Sound waves
SOURCES OF SOUND, TRANSMISSION OF SOUND, SPEED OF SOUND AND CHARACTERISTICS OF SOUND WAVES
Subject: Physics
Theme: Waves-Motion Without Material Transfer
Topic: Sound waves
Sub Topic:
Date: dd/mm/yyyy
Class: S.S.S 2
Duration: 40 Minutes
No of Learners: 40
Learning Objectives: By the end of the lesson learners should be able to:
1. Defined Sound and Sound Waves
Noise and music: Sound can be classified into noise and musical notes. Noise is due to vibrations of irregular frequency such as the rattling of a wheel on a rough road. While music is produced by vibrations of regular or constant frequency.Sound: is a form of energy produced by vibrating bodies, it is a vibration that propagates through a medium in the form of a mechanical wave. The medium in which it propagates can either be a solid, a liquid or a gas. Sound travels fastest in solids, relatively slower in liquids and slowest in gases.
Sound Waves: is the pattern of disturbance caused by the energy travelling away from the source of the sound. Sound waves are longitudinal waves. This means that the propagation of vibration of particles is parallel to the energy wave propagation direction. When the atoms are set in vibration they move back and forth. This continuous back and forth motion results in a high-pressure and a low-pressure region in the medium. These high-pressure and low-pressure regions are termed compressions and rarefactions, respectively. These regions are transported to the surrounding medium resulting in the sound waves travelling from one medium to another.
2. Explain Transmission/Propagation of sound waves
Sound waves are created by a disturbance that then propagates through a medium (e.g., crust, water, air). Individual particles are not transmitted with the wave, but the propagation of the wave causes particles (e.g., individual air molecules) to oscillate about an equilibrium position.Sound is transmitted through gases, plasma, and liquids as longitudinal waves, also called compression waves. It requires a medium to propagate.
sound waves are longitudinal waves i.e the air vibrates backwards and forwards in wave moving in a patterns known as compression (high pressure) and rarefaction (low pressure).
Acoustic transmission
Acoustic transmission is the transmission of sounds through and between materials, including air, wall, and musical instruments. The degree to which sound is transferred between two materials depends on how well their acoustical impedances match.Sound waves are created by a disturbance that then propagates through a medium (e.g., crust, water, air). Individual particles are not transmitted with the wave, but the propagation of the wave causes particles (e.g., individual air molecules) to oscillate about an equilibrium position.
Every object has a unique natural frequency of vibration. Vibration can be induced by the direct forcible disturbance of an object or by the forcible disturbance of the medium in contact with an object (e.g. the surrounding air or water). Once excited, all such vibrators (i.e., vibratory bodies) become generators of sound waves. For example, when a rock falls, the surrounding air and impacted crust undergo sinusoidal oscillations and generate a sound wave.
Vibratory bodies can also absorb sound waves. Vibrating bodies can, however, efficiently vibrate only at certain frequencies called the natural frequencies of oscillation. In the case of a tuning fork, if a traveling sinusoidal sound wave has the same frequency as the sound wave naturally produced by the oscillations of the tuning fork, the traveling pressure wave can induce vibration of the tuning fork at that particular frequency.
3. List and explain Characteristics of Sound
The sound produced by a guitar is different from the sound produced by a drum. This is because the sound produced by different sources have different characteristics. Some quantities that are used to explain characteristics of sound waves are: amplitude, frequency, time period, wavelength and velocity.Characteristics of sound waves are as follows: pitch, loudness, and quality.
1. The Loudness (amplitude):
Loudness depends on the amplitude of vibration of the body, it is the magnitude of the sensation resulting from a sound reaching the ear.The amplitude of a wave defines the maximum displacement of the particles disturbed by the sound wave as it passes through a medium. A large amplitude indicates a large sound wave.
If the amplitude of a sound wave is large, then the loudness of the sound will be more. If the amplitude is small, then the sound will be feeble.
Humans can hear a limited range of frequencies of sound. Physicists have identified the audio frequency spectrum of the human ear to be between 20 Hz and 20,000 Hz. Under ideal laboratory conditions, the human ear can detect frequencies that are as low as 12 Hz and as high as 20,000 Hz.
2. The Pitch (frequency):
Pitch depends on frequency, it is the degree of highness or lowness of sound as perceived by the listener.The frequency of a sound wave indicates the number of sound waves produced each second. Low-frequency sounds produce sound waves less often than high-frequency sounds.
The formula of the frequency of a wave is given as:
f = 1/T
Where,
• f is the frequency of a sound wave and
• T is the time period.
It can be shown that pitch depends on frequency by using:
(a) Method of Disc Siren: A Disc siren consist of a circle metal plate with whole punched at regular intervals as shown diagram
Thus frequency (f) of its giving by
f = No. of revolutions × No. of holes in the ring ÷ Time of revolution in seconds
(b) Method of Toothed wheel: A touched wheel mounted on a stand so that it can rotate freely. A piece of card board held against the rotating wheel as shown. The cardboard vibrate as the wheel rotate and produce a note of frequency (f) given by
f = No. of revolutions × No. of teeth in the wheel × Time in Seconds
Example 1
In an experiment to determine the frequency of a note it was insured that the counter records a revolution of 500 for a time 2.5 minutes if the disc siren has 60 holes in ring. What is the frequency of the note emitted?
