METHOD CONCEPT / EDITOR
Boom voortgezet onderwijs
AUTHORS
Ton van den Broeck
Sjef Buil
Freek Hoogeveen
Laurens Jeronimus
Peter Koopmans
Donald Staal
Arjen Wielemaker
POLARIS
PHYSICS
VWO / GYMNASIUM
SCHOOL YEAR 3
BOOM VOORTGEZET ONDERWIJS
Contents 1 Sound
1.1 Sound vibrations 8 1.2 Properties of sound vibrations 14 1.3 Resonance 20 1.4 Sound intensity 26 1.5 Hearing 32
Test preparation 38 2 Force and motion
2.1 Types of force 42 2.2 Measuring forces 48 2.3 Resultant force 54 2.4 Force and acceleration 60 2.5 Road safety 66
Test preparation 72 3 Energy
3.1 Energy conversions 76 3.2 Work and power 82 3.3 Heat 88 3.4 Renewable energy 94 3.5 Energy transition 100
Test preparation 106
4 Electricity
4.1 Electrical power and capacity 110
4.2 Resistance 116
4.3 Electrical circuits 122
4.4 Sensor circuits 128
4.5 Transmission of electrical energy 134
Test preparation 140
5 Radiation
5.1 Electromagnetic radiation 144
5.2 The atomic model 150
5.3 Radioactivity 156
5.4 Half-life and activity 162
5.5 Radiation hazards 168
Test preparation 174
6 Materials
6.1 Tensile strength and elasticity 178
6.2 Refraction of light 184
6.3 Heat conduction 190
6.4 Electrical conduction 196
6.5 Magnetism 202
Test preparation 208
Reference
A Practicals
A1 Quantities and units 212
A2 Range and precision 213
B Graphs
B1 Reading graphs 214
B2 Making graphs 215
C Calculation
C1 Prefixes 216
C2 Powers of ten 216
C3 Converting units 217
C4 Converting compound units 218
C5 Equations 218
C6 Calculating with equations 220
C7 A systematic approach to assignments 221
C8 Proportions 222
Research 224 E Design 226 Image credits 228 Register of terms 229
1 Sound
1.1 Sound vibrations 8
1.2 Properties of sound vibrations 14
1.3 Resonance 20
1.4 Sound intensity 26
1.5 Hearing 32 Test preparation 38
1.1 Sound vibrations
GoaL You learn what sound is and how sound moves.
Sound sources It’s very rarely completely quiet around you. You always hear sound from somewhere. Something that makes sound is called a sound source. Sound is caused by something in the sound source moving back and forth quickly; your vocal cords produce your voice, a guitar’s music comes from its strings, while a loudspeaker works by the movement of the diaphragm. This sort of back-and-forth motion is called a vibration
A tuning fork (Figure 1.1) is also a sound source. If you strike the tuning fork, both fork tips vibrate. A fork tip starts in the rest position, and then moves all the way to the left. Then it goes through the rest position all the way to the right and back to the rest position. This is one complete vibration.
Medium The vibration of a sound source causes the air around it to move. The molecules in the air close to the sound source start to vibrate and transmit the vibration to the molecules next to them (Figure 1.2). This is how sound vibration moves through air. Air is the medium (the substance in which something acts) needed for the sound vibration to propagate. A sound vibration that is transmitted is called a sound wave.
Receiver When a sound wave reaches your ear, you hear it because the air molecules make your eardrum vibrate. The vibration of the eardrum is converted into a signal that is sent to your brain. The ear is called the receiver of the sound. Sound therefore moves from the sound source through a medium to the receiver (Figure 1.3).
Sound source Medium Receiver
1.3 Sound travels from the sound source through a medium to the receiver.
Rest position
1.1 The vibration of a tuning fork
The direction of motion
1.2 Molecules in the air vibrate and transmit the sound wave.
Speed of sound When you talk to someone, your vocal cords vibrate. A sound wave travels from your vocal cords through the air to the other person’s ear. This wave takes a certain time to travel from the sound source to the receiver. That time depends on the speed at which the wave moves; the speed of sound. In air at 20 °C, the speed of sound is 343 m/s. The speed of sound is calculated using the equation for speed:
v = s t
v speed of sound in m/s
s distance travelled in m
t time in s
The speed of sound depends on the medium through which the sound wave travels. A sound wave moves faster in water than in air. Table 1.4 shows the speed of sound in various substances. The speed of sound depends on the air temperature (T). Table 1.4 shows the values for water and air.
Example [1] You see a firework in the distance. It is 0 °C. You see a flash of light and after 1.5 s you hear a bang. Calculate your distance from the firework in metres.
If: t = 1.50 s T = 0°C
Calculate: s in m
Calculation: 1 Calculate the speed of sound: See Table 1.4. The speed at 0°C is 0.332 km/s = 332 m/s.
2 Fill in the equation:
v = s t 332 = s 1.50
s = 332 × 1.50 = 498 m
Answer: You are 498 m from the firework.
1.4 Speed of sound in various substances
Exercises
1
a Give three examples of sound sources. T1
b What do you call the rapid back-and-forth motion that creates a sound source? R
2 True or False? R
a A sound comes from a sound source.
b A sound vibration moves as a sound wave.
c The speed of sound in air depends on the temperature.
d Your sense of hearing is a sound source.
