Unit 4 Physics 2007Chapter 10Sound Basics 1
SOUND BASICS NOTES 2007
SOUND
Sound is a form of energy that travels through any medium by causing parts of the medium to vibrate. Sound waves require a medium, they do not travel through a vacuum. All forms of wave motion allow the transfer of energy without the net transfer of matter. The energy travels by distorting the medium and so a measure of the wave's energy is the amount of distortion.
The sensation of sound is registered by our ears, vibrations of the air molecules adjacent to the eardrum cause tiny bones to move which is converted into an electrical signal which is interpreted as sound by the brain.
Vibrations also allow sounds to be detected, in microphones the sound vibration causes a diaphragm or crystal to vibrate slightly and produces an electrical signal that is sent to an amplifier.
If we place a candle in front of a speaker, the sound from the speaker causes the flame to move backwards and forwards.
Sound is a wave, and waves are forms of energy that can travel through a medium. Sound is a longitudinal or compressive wave, this means that the medium vibrates along the same direction in which the wave is travelling. Below is an example of a loudspeaker “L” and the air particles in front of it.
Example 1997 VBOS Exam
A loudspeaker consists of a cardboard cone that acts as a diaphragm which is moved backwards and forwards to produce a travelling sound wave in air. Such a speaker is mounted on a wall as shown below in Figure 1.
Figure 1
Figure 2 below shows the horizontal displacement of the speaker cone as a function of time.
Figure 2
In the air are fine dust particles which are floating at rest. This is also illustrated in Figure 1. After the loudspeaker is turned on, the particles will be forced to move by the pressure variations associated with the sound wave.
Example
1997 Question 2
For the dust particle at point P, directly in front of the loudspeaker, which of the statements (A–D) below best describes its motion?
A. It vibrates vertically up and down at the frequency of the sound wave.
B. It vibrates horizontally backwards and forwards at the frequency of the sound wave.
C. It moves horizontally forwards, travelling with the sound wave.
D. It remains at rest.
Question 3
Explain the reason for your choice in Question 2.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Transmission, Reflection and Absorption of Sound
Sound waves striking a solid object will be transmitted, reflected and absorbed to varying degrees. Thick materials transmit less sound than thin materials, this is why thick concrete sound barriers often line freeways passing through residential areas.
Porous materials such as carpet and curtains absorb sound. These materials reduce the amount of reflected sound and are often used in Concert Halls to limit reverberation of sound and so improve the sound quality. Classrooms that have carpet on the floor are a lot quieter than rooms with wooden floorboards.
Smooth polished surfaces such as glass and tiles reflect a great deal of sound. People who sing in the shower often have an enhanced opinion of the strength of their voice due to the reflections off the glass and tiles of the shower recess.
Sound like all other waveforms, obeys the laws of reflection, the angle of incidence is equal to the angle of reflection. Angles are always measured between the direction of travel of the wave and the normal to the media boundary.
WAVE TYPES
Longitudinal waves
When the vibration of the waves is in the same direction as the line of travel, then the wave is longitudinal. In a longitudinal wave the motion is in the same direction as the motion of the wave, but the particles do not move forward, they vibrate around an equilibrium position.
The distance between any two identical points is called the wavelength.
The movement of particle is in this direction.
compression rarefaction
It is possible to represent this particle movement as a pressure variation
+max. positive pressure variationnormal pressure
-max. negative pressure variation
Dots close together represent high pressure regions, where there is less shading there is a lower pressure region. At the two positions of maximum pressure variation (compressions and rarefactions) the molecules at these points are in their rest position. The minimum pressure variation occurs when particles are the furthest from their rest positions. The air pressure is normal, (pressure variation is zero) midway between the compressions and rarefactions. When sound travels through air it is a series of compressions and rarefactions
Frequency
The frequency of the wave determines the pitch of the sound. Frequency is a measure of how rapidly the source of the wave is vibrating. The frequency (f) is defined as the number of complete waves that pass a point in one second. The units for frequency are Hertz, Hz, which are cycles per second.
Wave equation
The wave equation links the velocity of the wave to the frequency and the wavelength.
v = f
where v is the velocity in m/s, f is the frequency in Hz and is the wavelength in metres.
Example
2000 VCAA Exam
Kylie is rehearsing for a concert. She sings a pure note of a single frequency. The time between successive compressions of the sound wave arriving at her ears is 5 ms.
Question 4
What is the frequency of the sound?
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Transverse waves
When waves vibrate up and down in a direction perpendicular to the direction of motion of the wave, it is referred to as a transverse wave. e.g. water waves, where the motion of the water particles is at right angles (up and down) to the direction of the wave (forward).