Solution
f = No. of revolution × No. of holes ÷ Time in Second
f = (500 × 60) ÷ (2.5 × 60)
f = 200 vibrations per sec or 200 Hz
Example 2
If in another experiment a note of the same pitch as in example 1 above is produced using a toothed wheel and it requires 400 revolutions of th wheel in 1.5 minutes to produce the note, calculate the number of teeth in the wheel.
Solution
Pitch is proportional to frequency. Since a note of the same pitch is produced, the frequency is 200Hz.
:. 200 = No. of revs × No. of teeth in wheel ÷ time in secs
200 = 400 × No. of teeth ÷ (1.5 × 60 secs)
:. No. of teeth = 200 × 1.5 × 60 ÷ 400
No. of teeth = 45
3. The Intensity (time):
Intensity is the rate of flow of sound energy per unit area perpendicular to the direction of propagation of the sound waves.PLEASE N/B: the word rate as used in mathematics and physics mean reciprocal of time i.e 1/time
The time period of a sound wave is the amount of time required to create a complete wave cycle. Each vibration from the sound source produces a wave’s worth of sound. Each complete wave cycle begins with a trough and ends at the start of the next trough.
4. The Quality or Timbre:
This is the property which enables us to distinguish the same note of the same pitch and loudness played on different instruments e.g a piano and an organ, the quality of a musical notes depends on the harmonies. When a note is produced, the strongest, audible frequency heard is the fundamental. All other frequencies present are harmonics or overtones.5. The wavelength:
The wavelength of a sound wave indicates the distance that wave travels before it repeats itself. The wavelength itself is a longitudinal wave that shows the compressions and rarefactions of the sound wave.The wavelength of the sound formula is given as follows:
𝜆 = 𝑣/f
Where, f is the frequency of the sound wave and
v is the velocity of the sound wave.
6. The velocity/Speed of sound in solid, liquid, and gas:
The velocity of a sound wave tells us how fast the wave is moving and is expressed as meters per second.Sound waves travel through a medium by alternately contracting and expanding the parts of the medium in which it travels. The rate at which the sound waves propagate through the medium is known as the speed of sound.
The speed of sound varies from substance to substance typically, sound travels most slowly in gases, faster in liquids, and fastest in solids. For example, while sound travels at 343 m/s in air, it travels at 1,481 m/s in water (almost 4.3 times as fast) and at 5,120 m/s in iron (almost 15 times as fast).
The velocity of sound in air is about (330 - 323) m/s in air at about 0˚c as against the velocity of light which is 3 x 108 m/s, this explains why we see the lighting flash before we hear the thunder strike even as both occur at the same time and from same source. At 0˚c, the velocity of sound in water is about 1500 m/s and about 5000 m/s in a steel rod. Velocity of sound also varies with the square root of the absolute temperature. Wind direction also affect the speed of sound in air.
The ormula used to calculate the speed of sound is given as:
v = d/t
But, d = λ and t = 1/f
:. V = fλ
Where,
d = distance travelled by sound
t = time taken to cover the distance.
V = velocity,
f = frequency, and
λ = wavelength
The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves. The wavelength of a sound is the distance between adjacent compressions or rarefactions. The frequency is the same as the source’s and is the number of waves that pass a point per unit time.
The table below lists the speed of sound in different media.
| S/n | Medium | Speed of sound |
| 1. | Water | 1481 m/s |
| 2. | Air | 343.2 m/s |
| 3. | Copper | 4600 m/s |
| 4. | Hydrogen | 1270 m/s |
| 5. | Glass | 4540 m/s |
Factors Affecting the Speed of Sound
Density and temperature of the medium in which the sound wave travels affect the speed of sound.
4. Calculate time, frequency, amplitude, speed/velocity on sound waves
Example 1.What is the wavelength of sound from a tuning fork which vibrates at a frequency of 250 hz in air? Take the velocity of sound in air to be 330 m/s
Solution
V = fλ
λ = v/f
λ = 330/250
λ = 1.32 m
Example 2.
A source of sound produces waves of wavelength 1.65m, if the speed of sound in air is 330 m/s. find the period of vibration in seconds?
Solution
V = fλ
f = v/λ
f = 330/1.65
but f = 1/T
:. T = 1.65/330
T = 0.005 seconds
Example 3.
How long does it take for a sound wave of frequency 2 kHz and a wavelength of 35 cm to travel a distance of 1.5 km?
Solution
v = λν
v = 0.35 m × 2000 Hz
v = 700 m/s
1.5 km = 1500 m
Time = Distance Travelled/ Velocity
Time = 1500 m/ 700 m/s
T = 2.1 s
5. Echoes of sound waves
Like light, sound waves can be reflected, refracted, diffracted and also show their interference phenomenon.Echoes occur as a result of reflection of sound waves. An echo is a sound heard after the reflection of sound waves from a plane surface.
Suppose the reflecting surface such as a wall that is producing the echo is d meters away, the time taken to hear the echo is given by
Time (t) = total distance travelled by sound /velocity of sound.
Since the sound waves travels to the obstruction or the reflecting surface and returns to the production source as echo, the total distance travelled is 2d, hence
t = 2d/v seconds,
v = speed of sound in air is 330 m/s
t = 2d/330 sec
d is measured in meters
t is second
v is meters/second.