3 Convert. T1
a 20 km = …… m
b 1,500 km/h = …… m/s
c 334 m/s = …… km/h
d 1 h 20 min = …… h
e 4,606 s = …… h
4 In a laboratory, scientists measure the speed of sound in a particular material by measuring the time it takes for sound to travel 1 m.
In material A, the sound takes 0.85 ms and in material B the sound takes 0.20 ms.
a Calculate the speed of sound in materials A and B.
b What materials are these? Use Table 1.4 to determine this. T2
5 Figure A
This athlete is listening to music on his phone while running.
a What is the sound source in this situation? And what is the receiver? T2
b Explain what the medium is. T2
6 A cannon fires blanks to celebrate a national holiday. Julian is 500 metres from the cannon. He sees a flash of light and hears the cannon go off 1.5 seconds later. Estimate the air temperature on that day. Use Table 1.4. T2
7 A building site supervisor is watching pile-driving on the construction site on a summer’s day from a distance of 250 m. The supervisor sees the piledriver hit the top of the pile.
a What is the speed of sound in air at 20 °C in m/s? T1
b Calculate how many seconds later the supervisor hears the blow on the pile. If necessary, see reference C6 – Calculating with equations. T1
c Dewi is also watching the pile-driving. She hears the blow of the piledriver after a delay of 1.2 seconds. Calculate how much further Dewi is from the work than the supervisor. T2
d The piledriver blows occur at regular intervals of 1 second. How far away do you have to stand from the piledriver to hear the sound of one blow at the same time as you see the next one? I
8 a Explain why the speed of sound in the mountains is lower than at sea level. T2
b There is no atmosphere on the Moon. An astronaut drops a hammer on the ground, which is made of rock. Does the astronaut hear the hammer hitting the ground? Explain your answer. I
HOME EXPERIMENT
Using both hands, hold an inflated balloon against your mouth. What do you feel on your fingers when you talk?
b
9 Figure B
During construction work, a hole is drilled in a 0.5-m thick concrete wall. A construction worker is 10 m away on the other side of the wall.
a Calculate how long it takes for the sound to get from the drill to the other side of the wall. Ignore the depth of the drill hole. T1
b It is 10 °C. Determine the speed of sound from Figure B. T1
c Calculate how long it takes for the sound from the drill to reach the construction worker. T2
10 Figure B
a Calculate the speed of sound at 50 °C and at –50 °C.
b Explain how to make a thermometer by measuring the speed of sound. I
Find out more
11 Read the text on the right about how ultrasound is used for injuries. Use the table to answer the following questions.
a What is the speed of sound in muscle? R
b Explain which tissue transition gives the clearest echo. T2
c After 0.04 ms, the sensor receives an echo of the sound that has only passed through muscle. Calculate the distance from the sensor at which the signal was reflected back. T2
d The sensor receives a very weak echo and, a little later, a very clear echo of the same signal. Explain why this occurs. I
12 A gel is applied between the sensor and the skin to ensure that there is no air between them.
a Explain why the sound cannot enter the body tissue when there is air between the sensor and the skin. T2
b What is the ideal speed of sound in the gel to avoid an echo at the transition from the sensor to the fatty tissue in the skin? I
13 The sensor cannot transmit and receive sound simultaneously, so the sensor emits a signal every 0.33 ms. The sensor waits for an echo between transmitting two consecutive signals.
a Calculate the maximum distance the sensor can measure. T2
b Explain what goes wrong if the sensor receives an echo of the previous signal after 0.33 ms. I
Have you achieved the learning goal?
R I know the meaning of the following terms:
Sound source
Vibration
Sound wave
Medium
Receiver
Speed of sound
T1 I can explain how sound waves are formed and how you can detect them.
T2 I can calculate using the speed of sound in various substances.
I I can explain how internal injuries can be visualised using sound.
Ultrasound is often used in sports for imaging knee or ankle injuries. To get a good picture of an injury, the doctor needs to know what is going on inside your knee or ankle. One method is to take an X-ray, but this mainly reveals the bone structure. To see the muscles, tendons and other soft tissues, the doctor uses ultrasound. Ultrasound works with sound waves. The sound waves are partially reflected back at the transition between different tissues. The greater the difference in the speed of sound between the tissues, the more sound is reflected. The reflected sound is called the echo, and is converted into an image on a monitor (Figure 1.5). This makes it possible to examine joints, muscles and tendons.
1.5 An ultrasound of a knee
1.2 Properties of sound vibrations
G oa L You learn about the properties of sound vibrations.
Oscillogram The vibration of a tuning fork can be seen on a computer screen via a microphone. The microphone converts the sound signal into an electrical signal. The screen shows a displacement-time or (u,t) graph of the vibration. This is called an oscillogram. The oscillogram shows the time on the horizontal axis and the displacement on the vertical axis. The oscillogram in Figure 1.6 shows one complete vibration.
Period and frequency Figure 1.7 shows an oscillogram of a vibrating tuning fork. It shows how long one vibration lasts. The duration of one vibration is called the period. The number of vibrations in one second is called the frequency. The frequency is expressed in the unit Hertz (Hz). For example, 2 Hz is equal to two vibrations per second; one vibration lasts 0.5 s. The frequency of a vibration is calculated as follows:
f = 1 T
f frequency in Hz T period in s
Example [2] Calculate the frequency of the vibration shown on the oscillogram in Figure 1.7.
