» WAVES, SOUND AND OPTICS
» LECTURER :
MR. N MANDIWANA
» OFFICE
:
NSB 208
»Waves and Sound
16.1 The nature of waves
16.2 Periodic waves
16.3 The Speed of a Wave on a String
16.5 The Nature of Sound Waves
16.6 The Speed of Sound Waves
16.7 Sound Intensity
16.8 Decibels
16.9 The Doppler Effect
16.10 Application of Sound in Medicine
16.11 Sensitivity of the human ear (Self Study)
When one end of a (stretched) string is given a single up-anddown jerk, a wave in the form a pulse travels along the string
1. A wave is a traveling disturbance.
2. A wave carries energy from place to place.
Transverse Wave
• A transverse wave is one in which the disturbance occurs
perpendicular to the direction of travel of the wave
• Examples of transverse waves are radio waves, light waves
and microwaves
Longitudinal Wave
A longitudinal wave is one in which the disturbance occurs
parallel to the line of travel of the wave
A sound wave is a longitudinal wave
• Some waves are neither transverse nor longitudinal
• Water waves are partially transverse and partially
longitudinal.
• Periodic waves consist of cycles or patterns that are produced
over and over again by the source.
• In the figures, the repetitive patterns occur as a result of simple
harmonic motions.
In the drawing, one cycle is shaded in color.
• The amplitude A is the maximum excursion of a particle of the
medium from the particles undisturbed position.
• The wavelength is the horizontal length of one cycle of the
wave.
• The period is the time required for one complete cycle.
• The frequency is related to the period and has units of Hz, or s-1.
1
𝑓=
𝑇
𝜆
𝑣 = = 𝑓𝜆
𝑇
Example 1 The Wavelengths of Radio Waves
AM and FM radio waves are transverse waves consisting of
electric and magnetic field disturbances traveling at a speed of
3.00x108m/s. A station broadcasts AM radio waves whose
frequency is 1230x103Hz and an FM radio wave whose
frequency is 91.9x106Hz. Find the distance between adjacent
crests in each wave.
v
AM
FM

T
 f
v

f
v 3.00 108 m s
 
 244 m
3
f 1230 10 Hz
v 3.00 108 m s
 
 3.26 m
6
f
91.9 10 Hz
• The speed of a wave depends on the properties of the medium
in which it travels.
• For a transvers wave (on a string) that has a tension 𝐹 and
mass per unit length 𝑚 𝐿, the speed is
tension
𝑣=
𝐹
𝑚 𝐿
linear density
Example 2 Waves Traveling on Guitar Strings
Transverse waves travel on each string of an electric guitar
after the string is plucked. The length of each string between
its two fixed ends is 0.628 m, and the mass is 0.208 g for the
highest pitched E string and 3.32 g for the lowest pitched E
string. Each string is under a tension
of 226 N. Find the speeds of the waves on the two strings.
Conceptual Example 3 Wave Speed Versus Particle Speed
Is the speed of a transverse wave on a string the same as the
speed at which a particle on the string moves?
LONGITUDINAL SOUND WAVES
Sound is a longitudinal wave that can be
created only in a medium, i.e. sound cannot
exists in a vacuum
The distance between adjacent condensations
(or rarefaction) is equal to the wavelength of
the sound wave.
• Individual air molecules are not carried along with the wave.
• A given molecule vibrates back and forth about a fixed point
THE FREQUENCY OF A SOUND WAVE
• The frequency is the number of cycles per second.
• A sound with a single frequency is called a pure tone.
• The brain interprets the frequency in terms of the subjective
quality called pitch.
• Human ear can detect sound of frequencies approximately
from 20 Hz to 20 000 Hz (20 kHz)
• Sound waves with frequencies below 20 Hz are inaudible
and are said to be infrasonic
• Frequencies above 20 kHz are ultrasonic
THE PRESSURE AMPLITUDE OF A SOUND WAVE
• Loudness is an attribute of a sound that depends
primarily on the pressure amplitude of the wave.
Sound travels through gases,
liquids, and solids at considerably
different speeds.
• In a gas, it is only when molecules collide that the
condensations and rarefactions of a sound wave can move
from place to place.
• For an ideal gas
𝑣𝑟𝑚𝑠 =
3𝑘𝑇
𝑚
𝑘 𝑖𝑠 𝑡ℎ𝑒 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛′ 𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 1.38 × 10−23 𝐽/𝐾
The speed of sound in an ideal gas is given
𝑣=
𝛾𝑘𝑇
𝑚
5
7
𝛾 = 𝑜𝑟
3
5
Conceptual Example 5 Lightning, Thunder, and a Rule of
Thumb
There is a rule of thumb for estimating how far away a
thunderstorm is. After you see a flash of lighting, count off the
seconds until the thunder is heard. Divide the number of seconds
by five. The result gives the approximate distance (in miles) to the
thunderstorm. Why does this rule work?
LIQUIDS
𝑣=
𝐵𝑎𝑑
𝜌
SOLID BARS
𝑣=
𝑌
𝜌
• Sound waves carry energy that can be used to do work.
• The amount of energy transported per second is called the
power of the wave.