Example 1.
A man standing 510 m away from a wall sounds a whistle, the echo from the wall reaches him 3seconds later. Calculate the velocity of sound in air.
Solution
Distance d = 510 m
echo t = 3 s
velocity v = ?
t = 2d/v
vt = 2d
v = 2d/t
v = 2 × 510/3
v = 1020/3
v = 340m/s
Example 2.
A sound note of frequency 250Hz and wavelength 1.3m is produced at a point near hill. If the end of the sound is received one second later at the pot, how far away is the hill from the point.
Solution
F = 250 hz
𝛌 = 1.3m
t = 1 sec
d = ?
t = 2d/v
But v = F𝛌
v = 250×1.3
v = 325m/s
:. d = v𝑡/2
d = 325×1/2
d = 325/2
d = 162.5m
Example 3.
A sound note is produced by a ringing bell and the echo of the note from a nearby wall is received 0.5sec later. If the frequency of the note is 400Hz and it wave length 1m calculate the distance between the bell and the wall.
Solution
t = 0.5sec
f = 400Hz
𝛌 = 1m
d = ?
Recall v = f𝛌
v = 400 x 1
V = 400m/s
d = 𝑣𝑡/2
d = 400×0.5/2
d = 200/2
d = 100 m
Example 4.
A man standing 510m away from a wall sounds a whistle, the echo from the wall reaches him 3seconds later. Calculate the velocity of sound in air.
Solution
Distance d = 510m
echo t = 3s
velocity v = ?
t = 2d/v
v = 2d/t
v = 2×510÷3 v = 1020÷3
v = 340m/s
6. Explain Multiple echoes and Reverberation
(a) Multiple echoes: Multiple echoes occur when the sound wave is propagated between two reflecting surface is made to reflect from one reflecting surface to the other a number of times.(b) Reverberation: Reverberation occurs as multiple echoes which are not heard as separate sound due to the closeness of the reflecting surface, but as a persistence and continuous sound which gives on idea of a long drawn sound.
Reverberation is the continuous occurrence of echoes due to multiple reflection of the original sound. It is the perseverance of sound (echoes) after the source have ceased. Reverberation can be reduced in an auditorium or hall by covering the ceiling and walls with soft perforated boards that minimizes reflection of sound waves by quickly absorbing the sounds. Other methods are hanging curtains around the walls and putting more holes in the walls.
7. Explain Application of echoes
Application of non-multiple echoes: The principle echoes applied is:1. Determination of the speed of sound in air i.e Distance of sound waves from source
2. Echo sounding: this is used to determine the depth of the sea bed.
3. Exploration of oil and gas.
4. Detection of submarines or a school of fish or Shoals of fish.
5. Under water object or obstacles.
Application of multiple echoes: The principle echo is applied in speaking tube, a device used in carrying voices from the bridge of a ship to the engine room.
8. Explain Factors that affect the velocity of sound
the factors that affects the velocity of sound are:I. Density of the medium
II. Wind
III. Temperature
8. Explain and Defined Force Vibration and Resonance
All types of vibrating systems have their own natural frequency of vibration. This is the frequency with which the body or system will vibrate when it is left undisturbed after being set into oscillatory motion.For example the natural frequency(f) of a simple pendulum at any place depends on the length of the pendulum according to the relation
T = 2π√(L/g)
where T = 1/f
If a vibrating body is placed in contact with another body, the latter will also be forced to vibrate at the same frequency with the former.
The vibration of the second body is called a forced vibration.
Examples of force vibrations in everyday life includes:
plucking a guitar string, striking a drumhead, or speaking into a microphone.
1. If tuning fork A is struck and stopped, you find that it will cause tuning fork B to vibrate, provided both forks have the same frequency. This is called forced vibration.
2. When you stuck the tuning fork to vibrate and then placed it on the table. The table produced sound higher than the tuning fork
3. A glass producing a loud sound when tapped with a spoon. The idea that this phenomenon is related to is what we refer to as Resonance.
4. A glass filled with water as a musical instrument. When tap lightly, it produces a sound. Tapping it with the right frequency creates resonance, making the sound louder. These are example of forced vibration and resonance
FORCED VIBRATION: Force vibration is the induction of vibrations in an object due to an external force acting upon it and they can result in the generation of sound waves. OR
Forced vibrations: vibrations that result from an external periodic force acting on a system and setting vibrating at the same frequency as the external periodic force.
RESONANCE: Resonance is a special case of forced vibration which occurs when a system is made to vibrate at its own natural frequency as a result of forced vibrations received from another sources of the same frequency. OR
Resonance is a phenomenon which occurs whenever a particular body or system is set in oscillation at its own natural frequency as a result of impulses or signals received from some other systems or body which is vibrating with the same frequency.
8. Explain Fundamental frequency, harmonics and overtones of a string, closed pipe at both ends, closed pipe at one end and opened pipe at both ends
Stationary/Standing waves can occur on a stretched string or wire fixed at both ends. This is obtained by varying the driving frequency of the string.If a string is fixed at both ends and plucked gently at the middle, a transverse wave travels along the string in its simple segment loop.