If: Determine from the graph: three vibrations last 7.5 ms.
Calculate: f in Hz
Calculation: 1 First convert the period from ms to s: T = 7.5 3 = 2.5 ms = 0.0025 s
2 Fill in the equation:
f = 1 T = f = 1 0.0025 = 400 Hz
Answer: The frequency of the vibration is 4.0 × 102 Hz.
Pitch The higher the frequency of a sound vibration, the higher the tone you hear. This is called the pitch. Figure 1.8 shows two vibrations. The vibration in the left-hand (u, t) graph has a longer period than the vibration in the right-hand (u,t) graph. The frequency of the vibration in the left-hand graph is lower, so the pitch is also lower. Musical tones are tones with specific frequencies. The frequency of the musical pitch A above middle C is 440 Hz. Table 1.9 shows an overview of musical tones and their frequencies.
Amplitude Not every tone sounds equally loud. The loudness of a tone depends on the maximum displacement of the sound vibration. The maximum displacement is called the amplitude of a vibration. The greater the amplitude, the louder the tone (Figure 1.10).
Compound tones Sounds that consist of a single frequency are called pure tones. A spoken word consists of different tones, and therefore different frequencies. These frequencies form a compound tone. Figure 1.11 shows how two pure tones form a compound tone. The compound tone can be made by adding together the displacements of the pure tones.
Exercises
14 What do you call a (u,t)-graph of a sound vibration? R
15 True or False? R
a A vibration with a period of 1 s has a frequency of 1 Hz.
b The amplitude of a sound vibration indicates how high the tone you hear is.
c The frequency of the musical tone A above middle C is 494 Hz.
d The higher the tone, the higher the frequency of the sound.
16 a The period is 25 ms. Calculate the frequency. T1
b The frequency is 15 kHz. Calculate the period. T1
17 Figure A Figure A shows a (u,t) graph.
a Determine the amplitude of the vibration. T1
b Explain how many vibrations are shown in the graph. T1
c Calculate the period in seconds.
d Calculate the frequency. T1
18 Figure B Figure B shows five different oscillograms.
a Explain which oscillogram shows the tone with the loudest sound. T1
b Explain which oscillogram shows the highest tone. T2
c Explain which oscillogram shows the tone with the lowest frequency. T2
d Explain which oscillogram shows a compound tone. T2
19 Figure C
a How many vibrations are shown on the oscillogram? Explain your answer. T1
b Calculate the period of this vibration. T2
c Calculate the frequency of this vibration. T2
20 Figure D
You make an oscillogram of a tone that gets weaker and weaker.
a Explain which graph shows your oscillogram. T2
b Explain why the tone is getting weaker. I
21 The wing of a mosquito goes back and forth very quickly. As a result, the mosquito produces a buzzing sound with a frequency of 1,400 Hz. Calculate the period of the mosquito’s wing in ms. T2
22 Figure E
The graphs show two tones.
a Use Table 1.9 to determine which two musical tones these are. T2
b Which of these two tones sounds higher? Explain your answer. T1
c Which of these two tones sounds louder? Explain your answer. T1
23 A musician strikes a drum.
a Why is the sound louder when the musician hits the drum harder? T2
b Explain why you no longer hear sound if the musician puts his hand on the drum immediately after striking it. T2
HOME EXPERIMENT
Download an app on your smartphone that you can use to create an oscillogram. Speak a few sentences into the microphone. What do you see? Whistle a nice continuous tone. What do you see?
Find out more
24 Figures F and G
Read the text on the right about speech recognition. Below is an oscillogram of a compound tone and the corresponding spectrogram.
a What are the two tones that make up the compound tone? T2
b Calculate the periods of the two tones. T2
C Explain how you can calculate these periods from the oscillogram. I
25 Figure H
The oscillogram shows a compound vibration. This vibration consists of two frequencies.
a Explain how you can see that the sound consists of two frequencies. T2
b Calculate both frequencies. T2
Have you achieved the learning goal?
R I know the meaning of the following terms:
Oscillogram
Period
Amplitude
Frequency
Pure tones
Compound tones
T1 I can recognise differences in the pitch and hardness of sound from oscillograms.
T2 I can determine the characteristics of sounds from an oscillogram.
I I can explain how to compare sounds with a spectrogram.
Speech recognition You can give instructions to your smartphone with your voice. Your smartphone features speech recognition to do this. Glasses are currently being developed that show you subtitles when someone speaks to you in a foreign language.
A spoken word consists of different tones, and therefore different frequencies. These frequencies form a compound vibration and make the word sound the way it does. In speech recognition, this vibration is split into the frequencies that occur in the vibration. The separate frequencies of a compound vibration are displayed in a spectrogram of the sound (Figure G on the left of the page). This spectrogram is a kind of fingerprint of the sound. Speech recognition software compares this spectrogram with the spectrograms of known sounds.
1.12 Your smartphone features speech recognition.
1.3 Resonance
G oa L You learn how tones can be created and strengthened.
