Mechanical Waves
• Represents the periodic motion of
matter e.g. water, sound
• Energy can be transferred from one
point to another by waves
• Waves are cyclical in nature and
display simple harmonic motion
• There are many types of waves, we
will look at two:
1. Transverse Waves: the particles
in the medium move perpendicular
to the direction of the wave motion
eg) earth during earthquake, water,
snapping a rope, light (special case)
crest of wave = highest point
rest position
trough of wave = lowest point
2. Longitudinal Waves: the particles
in the medium move parallel to the
direction of the wave
eg) earth during earthquake, sound
waves
wave direction
source
of wave
•
compression =
particles together
rarefaction =
particles apart
Definitions
1. Amplitude = maximum
displacement of wave from rest
position
• the greater the amplitude, the
greater the amount of energy
being transferred and the greater
the work that can be done
Rest position
2. Frequency (f) = number of cycles
(waves) per second
f = # cycles
time
unit = cycles per second or
the Hertz (Hz)
3.
Period (T) = time for one cycle
T=
time
# cycles
f = 1/T
T = 1/f
Unit of period (T) is second (s)
Example
A wave emits 20 cycles in 10 seconds.
Calculate the frequency and the period.
t = 10 s
f = # of cycles
time
=20/10
=2.0 cycles/s or 2.0 Hz
T = time = 10 = 0.50 s
# of cycles 20
***Note: to calculate period, you could
also use T = 1/f = 1/2.0 = 0.50 s
4. Wavelength (λ) = shortest
distance between two points where
the wave pattern repeats itself
• the unit is the meter (m)
1λ
1λ
1λ
• Two points on a wave are said to
be “in phase” if they are
travelling in the same direction
and are the same distance from
the rest position
G•
•A
B•
C•
•F
D•
L•
K•
•H
•E
•M
rest position
•I
•J
Points in phase: B&H, D&I, E&J, F&K, G&L
Wavelength is the distance between two
adjacent points in phase.
5. Speed (v) = distance travelled by
the wave in a certain time period
v = d/T
=λ/1/f
v =fλ
• This leads us to the universal
wave equation:
v = fλ
where: v = speed of wave in m/s
f = frequency of wave in Hz
(cycles per second)
λ = wavelength in m
Example 1
A wave has a speed of 340 m/s. If the
wavelength is 20 m, what is the
frequency?
v = 340 m/s
v=fλ
λ = 20 m
340=f·(20)
f = 17 Hz
Example 2
The distance between successive crests
in a series of water waves is 1.5 m and
the crests travel 6.5 m in 2.8 s. Calculate
the frequency of a surfer bobbing up and
down in the water.
d = 6.5 m
v = d/t
t= 2.8 s
=6.5/2.8
=2.32 m/s
v=fλ
2.32=f(1.5)
f=1.5 Hz
p. 435 – key terms
http://www.bchs.calgary.ab.ca/site.php?p=/curriculum/science/physics30/
Properties of Waves
• There are 4 fundamental properties of
waves:
1. Reflection
2. Refraction
3. Diffraction
4. Interference
1. Reflection of Waves
• waves are reflected when they hit a
barrier
• if waves hit a barrier straight on, they
are reflected in the opposite direction
barrier
reflected waves
incoming
(incident) waves
• if waves hit a barrier at an angle, they
will be reflected at an angle
barrier
incident wave
angle of angle of
incidence reflection
reflected wave
normal
• Normal = a line perpendicular to the
barrier
• angle of incidence = angle between the
normal and the incident wave
• angle of reflection = angle between the
normal and the reflected wave
• Law of Reflection = the angle of
incidence is always equal to the angle
of reflection
Example
If the angle between the incident wave
and the reflected wave is 120°, what is
the angle of reflection?
barrier
60°
reflected wave
120°
incident wave
120° ÷ 2 = 60°
2. Refraction of Waves
• Refraction is the change in direction
of waves at the boundary between two
different media (substances).