Stationary waves are produced when two waves with the same frequency and amplitude travelling in opposite directions are superposed. Nodes are points on a stationary waves which are at rest. Antinodes are points where the amplitude of vibration is maximum.
i. Fundamental frequency is the lowest frequency of any vibrating object. i.e fo.
ii. Harmonic is a resonance frequency equal to an integral multiple of the fundamental frequency. E.g 1fo, 2fo, 3fo, 4fo, 5fo, etc.
iii. Overtone is any resonanat frequency above the fundermental frequency. All harmonic resonance frequency are overtone, but not all overtone are harmonic. For example, the overtones of a bell are not harmonics, The resonant vibrational modes of an ideal membrane with length equal to 1.41 times its width, 1fo, 1.41fo, 1.73fo, 2.00fo, 2.38fo, 2.71fo, 3.00fo, 3.37fo are not harmonics,
The circular resonant vibrational modes of an ideal membrane, 1fo, 1.59fo, 2.14fo, 2.30fo, 2.65fo, 2.92fo, 3.16fo, 2.50fo, 3.60fo, 3.65fo, 4.06fo, 4.15fo are not harmonics, etc.
Since their frequencies are not integral multiples of the fundamental frequency, however, they are resonanat frequency above the fundermental frequency, hence they are called overtone.
1. Fundamental frequency, harmonics and overtones of a string and pipe closed at both ends
The distance between the two nodes is ʎ/2 to the length of the string L that is
L = ʎ/2
or ʎ = 2L ------(1)
but v = fʎ
:. f = v/ʎ
Substituting for ʎ from equ 1.
:. fo = v/2L, this is the fundamental frequency fo of a string fixed at both ends and a pipe closed at both ends
fo = v/2L
The distance between the three nodes is ʎ to the length of the string L that is
ʎ = L -----(2)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 2.
:. f1 = v/L
But fo = v/2L
2 × fo = 2 × v/2L
2 × fo = v/L
2 × fo = f1
:. 2fo = f1
The distance between the four nodes is 3ʎ/2 to the length of the string L that is
3ʎ/2 = L
ʎ = 2L/3 -----(3)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 3.
:. f2 = v ÷ 2L/3
f2 = 3v/2L
But fo = v/2L
3 × fo = 3 × v/2L
3 × fo = 3v/2L
3 × fo = f2
:. 3fo = f2
Higher frequencies which are integral or whole number multiples of the fundamental frequency fo can also be produced in the string. These are called harmonics or overtones. E.g fo, 2fo, 3fo, 4fo, 5fo, etc.
The fundamental frequency fo is the first harmonic, 2fo, 3fo etc. are the second and third harmonics.
Therefore, All harmonic are present in the modes of vibration of a string/pipe closed at both end. I.e both odd and even harmonics are present as overtones
The distance between a node and an antinode is ʎ/4 to the length of the string L that is
L = ʎ/4
or ʎ = 4L ------(1)
but v = fʎ
:. f = v/ʎ
Substituting for ʎ from equ 1.
:. fo = v/4L, this is the fundamental frequency fo of a pipe closed at one end and opened at the other end
fo = v/4L
The distance between a two nodes and an antinode is 3ʎ/4 to the length of the string L that is
L = 3ʎ/4
ʎ = 4L/3 -----(2)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 2.
:. f1 = v ÷ 4L/3
f1 = 3v/4L
But fo = v/4L
:. f1 = 3 × v/4L
f1 = 3 × fo
:. f1 = 3fo
The distance between the three nodes and an antinode is 5ʎ/4 to the length of the string L that is
5ʎ/4 = L
ʎ = 4L/5 -----(3)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 3.
:. f2 = v ÷ 4L/5
f2 = 5v/4L
But fo = v/4L
f2 = 5v/4L
f2 = 5 × v/4L
f2 = 5 × fo
:. f2 = 5fo
Higher frequencies which are integral or whole number multiples of the fundamental frequency can also be produced in the string. These are called harmonics or overtones. E.g fo, 3fo, 5fo, 7fo, 9fo, etc.
The fundamental frequency fo is the first harmonic, 3fo, 5fo etc, are the second and third harmonics.
Therefore, Only odd harmonic are present in the modes of vibration of a pipe closed at one end and opened at the other end.
The distance between an antinode, one node and an antinode is ʎ/2 to the length of the string L that is
L = ʎ/2
or ʎ = 2L ------(1)
but v = fʎ
:. f = v/ʎ
Substituting for ʎ from equ 1.
:. fo = v/2L, this is the fundamental frequency fo of a string fixed at both ends and a pipe closed at both ends
fo = v/2L
The distance between an antinode, two node and an antinode is ʎ to the length of the string L that is
ʎ = L -----(2)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 2.
:. f1 = v/L
But fo = v/2L
2 × fo = 2 × v/2L
2 × fo = v/L
2 × fo = f1
:. 2fo = f1
The distance between an antinode, three node and an antinode is 3ʎ/2 to the length of the string L that is
3ʎ/2 = L
ʎ = 2L/3 -----(3)
But v = fʎ
f = v/ʎ
Substituting for ʎ from equ 3.