Fundamental tone When you pluck a guitar string, it starts to vibrate. You then hear a tone with a certain frequency. That is called the fundamental tone. The fundamental tone of a string depends on the length, thickness and tension of the string.
The shorter the string, the higher the fundamental tone.
The thinner the string, the higher the fundamental tone.
The tighter the string, the higher the fundamental tone.
Harmonics The fundamental tone is not the only tone that a string produces. It also generates harmonics, which are whole multiples of the frequency of the fundamental tone. The A string of a guitar has a fundamental tone with a frequency of 110 Hz. At the same time, the A string vibrates at frequencies that are multiples of the 110 Hz fundamental tone. These are the harmonics. The first harmonic is therefore 2 × 110 = 220 Hz. The second harmonic is 3 × 110 = 330 Hz, and so on.
Figure 1.13 shows a fundamental tone and the first two harmonics on an oscillogram. The amplitudes of the harmonics are lower than those of the fundamental tone. The fundamental tone and the harmonics added together form the compound tone made by the guitar string.
Resonance A string does not move enough air to produce a good loud tone. To amplify the sound of the string, a guitar has a sound box. The air starts to vibrate in the sound box in unison with the string. This is called resonance. The air in the sound box vibrates at the same frequency as the plucked string. This amplifies the sound.
Timbre (sound colour) Not all harmonics are amplified equally by a sound box. This amplification is determined by the size and the shape of the sound box and the material it is made from. This gives each instrument its own characteristic sound. As a result, the A string of a violin sounds different from the A string of a guitar. The characteristic sound of an instrument is called the timbre
1.13 The fundamental tone and the first two upper tones of the A string.
Wind instruments The fundamental tone from a wind instrument is created by the air resonating inside it. The length of the vibrating air column determines the frequency of the fundamental tone. Different fundamental tones are created by changing the effective length of the air column. With a recorder, you do this by placing your fingers on the holes, and with a clarinet you close or open keys (Figure 1.14). The longer the air column, the lower the tone. A recorder therefore has the lowest fundamental tone when all its holes are covered.
Octave Music is based on twelve musical tones. Together, these form a scale. These musical tones all have a fixed frequency (Table 1.9). If you double the frequency of a musical tone, you get the same musical tone. The resulting musical tone is one octave higher. If you double the frequency of A above middle C (440 Hz), you get an A with a frequency of 880 Hz. A tone of 110 Hz is also an A, but this is two octaves lower than the A of 440 Hz.
1.14 Wind instruments: recorder and clarinet
Exercises
26 Fill in the missing words. Choose from: highest, higher, lowest, lower. R
a The fundamental tone is the tone of a string.
b The frequency of a harmonic is than the frequency of the fundamental tone.
c When two strings differ only in length, the shortest string gives the fundamental tone.
d The lower the frequency, the the fundamental tone.
27 Figure A
This is the oscillogram of a vibrating string.
a How can you tell that this is a compound tone? T1
b Calculate the frequency of the fundamental tone of this string. T2
c Calculate the frequency of the first harmonic of this string. T2
d Which musical tone is this? Explain your answer. See Table 1.9. T1
28 Figure B
Figure B shows a guitar. On each string, you can see the musical tone for the fundamental tone of that string. The corresponding frequencies are shown under the picture.
a Explain which string has the fundamental tone with the shortest period. T1
b What is the frequency of the first harmonic of the G string? T1
c The fundamental tone of both the top string and the bottom string is E. How many octaves are there between the upper E and the lower E? T2
d The frequency of the fundamental tone of the top E-string is equal to one of the harmonics of the bottom E-string. Which harmonic is that? T2
29 A guitar has metal bars on the fingerboard called frets.
a Explain how the frequency of the tone changes when you press a plucked string against a fret with your fingers. T1
b You tune a guitar string by tightening or slackening the string with the tuning peg. Explain whether you should tighten or slacken the string if the tone of the A string is a little too high. T1
c Guitar strings have different thicknesses. Explain which string is thickest.
See Figure B. T2
30 Figure C
Figure C shows two identical tuning forks placed on sound boxes. The sound boxes are placed with the openings facing each other.
a Explain the function of the sound box. T1
b If you strike the left tuning fork, the right tuning fork will vibrate in unison with it. What do you call this effect? R
c Explain whether you still hear the sound of the right tuning fork if you stop the left tuning fork from vibrating by holding your finger against it. I
d Why does the right tuning fork have to be identical to the left tuning fork in order to start vibrating in unison with it? I
31 Figure D
The pipes of an organ are all different lengths.
a Which organ pipe generates the highest tone: the longest or the shortest? T1
b Figure D shows an organ pipe. The relationship between the length of the organ pipe and the frequency of the fundamental tone it produces is given by:
f = v 4L
where L is the length of the vibrating air column in the organ pipe and v is the speed of sound in air.
Calculate the frequency of the fundamental tone made by this organ pipe at 20 °C.
c Which musical tone best matches this frequency? T1
d How can you tell from the equation that a longer organ pipe produces a lower tone? T2
e If the temperature drops, explain what happens to the pitch. Assume that the organ pipe stays the same length. I
HOME EXPERIMENT
Gently blow over the top of an empty bottle and over a bottle half filled with water. When is the tone lowest? Why is that?