• the frequency and period remain
constant
• the wavelength, speed and direction
of the waves change
eg) waves from deep to shallow water,
light through a prism, your legs in a
pool
normal
i
r
• Angle of incidence (i) = angle between
the normal and the incident wave
• Angle of refraction (r) = angle
between the normal and the refracted
wave
angle i ≠ angle r
Example
If the speed of waves in deep water is 2.5
m/s and in shallow water is 1.0 m/s, and
the wavelength is 10 m in deep water,
what is the wavelength in shallow water?
Deep
v = 2.5 m/s
λ = 10 m
Shallow
v = 1.0 m/s
λ=?
λ=4m
v1 = v2
λ1 λ2
3. Diffraction of Waves
• Diffraction is the spreading of waves
around the edge of a barrier
If there is a hole in the barrier, the wave
will go through and bend…the smaller
the hole, the more the waves bend
(eventually producing a circular wave)
The longer the wavelength of the
incident wave, the greater the diffraction
eg) sound diffracts more than light,
red light diffracts more than blue light
(this is one reason why the sky is blue)
diffracted wave
reflected wave
incident wave
4. Wave Interference
• Waves can combine to form larger or
smaller waves when they interfere
• Constructive interference = when two
or more waves in phase (same
direction above or below rest position)
combine to form a larger wave
wave 1
combination of
waves 1 and 2
wave 2
• Amplitude increases
• waves return to original size
and position relative to rest
position after they pass each
other
• Destructive interference = when two
or more waves out of phase (opposite
direction above or below rest position)
combine to form a smaller wave or
cancel each other out
combination of
waves 1 and 2
wave 1
wave 2
wave 1
combination of
waves 1 and
2..cancels out
wave 2
• amplitude decreases
• Principle of Superposition = when two
or more waves combine, the amplitude
of the resultant wave is the sum of the
amplitudes of the individual waves
Example
Two waves of amplitude 5.0 cm and 8.0
cm interfere. Calculate the amplitude of
the resultant if the waves are: a) in phase
b) out of phase
a) in phase
5.0 cm
8.0 cm + 5.0 cm = 13 cm
13 cm
8.0 cm
b) out of phase - 8.0 cm + 5.0 cm = - 3.0
cm
5.0 cm
3.0 cm
8.0 cm
http://www.glenbrook.k12.il.us/gbssci/ph
ys/mmedia/waves/swf.html
Nodes and Antinodes
• Node = point of zero amplitude (caused
by destructive interference)
• Antinode = point of maximum
amplitude (caused by constructive
interference)
• if you increase the frequency, you
increase the number of nodes
• the distance between two adjacent
nodes = ½ wavelength
node
antinode
Example
The distance between adjacent nodes on
a standing wave is 1.50 m. The
frequency is 50.0 Hz.
a) Calculate the wavelength
Distance between 2 adjacent nodes = ½
wavelength
∴λ = 1.50 m x 2
= 3.00 m
b) Calculate the speed of the wave
λ = 3.00 m
v = fλ
f = 50.0 Hz
= (50.0)(3.00)
= 150 m/s
Example The distance between the
second node and the sixth node is 50 cm.
Find the wavelength of the wave.