:. f2 = v ÷ 2L/3
f2 = 3v/2L
But fo = v/2L
3 × fo = 3 × v/2L
3 × fo = 3v/2L
3 × fo = f2
:. 3fo = f2
Higher frequencies which are integral or whole number multiples of the fundamental frequency can also be produced in the pipe. These are called harmonics or overtones. E.g fo, 2fo, 3fo, 4fo, 5fo, etc.
The fundamental frequency fo is the first harmonic, 2fo, 3fo etc. are the second and third harmonics.
Therefore, All harmonic are present in the modes of vibration of a pipe opened at both end. I.e both odd and even harmonics are present as overtones
:. The harmonic of a pipe closed at both ends is the same as the harmonic of the pipe opened at both ends.
9. Calculate the harmonic/overtone of musical instruments
Example 1Find the frequency of the first harmonic of a piano string of length 1.5m.
If the velocity is 120 m/s
SOLUTION
First harmonic of a string of length L = ʎ/2
ʎ = 2L
But v = fʎ
f = v/ʎ
f = v/2L f = 120÷(2×1.5) f = 120÷3 f = 40Hz
Example 2
Find the frequency of the first three harmonic of piano string of length 1.5, if the velocity of the waves on the string is 120m/s.
Solution
L = 1.5
V = 120m/s
First, Second and third Harmonic of a string are:
f0 = v/2L
f1 = v/L
f2 = 3v/2L
f0 = v/2L
f0 = 120/(2×1.5)
f0 = 40Hz
f1 = v/L
f1 = 120/1.5
f1 = 80Hz
f2 = 3v/2L
f2 = (3×120)÷(2×1.5)
f2 = 120Hz
Example 3
A pipe close at one end is 100cm long. If the air in the pipe is set into vibration and fundamental note produced, calculate the frequency of the note (velocity of sound in air = 340m/s).
Solution
Fundamental frequency of a pipe closed at one end f0 = v/4L
L = 100cm = 1m
f0 = v/4L
f0 = 340/(4×1)
f0 = 85Hz
Example 4
A turning fork of frequency 600HZ is sounded over a closed tube. If the first and second resonant position are 0.130m and 0.413m respectively, calculate:
(a) the speed of sound in air
(b) the wavelength of the first overtone of a note in a closed pipe of length 33cm
Solution
(a) Given that
f = 600Hz
L1 = 0.130m = ʎ/2
L2 = 0.413m = ʎ
L2 - L1 = 0.413m - 0.130m = ʎ - ʎ/2
L2 - L1 = 0.283 = ʎ/2
:. ʎ = 2 × 0.283 = 0.566m
V = ?
but f = v/ʎ
v = ʎf
v = 0.566×600
v = 339.6m/s
(b) L = 33cm = 0.33m
L = ʎ/2
ʎ = 2L
ʎ = 2×0.33 ʎ = 0.66m
Example 5
In a resonance tube experiment, a tuning fork of frequency 256Hz gave resonance when the water level was 32.5cm below the open end of the tube. If the next position of resonance was 100cm, what is the speed of sound in air?
Solution
1st position L1 = ʎ/4
2nd position L2 = 3ʎ/4
L2 - L1 = 3ʎ/4 - ʎ/4
L2 - L1 = ʎ/2
:. ʎ = 2(L2 - L1) -----(1) f = 256Hz
L1 = 32.5cm = 0.325m
L2 = 100cm = 1m
ʎ = 2(1 - 0.325)
ʎ = 2 × 0.676
ʎ = 1.35
But f = v/ʎ
:. v = ʎf
v = 1.35 × 256
v = 345.6m/s
Example 6
The length of air column at which the first resonance was observed, when a vibrating fork was placed on a resonance tube was 30cm. Calculate tne wavelength of the air column and the frequency of the fork. (Take speed of sound as 330m/s.)
Solution
Since the fork is placed on a tube, the tube must be one end closed and its top opened and sounding its fundamental frequency, hence,
(i) L = ʎ/4
ʎ = 4L
ʎ = 4 × 30cm
× = 120cm or 1.20m
(ii) f = v/ʎ
f = 330/1.20
f = 275Hz
10. Explain the modes of vibration of a stretched string (the Sonometer)
The Sonometer consists of a hollow sounding-box with a wire stretched along its length. The wire is attached to a string vibrator while the other end is passed over a grooved wheel frictionless pulley and kept taut by attaching weights W at its free end. We can vary the length of the vibrating part of the wire by adjusting the movable bridges.The Sonometer is used to determine the frequency of a note. It depends on the tension T, length L, and mass per unit length M.
You will notice that a long, thick and loose guider string produces a low frequency note, while a thin, short and taut string produces a high frequency note.
Therefore
(i) The frequency f0 is inversely proportional to the length at constant Tension(T) and Mass(M) i.e f0 α 1/L
(ii) The frequency f0 is proportional to the square root of the Tension at constant Length(L) and Mass(M) i.e f0 α √T
(iii) The frequency f0 is inversely proportional to the square root of the mass at constant Tension(T) and Length(L) i.e f0 α 1/√M
:. f0 α 1/L × √(T/M)
f0 = K/L × √(T/M)
(iv) It can be shown that the velocity v of a wave propagated along a string depends on it tension T, its mass M per unit length and given by:
V = √(𝑇/𝑀)
The two movable knife edges or bridges show that the string is fixed at both ends, hence
L = ʎ/2
ʎ = 2L
f0 = v/ʎ
f0 = v/2L
The fundamental frequency can therefore be written as
f0 = 1/2L√(𝑇/𝑀)
Example
An instrument has three identical strings which are subjected to the following tension: 5N, 10N, 20N. Which string will produce a note of the highest pitch when sounded? Give reasons for your answer.