Find out more
32 Read the text on the opposite page about breaking a glass. The resonant frequency of a crystal wine glass is between 500 and 1,000 Hz. The resonant frequency is determined by the vibrating air column in the glass.
This frequency is calculated with the following equation:
f = v 4L
where L s the length of the air column in the wine glass and v is the speed of sound.
a Calculate the resonant frequency of the glass at 20 °C. Estimate the length of the air column. T2
b A frequency of exactly twice the resonant frequency can also break the glass. Explain why this happens. I
c Explain whether the resonant frequency goes up or down when wine is poured into the glass. T2
33 The strings of a four-string bass guitar have the following musical tones as their fundamental tones: E, A, D and G.
The corresponding frequencies are 41.2 Hz, 55.0 Hz, 73.4 Hz and 98.0 Hz.
a Explain why a bass guitar has thicker strings than a normal guitar. T2
b Calculate the frequency of the second harmonic of the A string. T1
c The frequencies of the fundamental tones differ from those of a normal guitar. Show that these are nevertheless the same musical tones (see Assignment 28). I
34 Figure E
A clarinet is a wooden wind instrument. You can make it produce different tones by closing one or more of the holes. Figure E shows the oscillogram of the lowest tone of the clarinet.
a Explain why you need to close all the holes to produce this tone. T1
b The tone is a combination of the fundamental tone and the harmonics. Determine the frequency of the fundamental tone as accurately as possible from the graph. T2
Have you achieved the learning goal?
R I know the meaning of the following terms:
Fundamental tone
Harmonic
Resonance
Timbre
Octave
T1 I can describe what determines the frequency of a fundamental tone in string and wind instruments.
T2 I can explain what resonance is, and the role that resonance plays in creating the timbre of an instrument.
I I can explain when resonance occurs.
Can you break a glass by singing? There are singers who claim they can break a glass with their voice. Like a tuning fork, a glass has a natural frequency at which it vibrates. This is also called the resonant frequency. If you sing a tone precisely at this frequency, increasing resonance will cause the glass to vibrate more and more strongly. If you maintain the tone for long enough, the glass can shatter. So it really is possible. You do have to sing exactly the right tone, however, which takes a lot of practice.
1.15 A glass can shatter due to resonance.
1.4 Sound intensity
G oa L You learn what sound intensity is and how to measure it.
Sound intensity The amplitude of a vibration determines how loud the sound is. The greater the amplitude, the louder the sound. When two people talk over each other, they make twice as much sound as when one person talks. The conversation, however, does not sound twice as loud. How loud you hear the sound is called the sound intensity or sound level. To express the sound intensity, a special scale is used that suits the sensitivity of our hearing: the decibel scale. If the sound is twice as loud, the sound intensity increases by 3 decibels (dB).
Sound intensity is measured with a decibel meter (Figure 1.16). Figure 1.17 shows how many decibels common sound sources produce.
Example [3] The sound intensity of one scooter is 80 dB.
Calculate the sound intensity of four scooters.
If: One scooter has a sound intensity of 80 dB.
Calculate: The sound intensity of four scooters in dB.
Calculation: Each doubling of the number of scooters increases the sound level by 3 dB.
This involves doubling twice.
The sound intensity is therefore 80 + 3 + 3 = 86 dB.
Answer: The sound intensity of four scooters is 86 dB.
Example [4] You measure sound with an intensity of 55 dB. How many times as loud is sound with an intensity of 70 dB?
If: The sound intensity is 55 dB.
Calculate: How many times as loud is sound with an intensity of 70 dB?
Calculation: An 3-dB increase means that the sound is twice as loud. A 15-dB increase represents doubling 5 times. The sound is therefore 25 = 32 times louder.
Answer: The sound is 32 times louder.
1.16 Sound intensity is measured with a decibel meter.
Inverse square law The further away you are from the sound source, the quieter the sound is. If you stand twice as far away, the sound is 22 = 4 times as quiet. That represents halving the sound twice. Each halving is a decrease of 3 dB. The sound intensity therefore decreases by 2 × 3 = 6 dB. If you stand three times as far away, the sound is 32 = 9 times as soft. The decrease is therefore the square of the distance. This is why the relation to the distance of the sound source is known as the inverse square law.
Example [5] You measure a sound intensity of 90 dB at a distance of 1 m from a hairdryer. Calculate what the decibel meter would show at a distance of 4 m from the hairdryer.
If: The sound intensity at 1 m is 90 dB.
Calculate: The sound intensity at 4 m from the hairdryer.
Calculation: If you stand twice as far away, the sound is 42 = 16 times as quiet. This represents halving the sound four times. Each halving results in a reduction of 3 dB.
So the sound intensity at 4 m is 90 – 3 – 3 – 3 – 3 = 78 dB.
Answer: The decibel meter shows 78 dB.
1.17 The sound intensity produced by various sound sources
Exercises
35 True or False? R
a The unit for sound intensity is the decibel.
b As the distance from the sound source increases, the sound intensity also increases.
c If the number of sound sources doubles, they produce twice as many decibels.
d If you stand twice as far from a sound source, the sound intensity decreases by 3 dB.