2λ=50 cm 1λ=25 cm
Waves at Boundaries
• Speed of a wave depends on the
medium that it is passing through
eg) water → depends on depth
sound → depends on air
temperature
rope → depends on thickness,
tension (force)
• When a wave reaches a boundary
between two media, part of the energy
is transmitted into the new medium as
a wave of the same frequency
• the wavelength and speed change once
in the new medium
• the rest of the energy is reflected back
as a wave of the same frequency in the
original medium
reflected wave
medium 1
incident wave
transmitted wave
medium 2
• the amount of energy transmitted
depends on the difference between the
two media ie) small difference = most
transmitted
large difference = most
reflected
• the relative densities of the two media
affect the reflected and transmitted
waves:
Situation #1: Less Dense to More Dense
reflected transmitted
less dense
more dense
incident
http://www.physicsclassroom.com/mmed
ia/waves/fix.html
• characteristics:
1. reflected wave is inverted
2. transmitted wave (in more dense
medium) has shorter wavelength
and smaller speed
3. frequency (and period) remain
constant
Situation #2: More Dense to Less Dense
reflected wave
more dense medium
transmitted wave
less dense medium
incident wave
• characteristics:
4. reflected wave is upright(erect)
5. transmitted wave (in less dense
medium) has longer wavelength
and faster speed
6. frequency (and period) remain
constant
p. 698 #1-11 (Ch. 14)
Nodes and Antinodes
• Node = point of zero amplitude (caused
by destructive interference)
• Antinode = point of maximum
amplitude (caused by constructive
interference)
• if you increase the frequency, you
increase the number of nodes
• the distance between two adjacent
nodes = ½ wavelength
Sound
• Sound waves are longitudinal which
means they consist of a series of
compressions and rarefactions
direction
source of
sound wave
•
rarefaction
compression
• Sound requires a medium for
transmission
ie) it will not travel in a vacuum
• Travels through solids, liquids and
gases
• Travels fastest through solids
(particles closer together), slowest
through gases
• Speed of sound in air is affected by
temperature
Changes by 0.60 m/s per °C
Speed of sound at 0°C in air = 332 m/s
Example 1
Find the speed of sound at:
a) 18°C
v = 332 m/s + (18 x 0.60 m/s)
= 332 m/s + 10.8 m/s
= 342.8 m/s
= 3.4 x 102 m/s
b) -30°C
v = 332 m/s + (-30 x 0.60 m/s)
= 332 m/s + (-18 m/s)
= 314 m/s
= 3.1 x 102 m/s
Example
Calculate the time taken for an explosion
at Syncrude, 50 km away, to reach Comp
when the air temperature is -15°C.
d = 50 km
t=d
= 50 000 m
v
v = 332 m/s + (-15 x 0.60 m/s) = 323
m/s
t=50,000/323
= 154.8 s = 2.6 min
• Mach number = the number of times
the speed of an object is greater than
the speed of sound
mach # = vobject
vsound
Example
Find the mach number of a plane
travelling at 1650 km/h at 12°C.
vobject = 1650 km/h = 458.3333333 m/s
vsound = 332 + (12 x 0.60 m/s) = 339.2 m/s
Mach number = 458.333/339.2
= 1.351218553
= 1.4
• Doppler Shift = change in frequency
of sound caused by the motion of
either the source or the detector
• When a source generating waves
approaches an observer, the
frequency increases
• When the source moves away, the
frequency decreases
• You can’t tell the direction the source
is coming from but from the change
in pitch of the sound you can tell
when it passes you
away
λ longer
v same
f decrease
•
source
towards
λ- shorter
v-same
f increase
eg) radar, Doppler radar for weather,
ultrasound used by physicians, bats,
ambulance traveling toward an observer,
train approaching or moving away from
a railway station
Properties of Sound
1. Reflection
• sound waves can be reflected off of
hard surfaces
• angle of incidence = angle of
reflection
2. Refraction
• the bending of sound waves occurs
when they move through air at
different temperatures
3. Diffraction
• sound waves can go around objects
• they diffract more than light because
of longer wavelength
4. Interference
• Constructive and destructive
interference can make sounds louder
or softer
• Beats are produced at regular intervals
due to the interference of two sound
waves
• Beat frequency is simply the
difference in frequency of two sources
emitting a sound wave and is the
number of beats heard per second
Bf = f1 – f2
Where: Bf = beat frequency in Hz or beats
per second
f1 = frequency of source 1 in Hz
f2 = frequency of source 2 in Hz
Example : Calculate the beat frequency
for two tuning forks. One is 256 Hz and
the other is 250 Hz.
f1 = 256 Hz
Bf = f1 – f2
f2 = 250 Hz
= 256 – 250
= 6.00 Hz
Example
Two tuning forks produce 5.00 beats per
second. How many beats are heard in
1.00 minute?
5.00 beats/s x 60.0 s = 300 beats
p. 309 #1-4
p. 324 (key terms)
Example 3
A tuning fork with a frequency of 300
Hz is sounded with another tuning fork
to produce 5.00 beats per second. What
are the two possible frequencies for the
second tuning fork?