Solution
The string which has tension 20N will have the highest frequency (pitch) because the frequency is proportional to the square root of the tension.
Example 2
A metal wire of mass 1g and length 50cm is under a tension of 80N. Compute
(a) the fundamental frequency
(b) the first and second overtones
(b) the speed of a transverse wave in the string.
Solution
(a) f0 = 1/2L√(𝑇/𝑀)
L = 50cm = 0.5m
T = 80N
M = 1g = 0.001kg Mass per unit length M = 0.001kg/0.5m = 0.002kg/m
f0 = 1/2×0.5√(80/0.002)
f0 = 200 per sec or 200Hz
(b) 1st overtone = 2f0 = 2×200 = 400 Hz
2nd overtone = 3f0 = 3×200 = 600 Hz
(c) v = f0ʎ
ʎ = 2L
:. v = f0×2×L
v = 200×2×0.5
v = 200m/s
Example 3
Under a constant tension, a plucked string produces a note of 250 Hz when the length is 1.0m. At what length is the frequency 200Hz? What frequency will be prouced with a length of 0.5m?
Solution
f α 1/L at constant tension
f = k/L
f = 250Hz
L = 0.1m
:. K = fL
k = 250×1.0
k = 250
when f = 200Hz
L = k/f
L = 250/200
L = 1.25m
when L = 0.5m
f = k/L
f = 250/0.5
f = 500Hz
Rationale:
Sound is a form of energy produced by vibrating objects while vibration is a rapid back-and-forth movement of particles in a medium. Examples of vibrating objects includes, guitar strings, drumheads, and tuning fork, vocal cords.Prerequisite/ Previous knowledge:
Application of light wave, musical nstrumentsetc.Learning Materials:
Tunning fork, guitar strings, drumheads, and tuning fork, vocal cords.Reference Materials:
1. New School Physic for Senior Secondary Schools, by M. W. Anyakowa.2. Lamlas's SSCE and UTME, by O. Ajaja and H. B. Olaniyi.
3. Okeke P.N and etal (2011), Macminian, Senior Secondary School Physics, New Edition.
4. Farinde E.O and etal (2015), Essential Physics. 6th Edition,
5. New System Physics by Dr. Charles Chow.
Lesson Development:
| STAGE | TEACHER'S ACTIVITY | LEARNER'S ACTIVITY | LEARNING POINTS |
|---|---|---|---|
| STEP I: PREVIOUS KNOWLEDGE full class session |
The teacher asks learners questions based on previous knowledge. Explain the Relation between Frequency and Wavelength. |
Learners' response to teacher's question. Frequency is defined as the number of oscillations of a wave per unit time being, measured in hertz(Hz). The frequency is directly proportional to the pitch. Humans can hear sounds with frequencies ranging between 20 – 20000 Hz. Sounds with frequencies over the human ears range are known as ultrasound, and sounds with frequencies less than the audible range are known as infrasound. Wavelength is defined as the distance between two most near points in phase with each other. Hence, two adjacent peaks or troughs on a wave are separated by a distance of a single complete wavelength. Mostly, we use the letter lambda (λ) to describe the wavelength of a wave. Relation between Frequency and WavelengthYou can create a wave having some frequency on a rope and then gradually make it faster. During this, you will notice that as the wavelength becomes shorter. Therefore, the relation between frequency and wavelength does exist, so let’s find out what exactly this relationship is. Another quantity is Period (T) which can be used to describe a wave. It is defined as the time taken to complete one oscillation. Because frequency determines the number of times a wave oscillates and it can be expressed as, Every point on the wave returns to the same value after one period since a wave undergoes one oscillation during one period. This occurs because every oscillation session travels a wavelength distance from one period to finish. The wave speed (v) is defined as the distance travelled by a wave per unit of time. If considering that the wave travels a distance of one wavelength in one period, ν=λ/T As we know that T = 1/f, hence we can express the above equation as, V = f λ The wave speed is equal to the product of its frequency and wavelength, and this implies the relationship between frequency and wavelength. |
Confirming previous knowledge. |
| STEP II: INTRODUCTION full class session |
The teacher introduce the topic for discussion. Sound is a form of energy produced by vibrating bodies. For example, when we strike a body such as a tuning fork its prongs vibrate, the vibrations give rise to sound waves. Therefore, sound waves are produced by vibrating bodies as a result every object that vibrates is a source of sound. Examples are chairs, iron, drums, air column etc. almost any object can vibrate and hence a source of sound. Sound waves are transmitted in air by the displacement (vibrations) of air molecules in longitudinal form as earlier discussed. Sound is also transmitted through liquids and solids. Sound does not readily travel from one medium to another due to the different densities of the media. For example, the heartbeat of a man is not heard by listening in the air close to his chest but by putting ones ear directly on his chest( a solid of similar density) or a stethoscope(connected to the listeners ear) to touch the mans chest. |
Learners listen to teacher, ask questions and make contribution. | Developing the idea of the topic. |
|
|
|||
| STEP III: EXPLORATION Grouping |
The teacher presents the instructional material to the students and divides them into four groups. The representative of a group is asked to discuss the uses of the material. Aim is to show Propagation of sound waves in air. HINT: Sound is a form of energy produced by vibrating bodies. For example, when we strike a body such as a tuning fork its prongs vibrate, the vibrations give rise to sound waves. Therefore, sound waves are produced by vibrating bodies as a result every object that vibrates is a source of sound. Examples are chairs, iron, drums, air column e.t.c. almost any object can vibrate and hence a source of sound. The teacher asks the students to strike the tunning fork against their desks place very to their hears and state their observations. The teacher leads the students into the discussion from the observations gathered from the exploration, the Students equally gives their contribution's and the teacher gives the students note on the topic. |