• The sound intensity is defined as the power that passes
perpendicularly through a surface divided by the area of that
surface.
𝑃𝑜𝑤𝑒𝑟 (𝑃)
𝐼=
𝐴𝑟𝑒𝑎 (𝐴)
Example 6 Sound Intensities
In the figure below12x10-5W of sound power passed through the
surfaces labeled 1 and 2. The areas of these surfaces are 4.0m2
and 12m2.
Determine the sound intensity at each surface.
𝑃
𝐼1 =
𝐴1
12 × 10−5 𝑊
−2
=
=
3.0
𝑊.
𝑚
4𝑚2
𝑃
12 × 10−5 𝑊
−5 𝑊. 𝑚−2
𝐼2 =
=
=
1.0
×
10
𝐴2
12𝑚2
• For a 1000 Hz tone, the smallest sound intensity that the
human ear can detect is about 1 × 10−12 𝑊. 𝑚−2 . This
intensity is called the threshold of hearing.
• On the other extreme, continuous exposure to intensities
greater than 1𝑊. 𝑚−2 can be painful.
• If the source emits sound uniformly in all directions, the intensity
depends on the distance from the source in a simple way.
power of sound source
𝑃
𝐼=
4𝜋𝑟 2
area of sphere
Conceptual Example 8 Reflected Sound and Sound Intensity
Suppose the person singing in the shower produces a sound power P.
Sound reflects from the surrounding shower stall. At a distance r in
front of the person, does the equation for the intensity of sound
emitted uniformly in all directions underestimate, overestimate, or
give the correct sound intensity?
𝑃
𝐼=
4𝜋𝑟 2
Ex 16-61
Deep ultrasonic heating is produced by applying ultrasound
over the affected area of the body. The sound transducer is
circular with a radius of 1.8 cm, and it produces a sound
intensity of 𝟓. 𝟗 × 𝟏𝟎𝟑 𝑾/𝒎𝟐 .
How much time is required for the transducer to emit 𝟔 𝟎𝟎𝟎 𝑱
of energy?
Solution
3
2
−2
𝐼 = 5.9 × 10 𝑊 𝑚
𝑃 = 𝐼𝐴
𝐸
∴ 𝐼𝐴 =
𝑡
𝐸
𝑃=
𝑡
𝑟 = 1.8 × 10 𝑚
𝐸 = 6000𝐽
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2
𝐸
𝑡=
𝐼𝐴
6000𝐽
𝑡=
5.9 × 103 𝑊 𝑚2 𝜋 1.8 × 10−2 𝑚
𝐸
𝑡=
𝐼𝜋𝑟 2
2
𝑡 = 999𝑠
• The decibel (dB) is a measurement unit used when comparing
two sound intensities.
• Because of the way in which the human hearing mechanism
responds to intensity, it is appropriate to use a logarithmic scale
called the intensity level:
𝐼
𝛽 = 10𝑑𝐵 𝑙𝑜𝑔
𝐼0
𝐼0 = 1 × 10−12 𝑊. 𝑚−2
Note that log(1)=0, so when the intensity of the sound is equal to the
threshold of hearing, the intensity level is zero.
Example 9 Comparing Sound Intensities
Audio system 1 produces a sound intensity level of 90.0 dB, and
system 2 produces an intensity level of 93.0 dB.
Determine the ratio of intensities.
𝛽(𝑑𝐵) = 10 log
𝛽1 = 10 log
𝐼1
𝐼0
𝐼
𝐼0
𝛽2 = 10 log
𝐼2
𝐼1
𝛽2 − 𝛽1 = 10 log
− 10 log
𝐼0
𝐼0
𝐼2
𝐼0
𝐼2 𝐼0
𝛽2 − 𝛽1 = 10 log
𝐼1 𝐼0
𝐼2
= 10 log
𝐼1
𝐼2
3.0 𝑑𝐵 = 10 log
𝐼1
𝐼2
0.3 = log
𝐼1
𝐼2
= 100.30 = 2.0
𝐼1
The Doppler effect is the change in frequency or pitch of the
sound detected by an observer because the sound source and
the observer have different velocities with respect to the
medium of sound propagation.
MOVING SOURCE
𝜆′ = 𝜆 − 𝑣𝑠 𝑇
𝑣
𝑣
𝑓0 = ′ =
𝜆
𝜆 − 𝑣𝑠 𝑇
𝑣
𝑓0 =
𝑣 𝑓𝑠 − 𝑣𝑠 𝑓𝑠
𝜈
𝑓𝑜 = 𝑓𝑠
𝜈 − 𝑣𝑠
𝑓0 = 𝑓𝑠
1
𝑣𝑠
1−
𝑣
𝜈 = 𝑓𝑠 𝜆
𝑇=
1
𝑓𝑠
source moving toward a
stationary observer
𝑓0 = 𝑓𝑠
1
𝑣𝑠
1−
𝑣
𝜈
𝑓𝑜 = 𝑓𝑠
𝜈 − 𝑣𝑠
source moving away from a
stationary observer
𝑓0 = 𝑓𝑠
1
𝑣𝑠
1+
𝑣
𝜈
𝑓𝑜 = 𝑓𝑠
𝜈 + 𝑣𝑠
Example 10 The Sound of a Passing Train
A high-speed train is traveling at a speed of 44.7 m/s when the
engineer sounds the 415-Hz warning horn. The speed of sound is
343 m/s. What are the frequency and wavelength of the sound, as
perceived by a person standing at the crossing, when the train is
(a) approaching and
(b) leaving the crossing?