Bf = 5.00 beats per second Bf = f1 – f2
f1 = 300 Hz
f2 = f1 ± Bf
= 300 Hz ± 5.00
beats per second
= 305 Hz or 295 Hz
Example 4
Some plasticine is placed on the
300 Hz tuning fork. Now only 3 beats
are heard. What is the actual frequency
of second tuning fork?
The plasticine decreases frequency.
Since only fewer beats are
heard, the frequency of the tuning fork
is 295 Hz.
Example
A 200 Hz tuning fork is sounded with
another tuning fork and 7 beats are
heard. If some plasticine is placed on the
200 Hz tuning fork and 9 beats are
heard, what is the actual frequency
of the second tuning fork?
Your Assignment:
34-38,40
Vibrating Strings
1. 1st harmonic (Fundamental
frequency) = lowest frequency
making up a sound. The string
vibrates as a whole segment
2. 2nd harmonic (1st overtone) =
string vibrating in two segments
3. 3rd harmonic (2nd overtone) =
string vibrating in three segments
Mode of
# of frequency wavelength
Vibration loops
1st harmonic
1
f
½λ
(fundamental
frequency)
2nd harmonic
(1st
overtone)
3rd harmonic
(2nd
overtone)
2
2f
1λ
3
3f
1½λ
Example
Find the frequency of the second
overtone if the fundamental frequency of
the vibrating string is 150 Hz.
f = 150 Hz
2nd overtone = 3f
= 3(150 Hz)
= 450 Hz
Factors Affecting the Frequency of
Vibrating Strings
1. Tension of String (T):
• frequency is directly proportional to the
square root of the tension
f1 = √T1
f2 √T2
Equation #40
where: f1, f2 = frequencies in Hz
T1, T2 = tensions in N
Example
The frequency of a string is 200 Hz.
What is the frequency if the tension is
tripled?
f1 = 200 Hz
f1 = √T1
T1 = 1
f2 √T2
T2 = 3
= (200 Hz)(√3)
√1
= 346 Hz
2. Length of String (l)
• frequency is inversely proportional to
the length of the string.
fα1/l
•
f1 = l2
f2 l1
Equation #39
where: f1, f2 = frequencies in Hz
l1, l2 = lengths in m
Example
The frequency of a 2.0 m long vibrating
string is 400 Hz. If the length is
increased to 2.5 m, what is the new
frequency?
f1 = 400 Hz
f1 = l2
l1 = 2.0 m
f2 l1
l2 = 2.5 m
f2 = f1l1
l2
= (400 Hz)(2.0 m) =
2.5 m
=3.2x102 Hz
3. Diameter of String (D)
• frequency is inversely proportional to
the diameter of the string
f1 = D2
f 2 D1
Equation #41
where: f1, f2 = frequencies in Hz
D1, D2 = diameters in m
4. Density of String (d)
• frequency is inversely proportional to
the square root of the density
f1 = √d2
f2
√d1
where: f1, f2 = frequencies in Hz
d1, d2 = densities(kg/m3)
Resonance
• two objects are said to be in resonance
if they vibrate at the same frequency
eg) cordless phones, remote and TV,
garage door openers
• all objects have a natural frequency
• when an object is affected by a wave
of the same frequency (sound, wind
etc.), resonance causes an increase in
the amplitude of the wave ∴increase
in amount of energy eg) louder sound,
higher swing
• Soldiers are asked to “break step”
when they march on bridges.
The bridge will otherwise reach
its natural frequency.