36 Figure A Figure A shows a fridge label.
a What is the sound intensity of this fridge? T1
b What is the sound intensity produced by two of these fridges next to each other? T1
37 Use Figure 1.17 for the following questions.
a What is the sound intensity of a hairdryer at a distance of 1 m? R
b What is the sound intensity of four hairdryers at a distance of 1 m? T1
c How many hairdryers together would produce a sound intensity of 102 dB at a distance of 1 m? T2
d Why is the distance always included in the above questions? I
38 The maximum permitted sound intensity for fireworks in the Netherlands is 116 dB at a distance of 2 metres.
a Calculate this maximum sound intensity at 4 metres. T1
b Calculate this maximum sound intensity at 8 metres. T2
c Calculate the minimum distance you would need to stand at to hear a sound intensity of less than 98 dB. I
39 Figure B The figure below shows a decibel meter.
a Read how many decibels the meter shows. T1
b The meter has two settings. In setting A, the actual sound intensity is displayed, in setting B it shows the maximum sound intensity since measurements started. Explain which setting you would use if you want to know when the sound intensity in the class exceeds 60 dB. T2
c Explain which setting you would use to measure the sound intensity of a banger at a distance of 10 metres. T2
40 Figure C
To measure the noise along a railway line, a sound sensor (microphone) is placed 50 m from the track. The graph shows how the voltage generated by the sensor varies with the sound intensity.
a If the train produces a sound intensity of 92.5 dB, what voltage does the sensor generate? T1
b The sensor measures a sound intensity of 80 dB. How much voltage does the sensor generate when the sound intensity doubles? T2
41 A game is played in a stadium in front of 1,000 supporters. When a goal is scored, all 1,000 of them cheer. The sound intensity measured in the middle of the stadium is 77 dB. A week later, a sound intensity of 89 dB is measured in the middle of the stadium during a game. Calculate how many supporters were cheering. T2
42 To reduce the nuisance from noise, you can stand further away from the sound source. Doubling the distance gives a reduction of 6 dB. This rule does not apply next to a busy road, as the sound comes from many different directions. If you stand twice as far from the road, the sound intensity ‘only’ decreases by 3 dB.
a You are 10 m away from the road and measure a sound intensity of 80 dB. Calculate how far you need to stand from the road to measure less than 65 dB. T2
b 80,000 cars a month use this road. A year later, there are twice as many cars per month on this road. Calculate how far you need to be from the road to still hear a sound intensity of 80 dB. T2
c Because of the increasing number of cars, road noise barriers are installed alongside the road. This makes the sound 64 times softer. Calculate the sound intensity that you then measure 10 m from the road. T2
HOME EXPERIMENT
Stand 1 metre away from a loudspeaker or radio. Measure the sound intensity using the decibel meter on your smartphone. Now stand 2 metres away from the sound source. What sound intensity do you measure now?
Find out more
43 Figure D to H
Read the text on the opposite page about noise cancellation. The following (u,t) graph shows a sound vibration.
44 Noise-cancelling headphones are fitted with a microphone that records the ambient sound. The ambient sound is ‘inverted’ and transmitted with the same amplitude.
a Explain what happens if the amplitude of the anti-noise signal is not the same as that of the ambient noise. I
b Explain why it is difficult to use an anti-noise signal to reduce traffic noise in a residential area. I
Have you achieved the learning goal?
a Explain which of the (u, t) graphs E, F, G and H represents the correct anti-noise signal. I
b Calculate the frequency of the sound coming from the sound source. T1
c Explain whether an anti-noise signal has the same speed as ‘ordinary’ sound. I
d What frequency must the anti-noise signal have? I
R I know the meaning of the following terms:
Sound intensity
Decibel
Inverse square law
T1 I can calculate using decibels.
T2 I can use the inverse square law to perform calculations with sound intensity.
I I can explain what an anti-noise signal is.
1.18
Noise-cancelling headphones produce an anti-noise signal.
Noise cancellation Sometimes you want to hear as little as possible of the sound in your surroundings. You could stand further from the sound sources, so that the amplitude of the sound vibrations falls and the sound intensity decreases. You can also use headphones with noise cancelling technology. Noise-cancelling headphones do not block ambient noise, but make sound themselves. They produce a sound wave with the same amplitude as the ambient sound. But this soundwave vibrates the other way around to the vibration of the ambient sound. This inverted sound is called an anti-noise signal. Adding the vibrations of the sound and the anti-noise signal results in a vibration with a small amplitude (Figure 1.18). This reduces your perception of ambient noise.
Softened sound
Microphone
Ambient sound
Anti-noise signal
1.5 Hearing
GoaL You learn how you perceive sounds.
Hearing A sound wave reaching your ear is directed through the ear canal to your eardrum (Figure 1.19). Your eardrum vibrates and transmits this vibration to the inner ear via three tiny bones called auditory ossicles. The inner ear contains your hearing organ, the cochlea. The fluid inside the cochlea is moved back and forth by the ossicles. This causes small sensory hairs, called stereocilia, to move in the liquid. These hairs convert the movement into electrical signals that are sent to the brain via the auditory nerve.
Hearing limits Human hearing is sensitive to sound with frequencies between about 20 Hz and 20,000 Hz. As you get older, you hear the high frequencies less well. Your hearing is most sensitive to frequencies between 500 and 5,000 Hz. Voice sounds also fall within this frequency range.