Solution
𝑣𝑠 = 44.7𝑚/𝑠
𝑓𝑠 = 415𝐻𝑧
𝜈 = 343𝑚/𝑠
𝑓𝑜 = 𝑓𝑠
𝜆′
1
𝑣
1 − 𝜈𝑠
1
𝑓𝑜 = 415𝐻𝑧
44.7𝑚/𝑠
1−
343𝑚/𝑠
𝜈 343𝑚/𝑠
= =
= 0.719𝑚
𝑓𝑜
477𝐻𝑧
𝑓𝑜 = 477𝐻𝑧
MOVING OBSERVER
𝑣0
𝑣0
𝑓0 = 𝑓𝑠 +
= 𝑓𝑠 1 +
𝜆
𝑓𝑠 𝜆
𝑣0
𝑓0 = 𝑓𝑠 1 +
𝑣
𝜈 + 𝑣𝑜
𝑓𝑜 = 𝑓𝑠
𝜈
Observer moving towards
stationary source
𝑣0
𝑓0 = 𝑓𝑠 1 +
𝑣
𝜈 + 𝑣𝑜
𝑓𝑜 = 𝑓𝑠
𝜈
Observer moving away
from stationary source
𝑣0
𝑓0 = 𝑓𝑠 1 −
𝑣
𝜈 − 𝑣𝑜
𝑓𝑜 = 𝑓𝑠
𝜈
GENERAL CASE
𝑓0 = 𝑓𝑠
𝑣0
1±
𝑣
𝑣𝑠
1∓
𝑣
Numerator: plus sign applies
when observer moves towards
the source
Denominator: minus sign applies
when source moves towards
the observer
Exercise 16-86
The siren on ambulance is emitting a sound whose frequency is
2450 Hz. The speed of sound is 343 m/s
a) If the ambulance is stationary and you are sitting in a
parked car, what are the wavelength and frequency of the
sound you hear?
b) Suppose the ambulance is moving toward you at the speed
of 26.8 m/s, determine the wavelength and frequency of
the sound you hear
c) If the ambulance is moving toward you at a speed of
26.8m/s and you are moving toward it at a speed of 14m/s,
find the wavelength and frequency of the sound you hear
a) If the ambulance is stationary and you are sitting in a parked car,
what are the wavelength and frequency of the sound you hear?
𝑣 343𝑚/𝑠
𝜆= =
= 0.140𝑚
𝑓𝑠 2450𝐻𝑧
𝑓0 = 𝑓 = 2450𝐻𝑧
b) Suppose the ambulance is moving toward you at the speed of 26.8
m/s, determine the wavelength and frequency of the sound you
hear
𝜆′
= 𝜆 − 𝑣𝑠 𝑇
𝑓0 = 𝑓𝑠
1
1−
𝑣𝑠
𝑣
𝑣𝑠
𝜆 =𝜆−
𝑓𝑠
′
𝜆′
26.8𝑚/𝑠
= 0.140 −
= 0.129𝑚
2450𝐻𝑧
1
𝑓0 = 2450𝐻𝑧
26.8𝑚/𝑠
1−
343𝑚/𝑠
𝑓0 ≈ 2660𝐻𝑧
c) If the ambulance is moving toward you at a speed of 26.8m/s
and you are moving toward it at a speed of 14m/s, find the
𝑣𝑜 = 14𝑚/𝑠
wavelength and frequency of the sound you hear
𝑓0 = 𝑓𝑠
𝑣𝑜
1±
𝑣
𝑣
1∓ 𝑠
𝑣
𝑓𝑜 = 𝑓𝑠
𝑣
1 + 𝑣𝑜
𝑣𝑠
1−
𝑣
𝑣𝑠 = 343𝑚/𝑠
14𝑚/𝑠
343𝑚/𝑠
𝑓𝑜 = 2450
26.8𝑚/𝑠
1−
343𝑚/𝑠
1+
𝑓𝑜 ≈ 2770 𝐻𝑧
By scanning ultrasonic waves across the body and detecting the
echoes from various locations, it is possible to obtain an image.
Cavitron Ultrasonic Surgical Aspirator (CUSA)
• Neurosurgeons use CUSA to
remove brain tumors once thought
to be inoperable
• Ultrasonic sound waves cause the
tip of the probe to vibrate at 23
kHz and shatter sections of the
tumor that it touches.
Doppler flowmeter
• The device measures the speed of blood flow, using a
transmitting and receiving elements placed directly on the skin
• When the sound is reflected from the red blood cells, its
frequency is changed in a kind of Doppler effect because the
cells are moving.
The Doppler flowmeter can be used to detect motion of a
fetal heart as early as 8-10 weeks after conception