Characteristics of Musical Sound
• pitch: depends on frequency ie)
↑frequency = ↑pitch
• loudness: depends on amplitude
(energy) ie) ↑amplitude = ↑loudness
• quality of sound = depends on
overtone
Vibrating Air Columns
• if a tuning fork is held above a column
of air of a particular length, the column
will vibrate at the same frequency as
the fork ie) resonance occurs
• the lengths of the air columns where
resonance occurs are called the
resonant lengths
• changing the length changes the pitch
(frequency)
1. Closed Pipe Resonator
• a cylindrical tube of air with one end
closed and a sound source at the other
end
• a standing wave is set up with nodes
and antinodes
• the shortest resonant length is ¼ λ
Equation #33
l1=1/4λ
where: l1 = first resonant length in m
λ = wavelength in m
• each additional resonant length is
½ λ further down the air column
1st resonant length:
l1 = λ
4
2nd resonant length: l2 = 3λ
4
3rd resonant length: l3 = 5λ
4
Example 1
If the shortest resonant length for a
closed pipe resonator is 16.5 cm when a
500 Hz tuning fork is used, what is the
speed of sound?
f = 500 Hz
λ = 4l
l1 = 16.5 cm
= (4)(0.165 m)
= 0.165 m
= 0.660 m
v = fλ
= (500 Hz)(0.660 m)
= 330 m/s
Example 2
If the first resonant length in a closed
air column is 18.5 cm
when sounded with a 480 Hz
tuning fork, what is the speed of
sound?
v=355 m/s.
Example 3
The shortest resonant length for a 470 Hz
tuning fork over a closed pipe resonator
is 18 cm. Find the speed of sound.
f = 470 Hz
λ = 4l
l1 = 18 cm
= (4)(0.18 m)
= 0.18 m
= 0.72 m
v = fλ
= (470 Hz)(0.72 m)
= 338.4 m/s
= 3.4 x 102 m/s
Example 3
Find the first three resonant lengths of a
closed air column that will resonate with
a 300 Hz tuning fork at 20°C
v = 332 m/s + (0.60 m/s x 20)
= 344 m/s
f
f = 300 Hz
v=fλ
344=300λ
λ = 1.14666m
l1=1/4λ=1/4x1.1466=0.29 m
l2=3/4λ=3/4x1.146=0.86 m
l3=5/4λ=5/4x1.1466=1.4 m
2. Open Pipe Resonator
• cylindrical column of air open at both
ends and a sound source at one end
• a standing wave is set up with nodes
and antinodes
• the shortest resonant length is ½ λ
l1 = λ
2
where: l1 = first resonant length in m
λ = wavelength in m
• each additional resonant length is
½ λ further down the air column
1st resonant length:
l1 = λ
2
2nd resonant length: l2 = 2λ
2
3rd resonant length: l3 = 3λ
2
Example 1
Find the shortest length of an open tube
air column that will resonate with a 400
Hz tuning fork at 15°C.
v = 332 m/s + (0.60 m/s x 15)
λ=v
= 341 m/s
f
f = 400 Hz
= 341 m/s
400 Hz
= 0.8525 m
1. l1 = λ = 0.8525 m = 0.42625 m =
2
2
0.43m
Simple Harmonic Motion
• simple harmonic motion is motion
where there is a restoring force that
varies linearly with displacement
• objects undergo a change in velocity,
acceleration, and force
equilibrium
position
Equilibrium:
mass at rest
F = max
a = max
v=0
F=0
a=0
v = max
Inertia makes the
mass continue to
move up
F = max
a = max
v=0
Spring now
compressed then
gravity pulls it
down
• to calculate the period of a mass
suspended on a spring:
T = 2π√m/k
where: T = period in s
m = mass in kg
k = spring constant in N/m
π = 3.14
Example
Calculate the period for a spring with a
force constant of 15 N/m if the mass
suspended is 1.0 kg.
k = 15 N/m
m = 1.0 kg
π = 3.14
F=0
a=0
v = max
F = max
a = max
v=0
= (2)(3.14) (1.0 kg)
√ (15 N/m)
= 1.621489028 s
= 1.6 s
= max
a = max
v=0
F
F=0
a=0
v = max
rest position
Assignment
1.A 0.23 kg object vibrates at the end of
a horizontal spring (k=32 N/m) along
a frictionless surface. What is the
period of the vibration?
2. An object vibrates at the end of a
horizontal spring(k=115 N/m) along a
frictionless surface. If the frequency
of vibration is 1.50 Hz, what is the
mass of the object?