
Show the solution
Given
$t_1 = 4.0\,\text{h}$ at a level $L_1 = 88\,\text{dB}$, $t_2 = 4.0\,\text{h}$ at a level $L_2 = 82\,\text{dB}$; reference $L_0 = 85\,\text{dB}$ and $t_0 = 8\,\text{h}$
Find
the day's dose as a percentage of the reference dose
Principle
The dose is the energy accumulated in the ear, so the shares of the different periods are added. The share of each period is the ratio of its duration to the permitted time at that level.
Formula
$t_{\text{perm},i}$$= \dfrac{t_0}{2^{\,n_i}}$where $n_i = \dfrac{L_i - L_0}{3\,\text{dB}}$, and dose $= \left(\dfrac{t_1}{t_{\text{perm},1}} + \dfrac{t_2}{t_{\text{perm},2}}\right) \cdot 100\,\%$
Conversions
Not needed: the times are in hours, and the unit cancels.
Substitution
$n_1 = 1$, so
$t_{\text{perm},1}$$= 8\,\text{h} : 2$$= 4.0\,\text{h}$$n_2$$= \dfrac{82\,\text{dB} - 85\,\text{dB}}{3\,\text{dB}}$$= -1$so
$t_{\text{perm},2}$$= 8\,\text{h} \cdot 2$$= 16\,\text{h}$Dose $= \left(\dfrac{4.0\,\text{h}}{4.0\,\text{h}} + \dfrac{4.0\,\text{h}}{16\,\text{h}}\right) \cdot 100\,\% = (1 + 0.25) \cdot 100\,\% = 125\,\%$
Answer
$125\,\%$
The times are given to two figures, so the result is an estimate, about $1.3$ times the dose. The upper action value is exceeded, so protectors must be used.
Reasonableness
The $4\,\text{h}$ at the loading dock alone fill the whole dose, so the result must be over $100\,\%$. The share of the packing room is small, because there the intensity is only a quarter of the intensity at the dock.
- Exercise 16-11
Two terraced houses each have an identical air-source heat pump outdoor unit in the yard. The level of each unit is $55\,\text{dB}$ at a distance of $3.0\,\text{m}$. The bedroom window of a third house is $12\,\text{m}$ from each unit.
What is the level at the window when both units are running? Treat the units as point sources in an open space.
Answer
$46\,\text{dB}$
Show the solution
Given
two identical units, each $L_1 = 55\,\text{dB}$ at a distance $r_1 = 3.0\,\text{m}$; the window's distance from each $r_2 = 12\,\text{m}$; point sources in an open space
Find
the level $L$ at the window when both are running
Principle
First calculate the level of one unit at the window: every doubling of the distance lowers the level by $6\,\text{dB}$. The intensities of two equally strong sources add, so the combined level is $3\,\text{dB}$ higher.
Formula
$L_\text{one} = L_1 - k \cdot 6\,\text{dB}$, where $2^{\,k} = \dfrac{r_2}{r_1}$, and $L = L_\text{one} + 3\,\text{dB}$
Conversions
Not needed:
$r_2 : r_1$$= 12\,\text{m} : 3.0\,\text{m}$$= 4$so $2^{\,k} = 4$ and $k = 2$.
Substitution
$L_\text{one}$$= 55\,\text{dB} - 2 \cdot 6\,\text{dB}$$= 43\,\text{dB}$$L$$= 43\,\text{dB} + 3\,\text{dB}$$= 46\,\text{dB}$Answer
$46\,\text{dB}$
In whole decibels like the given value.
Reasonableness
Because of the distance the level of one unit is clearly lower than close by, and the second unit raises it only a little. The result is below a quiet conversation, but in a bedroom a steady hum can still disturb sleep.
- Exercise 16-12
Noise-cancelling headphones dampen the low hum of an aircraft engine, whose frequency is about $100\,\text{Hz}$, well. The speech of passengers, whose frequencies are about $1000$–$3000\,\text{Hz}$, can be heard through clearly better. Explain the difference using wavelengths and phase, when the speed of sound in air is $343\,\text{m/s}$.
Answer
The wavelength of the hum is about $3.4\,\text{m}$ and its period $10\,\text{ms}$, so the anti-noise stays in the opposite phase at the ear; the wavelength of speech is only about $11$–$34\,\text{cm}$, and even a small difference in timing or position spoils the cancellation.
Show the solution
Find
why anti-noise dampens a low hum better than speech
Principle
Anti-noise cancels noise by destructive interference only if it is in the opposite phase at the ear. The wavelength $\lambda = \dfrac{v}{f}$ and the period $T = \dfrac{1}{f}$ tell how precisely the timing and the position must match.
Model answer
The wavelength of the hum is $\lambda = \dfrac{343\,\text{m/s}}{100\,\text{Hz}} \approx 3.4\,\text{m}$ and its period $10\,\text{ms}$. The whole ear canal and the inside of the cup are small compared with the wavelength, and the electronics have time to form the opposite signal, so the anti-noise stays in the opposite phase at the ear. The wavelengths of speech are $\dfrac{343\,\text{m/s}}{3000\,\text{Hz}} \approx 11\,\text{cm}$ and $\dfrac{343\,\text{m/s}}{1000\,\text{Hz}} \approx 34\,\text{cm}$, and the period is at most $1\,\text{ms}$. Even a small timing error or a difference in position of a few centimetres shifts the phase a lot, and then the waves no longer cancel each other. In addition, speech varies quickly, whereas a hum is steady. High sounds are dampened mainly by the cups of the headphones.