A transverse wave is modelled below
e
amplitudewave velocity
the movement of
the particles is
in this direction
The amplitude of a wave is the distance from the rest position to the limit of a crest or trough; the total from crest to trough is twice the amplitude.
Example
1998 VBOS Exam
A student produces two different types of travelling waves by shaking the end of a long slinky spring. The figure below shows snapshots of parts of the waves at an instant of time.
Question 1
Estimate the wavelength of travelling wave A.
Question 2
Which of the waves (A or B) is a better model for the propagation of sound waves? Your answer must mention the similarities between the model you choose and sound waves.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Loudness
The louder a sound is the greater the magnitude of the variation from normal pressure, this is represented in the following way. The loudness of a sound is related to the amplitude of the sound wave. A smaller amplitude wave sounds quieter than a larger amplitude sound wave.
Another way of representing this is:
Low frequency sound waveHigh frequency sound wave
Example
1998 VBOS Exam
A loudspeaker is emitting sound of a fixed intensity which travels equally in all directions. The figure below shows the pressure variation plotted against distance from the speaker, at a particular instant of time.
Question 3
Explain why the amplitude of the pressure variation decreases as the distance from the speaker increases.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Question 4
Circle the letter (A–D) of the graph below which shows the pressure variation as a function of distance from the speaker, at a time that is ¼ cycle later than shown below.
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
Sound Intensity and Sound Level
A measure of the amount of sound energy passing through an area of one square metre each second is SOUND INTENSITY. Sound Intensity (I) is measured in Watts per square metre (W/m2). The ear can detect intensities over a range from 1 to 1012. In order to work with such a large range a logarithmic scale is used.
Even though the range of pressure changes is huge, what we sense as loudness is not proportional to intensity. The doubling of the loudness of a quiet whisper would be very noticeable, but adding the same sound to a louder source would not be noticed at all. This is because the human ear compares the loudness of two sounds in terms of the ratio of the two amplitudes rather than the difference in amplitudes.
A logarithmic scale is used to indicate sound intensity levels. The unit is the bel, or, more commonly the decibel (dB). 1dB = 0.1 bel.
The sound loudness scale measures the ratio of the intensity of a sound wave against a standard intensity, called the "threshold of hearing" and = 10-12 Wm-2, I0.
The sound intensity level, (in dB) = 10 logwhere I0 = threshold of hearing = 10-12 Wm-2.
A tenfold increase in intensity will correspond to a level increase of 10 dB, which is heard as a doubling of the loudness. A soundwave of intensity 10-8 Wm-2 sounds twice as loud as a wave of intensity 10-9 Wm-2.
A one hundred fold increase will correspond to a level increase of 20 dB, which is heard as an increase of ×4 in the loudness. This is a doubling, followed by another doubling.
The human ear can only detect changes as small as 3 dB. A sound level increase of 3 dB is equal to the doubling in the sound's intensity. A 3 dB decrease will halve the sounds intensity. This means that to make the least noticeable increase in the volume of a stereo, we need to double the power supplied.
The loudness of a sound depends on how much energy is carried by the wave. The intensity of a sound wave is used to give a measurement of the rate at which the sound wave is carrying energy.
Power is energy per second, and intensity is power per unit area,
the units for sound Intensity are Wm-2.
As sound moves away from the source it spreads out over an increasingly larger area. As the energy is fixed, then the intensity must decrease.
The intensity variesI
This means that if you double the distance the intensity will decrease by a factor of 4. This is known as an "Inverse Square Law".
source Intensity Intensity
I
2000 VCAA Exam
The intensity of sound at a microphone situated 0.10 m from a singer’s mouth is 1.0 x 10-6 W m-2. Her manager is sitting 2.0 m away from the mouth of the singer, in the first row of the theatre. At this time the sound system is not turned on.
Question5
What is the sound intensity at the manager's location? Show your working.
The sound system is now turned on, and the sound intensity at the position of the manager increases by a factor of 1000.
Question 6
By how many decibel does the sound level increase for the manager, when the sound system is turned on?
Reproduced by permission of the Victorian Curriculum and Assessment Authority, Victoria, Australia.
For background information only, this is not directly on the course any more.
Sound intensities and sound intensity levels of common sounds
Sound / Intensity(W m-2) / Intensity level
(dB)
Threshold of hearing / 10-12 / 0
Rustle of leaves / 10-11 / 10
Whisper / 10-10 / 20
Quiet radio / 10-8 / 40
Normal conversation / 10-6 / 50 - 65
Busy street traffic / 10-5 / 70
Loud radio / 10-4 / 80
Noisy factory / 10-4 / 85
Motor car horn / 10-3 – 10-2 / 90 - 100
Jackhammer / 10-2 / 100
Thunder / 10-1 / 110
Rock concert / 1 / 120
Threshold of pain / 1 / 120
Jet aircraft / 100 / 140
Examples Solutions
1997 Question 2 solution
B
Don't look at the answers provided in the question, work your own out first and then see which one agrees with you.