Learners' expected answer The students carry out the instructions accordingly and state their observations. A material medium is required for sound to be transmitted or else sound will not be transmitted from the source to the hearer. Sound cannot travel through a vacuum as light does. Sound waves are transmitted in air by the displacement (vibrations) of air molecules in longitudinal form as earlier discussed. Sound is also transmitted through liquids and solids. Sound does not readily travel from one medium to another due to the different densities of the media. For example, the heartbeat of a man is not heard by listening in the air close to his chest but by putting ones ear directly on his chest( a solid of similar density) or a stethoscope(connected to the listeners ear) to touch the mans chest. |
Sound is transmitted through gases, plasma, and liquids as longitudinal waves, also called compression waves. It requires a medium to propagate. |
| STEP IV: DISCUSSION | The teacher explain vibration in pipe and guides the students to list and explain the characteristics of sound waves. Sounds are produced when an air column in a pipe is set into vibration an external disturbance from one end of it with the other end opened or closed. The type of sound waves produced in a stationary longitudinal wave. In an open pipe, the other end is an antinode, while in a closed pipe the other end is a node. The frequency of vibration of air in these pipes increases as the length of air column decreases. If (f) is frequency and(L) is length of air, then f is inversely proportional to L Usually we can vary the length of air column either by shortening the tube or by filling it with water and gradually running off water to create more space for air. The Teacher tells the students that vibration in pipes are used in flutes and other musical instruments. |
The students listed and explained the characteristics of waves: A. PITCH: This depends on the frequency of the sound waves. If the frequency is increase, the pitch of the sound also increases. B. LOUDNESS: The loudness of the sound depends on its intensity. The intensity of the sound of the wave is the rate of the flow of energy per unit area, perpendicular to the direction of the wave. Intensity is proportional to the square of the amplitude. The greater the intensity, the louder the sound. C. QUALITY: This is the property which enables us to distinguish the same note played on different instruments e.g a piano and an organ, the quality of a musical notes depends on the harmonies. When a note is produced, the strongest, audible frequency heard is the fundamental. All other frequencies present ar harmonics or overtones. |
Characteristics of sound waves and vibration in pipe. |
| The teacher explains the relationship between frequency and (1) Tension (2) mass per unit length and (3) length of the string. Using the sonometer. The sonometer is an instrument with which the frequency of a note produced by a plucked string can be studied. It consists of a hollow sounding box, along whose length is a string stretched. The string is attached to a peg at one end and the free end is passed over a grooved smooth pulley. To the free end, is an attached weight by which the string is stretched taut. Two moveable knife edges function as fixed ends and as well used to vary the length of the string between the fixed ends. The weight, W. equals the tension T in the string. When the weight (Tension) in the string is kept constant, and length are varied, the relationship below is obtained. f α 1/L The shorter the length, the higher the frequency When the length is kept constant and the frequency and tension are varied then we have: f α √T The higher the tension, the higher the frequency. When both the length of the string and the tension are kept constant and strings of different mass per unit length are used with varying frequencies then: f α 1/√m The smaller the mass per unit length (thickness) the higher the frequency. The varying frequencies are obtained with the help of tuning forks of known frequencies. When the three relationships above are combined, it gives f α 1/l√T/m therefore f = k/l(√T/m) Where: f = frequency of vibration of string L = length of string T = tension in the string m = mass per unit length of string k is a constant which depends on the mode of vibration (number of loops) Therefore fo = 1/2L√T/m F1 = 2fo = 1/L√T/m F2 = 3fo = 3/2L√(T/m) and so on. |
The Students gives their contribution's based on their Understanding. | ||
| STEP V: APPLICATION | The teacher guides learners to explain resonance and force vibration Compare resonance to a playground swing. Pushing the swing at its natural frequency makes it go higher and higher. Similarly, resonance occurs when an object is exposed to its natural frequency, causing it to vibrate more strongly. Resonance is the reinforcement or amplification of vibrations when an object is exposed to its natural frequency. Imagine a glass filled with water as a musical instrument. When you tap it lightly, it produces a sound. Tapping it with the right frequency creates resonance, making the sound louder. So, just as a swing goes higher when pushed at its natural frequency, objects resonate more strongly when exposed to their natural frequencies. Force vibration explains how external forces induce vibrations and influence sound production. Force vibration is about inducing vibrations in an object through external forces, affecting various sound characteristics or deals with the initiation of vibrations through external forces, while resonance involves the enhancement of vibrations when the applied force matches the natural frequency of an object. It focuses on the reinforcement or amplification of vibrations when an object is exposed to its natural frequency, leading to a stronger and more sustained sound. The teacher leads the students into the discussion from the observations gathered from the exploration, the Students equally gives their contribution's and the teacher gives the students note on the topic. |