Hearing threshold and pain threshold A tone of 1,000 Hz that you can just hear is equal to 0 dB. This is the hearing threshold at 1,000 Hz. Because your hearing is not equally sensitive to all tones, the hearing threshold depends on the frequency of the tone. This is shown by the lower black line in Figure 1.20. For example, a tone of 100 Hz can only be heard at a sound intensity of 20 dB. A tone of 1,000 Hz with an intensity of more than 120 dB is painful. This is called the pain threshold for 1,000 Hz. The pain threshold also depends on the frequency. You can see this from the top black line in Figure 1.20.
Hearing damage and risk threshold Most hearing damage is caused by damage to the sensitive sensory hairs in the inner ear. If these hairs are damaged, they can no longer send a signal to the brain. As the damage to these hairs is irreparable, hearing damage is permanent.
The sensory hairs wear out as you get older. As a result, almost all elderly people suffer impaired hearing loss. From the age of fifty, you notice that your hearing declines, especially with high frequencies. Direct damage to your hearing can occur from sounds near to the pain threshold. However, sounds above 80 dB already present a risk to your hearing. 80 dB is the risk threshold. The extent of hearing damage depends on how often and for how long you have listened to loud sounds. This is shown by the histogram in Figure 1.21.
Noise protection There are several ways to reduce noise. The most effective approach is to deal with the sound source itself. So do not set the volume of music too high. Listening to music with headphones or earphones is especially harmful. Noise pollution from traffic is counteracted by installing road noise barriers alongside highways in residential areas. Houses in urban areas are often additionally insulated with materials that absorb sound. Finally, you can protect your ears with hearing protectors or ear plugs. This is especially useful for workers in factories. These hearing protectors can attenuate noise by up to 15 dB.
Exercises
45 a In which part of the ear is the cochlea? R
b Which frequencies can healthy hearing detect? R
c What causes the sensitivity of your hearing to decrease after the age of fifty? R
d What is the risk threshold? R
e Write down three ways to reduce noise pollution. R
46 Use Figure 1.20 for the following questions.
a What is the frequency range of human speech? T1
b Above how many dB can a healthy ear detect sound at a frequency of 10,000 Hz? T1
c At what frequency is the pain threshold the lowest? T1
d Explain whether you can hear high or low tones better. T1
47 You cannot hear sound with a frequency of 100 Hz and an intensity of 5 dB. How much louder does the sound have to be for you to be able to hear it? Use Figure 1.20. T2
HOME EXPERIMENT
Listen to quiet music on your music system. Adjust the volume so that you can just about hear the music. Measure the sound intensity using the decibel meter on your smartphone. Now put in earplugs. If you don’t have any, make some by wetting a piece of toilet paper and putting it in your ear. Turn up the volume up until you can just hear the music again. Measure the sound intensity again. By how many dB do the earplugs attenuate the sound?
48 The ear canal is about 28-mm long in adults. It acts as an amplifier for sound with a certain frequency.
a What is this effect called? T1
b Use a calculation to show that sound with a frequency of 3 kHz is amplified in the ear canal. Use the equation from Assignment 31. T2
c Use Figure 1.20 to explain why the ear hears this frequency better. T2
d Is the frequency amplified more or less in babies than in adults? Explain. I
49 You attend an open-air pop concert and are 10 m from the stage. The sound intensity at that position is 92 dB. The harm caused by the sound is determined by the sound intensity and the time you are exposed to it. Use Figure 1.21.
a How long can you stay at the concert without incurring hearing damage? T1
b You did not bring any earplugs. The concert lasts 2 hours. What is the minimum distance you need to be from the stage to avoid hearing damage? T2
c Explain how long you can stand 10 m from the stage if you are using earplugs. I
50 Figure A
A company wants to use glass as sound insulation for a meeting room. The graph shows the attenuation produced by a certain type of glass at a range of frequencies.
a Use Figure 1.17. What is the sound intensity of a typical conversation? R
b What is the sound intensity of a typical conversation behind 8 mm of glass? T2
c A drill makes sound at a frequency of 4,000 Hz. Explain which glass thickness blocks this sound most effectively. I
d Explain which glass thickness would be best for a glass meeting room in the office. I
51 Figure B
Our ears can also determine which direction a sound is coming from. This is because the sound from one source does not reach both ears at exactly the same time. There is also a difference in sound intensity. To test the direction, Hamida and Livia attach two funnels to a 1-m garden hose. Hamida puts the funnels against her ears. Livia taps the hose at increasing distances from the centre and asks Hamida if she can tell which side the sound is coming from.
a Our hearing can detect time differences of as little as 30 μs. Calculate how far from the centre Livia needs to tap to make sure Hamida can just about detect whether the sound comes from left or right. T2
b Sound intensity also plays an important part in determining the direction a sound comes from. Explain why the frequency of the sound also affects the ability to determine the direction correctly. I
Find out more
52 Figure C
Read the text on the right about audiograms. Figure C shows two audiograms showing hearing loss for men (left) and women (right) of different ages.