1997 Question 3 solution
The dust particles are going to vibrate about their mean position. They are going to vibrate in a horizontal plane, because sound is a longitudinal wave. The actual molecules are not going to move in the direction of the wave, they will only vibrate about their mean position.
So B is the best answer.
2000Question 4 solution
The higher the frequency, the lower the period. This is summarised by: f = .
f = = 200 Hz
1998 Question 1 solution
The definition of wavelength is the distance that it takes for the waveform to repeat itself.
I.e. it is the distance between identical points on the wave train.
It is probably easier to deal with the crests in the top diagram, less confusion as to where the point actually is. So the distance between adjacent crests is 3 divisions on the horizontal scale.
= 3 × 20 cm = 60 cm ( 4 cm, because you are asked to estimate)
You don't need to do this, but for the second wave, which is a longitudinal representation, it is best to try and find the distance between adjacent compressions.
It looks like there were 4 compressions in a distance of about 7.5 divisions
This corresponds to 3 wavelengths (count them to see how this works)
So = (7.5 × 20)/3= 150/3= 50 cm.
1998 Question 2 solution
B
This is an explain question, they are very important on this exam. As a rule they are worth more marks than the other questions, so they require more time and effort.
Model B is a better representation of a sound wave, because the slinky is demonstrating a longitudinal wave, the same as sound. The slinky is an excellent model to demonstrate sound and it clearly shows that the particles only vibrate around their mean position and that there isn't a net movement of material (be it air molecules or whatever). If sound did create a net movement of material, then you would expect all the slinky to end up and one end. This does not happen.
The vibration of the particles is parallel to the direction of travel of the propagation of the wave energy, this is the same as sound.
This model demonstrates the compressions as a bunching up of the slinky and the rarefactions as a stretching of the slinky.
The examiners were looking for the following response
- Sound is an example of a longitudinal wave, with compressions and rarefactions.
- Wave B represents a model of a longitudinal wave.
- Longitudinal waves are characterised by particle vibrations in the same direction as wave velocity.
- Wave A is an example of a transverse wave.
1998 Question 3 solution
The amplitude of the pressure variation is a measure of the loudness if the signal. If the pressure variations are extreme, we hear this as a very loud sound. If the pressure variations are minimal, then we hear this as a soft sound. The wavelength must not change, because we interpret wavelength as the pitch of the sound, the shorter the wavelength, (the closer the crests are together) then the higher the frequency, because the eardrum is made to vibrate more often.
So the decrease in amplitude is showing that the sound level decreases with distance from the source. We know this from common experience, the further you are from a sound source the softer that it sounds. But we also know that even as it gets softer its pitch does not change.
The reason this happens is that the sound wave is spreading out in a three dimensional form (i.e. as an expanding spherical wave). The energy is being spread over a greater surface area, so the intensity decreases. The area of an expanding sphere is proportional to r2.
You could also draw a diagram to show the sound waves moving out in a circular pattern.
Direction of travel
You could also state that the intensity drops off as so the pressure variation should decrease as the wave energy moves away from the source.
1998 Question 4 solution
B
One quarter of a cycle later, the initial crest (at zero) will move of a wavelength to the right. The wavelength of this wave is 0.6 m, so the first crest has moved = 0.15 m to the right.
2000 Question 5 solution
2.0 10-9 Wm-2.
The intensity variesI Since the distance has increased from 0.10 m to 2.0 m,
this is a factor of = 20, then the intensity must drop off by a factor of (20)2.
the new intensity will be = = 2.0 10-9 Wm-2.
2000 Question 6 solution
A 1000 fold increase in the sound intensity will give rise to an increase of 30 dB in the sound level.
This comes from 1000 = 103 3 10 (30 dB) increase in sound level.
A 100 fold increase = 102 2 10 (20 dB) increase in sound level.
The other way of doing this question is to actually calculate the sound levels that correspond to the sound intensities I0 = 2.0 10-9 Wm-2 = 10log = 10 log2000 = 33 dB
In = 1000 I0= 1000 2.0 10-9 Wm-2=10log = 10 log2 000 000 = 63 dB
Which is an increase of 30 dB, but not a very efficient way of getting to the answer.
This second method would be a great way to check your answer when you have finished the paper and are going through checking all your answers.