The students listen to teacher, ask questions and make contribution. Force vibration is the induction of vibrations in an object due to an external force acting upon it and they can result in the generation of sound waves. Examples of force vibrations in everyday life includes plucking a guitar string, striking a drumhead, or speaking into a microphone. In the class, we stuck the tuning fork and then while it vibrated, we placed it on the table. Did you remember how the table produced sound higher than the tuning fork? That is an example of forced vibration. A glass producing a loud sound when tapped with a spoon? The idea that this phenomenon is related to is what we refer to as Resonance. All types of vibrating systems have their own natural frequency of vibration.This is the frequency with which the body or system will vibrate when it is left undisturbed after being set into oscillatory motion. For example the natural frequency(f) of a simple pendulum at any place depends on the length of the pendulum according to the relation. T = 2π√L/g where T = 1/f If a vibrating body is placed in contact with another body, the latter will also be forced to vibrate at the same frequency with the former. The vibration of the second body is called a forced vibration. |
Force Vibration are those vibration that result from an external periodic force acting on a system and setting the system vibrating at the same frequency as the external periodic force. Such vibrating body usually has its own amplitude of vibration greatly reduced by air resistance. This process of reduction is called DAMPING. Example, the setting of a table to vibrate when a vibrating tuning fork is placed on it. |
|
|
|||
| STEP VI: EVALUATION | The teacher asks the learners questions. 1. Why do we see a flash of lightning before the accompany sound of thunder? 2. List and explain the characteristics of sound waves. 3. Why are sound waves called mechanical waves? 4. In which medium does the sound wave travel the fastest? 5. In which medium is the speed of sound the lowest? 6. What is the frequency of a sound wave? 7. What is the reverberation of sound? |
The learners expected answers. 1. We see a flash of lightning before the accompany sound of thunder because the velocity of light is about (f = 3X10⁸ m/s) is far greater than the velocity of sound about (330m/s). Thus suppose we hear the sound of thunder 10 seconds after we have seen the lighting flash then the distance between us and the storm center or the point of the lighting and thunder is giving by S = Vt S = (300 X 10)m S = 33km. 2. The characteristics of sound waves include the following (a) Pitch: The pitch of a note is its position on the musical scale it is a characteristic of a note which enables one to differentiate a high note from a low pitch and the left hands side keys produce note of low pitch. (b) Loudness and Intensity: I. Intensity of sound is the rate of low of energy per second per unit area perpendicular to the direction of propagation of the sound wave. II. Loudness is the magnitude of the sensation resulting from a sound reaching the ear. (c) Quality or tone (timber): Quality is a characteristic of a note of a musical instrument which distinguishes it from another note of the same pitch and loudness produced by another instrument. 3. Mechanical waves are waves that require a medium to transport their energy from one location to another. Sound is a mechanical wave and cannot travel through a vacuum. 4. The sound waves travel the fastest through solids. 5. The speed of sound is lowest in gases. 6.The number of rarefactions and compressions that occur per unit of time is known as the frequency of a sound wave. 7. If the area is large and has multiple reflecting surfaces such as valleys and huge empty rooms, we hear multiple echoes from a source of sound and we call this phenomenon reverberation. |
Ask the learners questions to assess the achievement of the set objectives. |
| CONCLUSION | The teacher concluded the lesson with the students. Almost any object can vibrate and hence be a source of sound; we can demonstrate the production of sound in the laboratory using a tuning fork two steel prong which vibrate when struck with a hard rubber giving some sound. An important aspect of sound transmission is that a material medium is always required for the passage of sound can be demonstrated with an apparatus that consists of an electric bell covered with a bell jar placed on the plate of a vacuum pump. The clapper is still seen to be vibrate even through no sound is heard, showing that sound waves cannot travel through vacuum. |
The students listen and make contribute | Transmission of sound |
|
|
|||
| ASSIGNMENT | The teacher gives learners a take home. | Objectives 1. Sound wave differs from water wave_____ (a)energy transfer is involve (b) they can be refracted and reflected (c)no transfer of the medium is involved (d)They are longitudinal wave. 2. A source of sound produces waves in air if wavelength 1.65m. if the speed of sound in air is 330m/s, the period of vibration is. (a) 200 (b) 0.005 (c) 0.5 (D) 0.02 3. The speed of sound traveling in various media increases in the following correct order. (a) Iron bar, air, water (b) air, iron bar, water (c) air, water, iron bar (d) water, iron bar, air. 4. Why does the sound from an enclosed bell jar gradually fade away as the jar is gradually evacuated? (a) the sound is forced out (b) the pressure within the jar is reduced (c) there is no more material medium (d) the temperature is reduced. 5. A noise of frequency 2000Hz has a velocity of 400m/s. What is the wavelength of the noise? (a) 0.02m (b) 0.25m (c) 0.2m (d) 2m Theory 1. Briefly explain closed and Open pipe vibration. 2. Define a musical note 3. Mention the three characteristics of sound and the factors on which they depend |
Improving their level of understanding on sound waves. |