a Are men or women most affected by hearing impairment later in life? Explain your answer. T1
b Which frequencies are less audible for older people? Explain your answer. T1
c What is the hearing loss of a 60-year-old woman at 2,000 Hz? T2
d What is the hearing threshold of a 70-yearold man at 4,000 Hz. I
53 Figure D
An audiogram is made for a 20-year-old man who often goes to music parties. Two years later, he has another audiogram made. Figure D shows both audiograms. Doctors call this hearing loss the ‘disco dip’.
a Which frequency has been most damaged by the ‘disco dip’. T2
b Explain how you can see that this is not simply the ‘age-related dip’ from Assignment 52. I
c By how many decibels has this visitor’s disco dip increased in two years? T2
54 The extent of hearing loss is determined by the Fletcher index. This is calculated by averaging the hearing loss at the frequencies 1,000 Hz, 2,000 Hz and 4,000 Hz. This leads to the following categories:
Normal hearing: –10 to 15 dB
Mild hearing loss: 16 to 40 dB
Slight hearing loss: 41-55 dB
Severe hearing loss: 56-70 dB
Very severe hearing loss: 71-90 dB
Deafness: 90 dB
a What degree of hearing loss does the music party visitor in Question 53 suffer after two years? T2
b Determine the degree of hearing loss for an average 70-year-old male. T2
Have you achieved the learning goal?
R I know the meaning of the following terms:
Hearing threshold
Pain threshold
Hearing damage
Risk threshold
T1 I can determine whether a sound is perceptible to human hearing.
T2 I can determine when noise presents a risk to health.
I I can explain how you can determine hearing damage by using an audiogram.
1.22 Making an audiogram
Audiograms An audiogram is made during a hearing test. You hear beeps at different frequencies. At first. these are so quiet that you cannot hear them. The volume is increased until you hear the beep. You react by raising your hand. This determines your hearing threshold at different frequencies.
Test preparation
Check whether you have achieved the learning goals for each section of this chapter. If not, please read the explanation again or watch the explanatory videos. Then complete the following assignments.
1 A girl is listening to the radio in her bedroom. The music can be heard downstairs in the living room.
a What is the sound source? T1
b What is the receiver? T1
c What is the medium through which the sound propagates? T2
2 Figure A
You are on holiday in the mountains. There is a flash of lightning. You hear thunder 3 seconds after seeing the flash. One second later, you hear the echo of the thunderclap. Figure A is a drawing of the situation
a Calculate your distance from the storm. T1
b Calculate your distance from the mountainside. T2
c Explain why you cannot calculate your exact distance from the mountainside. I
d The light from the flash also takes time to reach you. The speed of light is 300,000 km/s. Explain why you can neglect the time it takes to get to you. I
3 Figure B
An oscillogram of a tone is shown below.
a Is this a ‘pure’ tone? Explain your answer. T2
b Calculate the frequency of this tone. T1
c Is this tone a musical tone? Explain your answer. T2
4 Figure C
You change the pitch of the tone of a trombone by changing the length of the tube. You do this by moving the slide.
a Explain whether the tone rises or falls as you lengthen the tube by moving the slide away from you. T2
b Explain how a trombone can be used to create tones that are not musical tones. T2
c The trombone produces a sound intensity of 110 dB at a distance of 1 m. Calculate the sound intensity of two trombones at a distance of 4 m.
5 Figure D
Figure D shows an oscillogram of three tones: a fundamental tone, a tone one octave above the fundamental tone, and a compound tone.
a Which tone is the fundamental tone? Explain your answer. T2
b Which tone is an octave higher than the fundamental tone? Explain your answer. T2
c Which tone is the compound tone? Explain your answer. T2
6 The viola d’amore is a special instrument. It is a type of violin, with seven ‘sympathetic strings’ below the seven played by the musician. The sympathetic strings are not touched, but vibrate in unison with the strings played by the musician.
a What do you call this effect? T2
b Explain the range of frequencies a sympathetic string can have when a 440-Hz tone is played. I
c When a tone of 440 Hz is played, a sympathetic string with a fundamental tone of 110 Hz also vibrates slightly. Explain why that happens. I
7 You measure sound intensity in a meadow. You make a tone with a frequency of 1,000 Hz. At a distance of 1 m, you measure a sound intensity of 64 dB.
a Calculate how many decibels you would measure at a distance of 2 m.
b You make the tone four times louder. Calculate how many decibels you would now measure at a distance of 2 m.
c You now make a tone at 50 Hz. You measure a sound intensity of 64 dB at a distance of 1 m. At what distance would you then need to stand in order to just about be able to hear the tone? Use Figure 1.20. I
d Explain why you can hear a higher frequency tone at a greater distance. I
8 Figure E
This is the audiogram of an 18-year-old boy. The red line is this boy’s audible threshold. The orange area represents the frequency range of vowels and consonants in normal speech.
a Explain whether this boy can hear a 1,000-Hz 25-dB tone. T2
b Explain the minimum sound intensity of a 4,000-Hz tone for the boy to be able to hear it. Also use Figure 1.20. I
c Does this boy hear vowels or consonants better? Explain your answer. T2
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Polaris is an RTTI-certified teaching method and distinguishes four types of questions:
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Translation
Paul Smith
Richard Purdom
Marjan Borger (Easy Translation)
Book design & cover
René van der Vooren, Amsterdam
Layout & technical drawings PPMP, Wolvega