AP Physics B
Giancoli 11 & 12
Waves and Sound
Assignments
• Reading: 11.1-4,7-9,11-13 and 12.1,2,4-7
• Problems: Waves: 11.42,43,55,56
•
Sound: 12.4,5,10,11
•
Strings/AirColumns12.29,30,35
•
Interference 12.42,43
•
Doppler 12.51,52
•
SHM: 11.4,5,20, 25
•
Pendulum: 11.31,32
Preview
• What are the two categories of waves with
regard to mode of travel?
– Mechanical
– Electromagnetic
• Which type of wave requires a medium?
– Mechanical
• An example of a mechanical wave?
– Sound
Velocity of a Wave
• The speed of a wave is the distance traveled by
a given point on the wave (such as a crest) in a
given internal of time.
• v = d/t
d: distance (m)
t: time (s)
• v=fl
v: speed (m/s)
l : wavelength (m)
f : frequency (s-1, Hz)
Period of a Wave
• T = 1/f
• T : Period = (s)
• F : frequency (s-1, Hz)
Problem: Sound travels at approximately
340 m/s, and light travels at 3.0 x 108 m/s.
How far away is a lightning strike if the sound
of the thunder arrives at a location 5.0
seconds after the lightning is seen?
Light travels almost instantaneously from
strike location to the observer.
The sound travels much more slowly:
d = vs t = (340 m/s)(5.0 s) = 170m
Problem: The frequency of a C key on the
piano is 262 Hz. What is the period of this
note? What is the wavelength? Assume speed
of sound in air to be 340 m/s at 20 oC.
T = 1/f = 1/262 s-1 = 0.00382 s
V=fl
l = v/f
l = 340 m/s / 262 /s = 1.30 m
Problem
• A sound wave traveling through water has
a frequency of 500 Hz and a wavelength
of 3 m. How fast does sound travel
through water?
• v = l f = 3m (500 Hz) = 1500 m/s
Wave on a Wire
v=
FT
m/L
v, velocity, m/s
FT, tension on a wire, N
m/L mass/unit length, kg/m
m/L may be shown as m
Problem Ex. 11-11
A wave whose wavelength is 0.30 m is traveling down a
300 m long wire whose total mass is 15 kg. If the tension
of the wire is 1000N, what are the speed and frequency
of the wave?
Using equation on prior slide:
v = √[( 1000N) / (15kg)(300m)]
= 140m/s
f = v / l = 140 m/s / 0.30 m = 470 Hz
Types of Waves
• A transverse wave is a wave in which particles
of the medium move in a direction
perpendicular to the direction which the wave
moves.
– Example: Waves on a guitar string
• A longitudinal wave is a wave in which
particles of the medium move in a direction
parallel to the direction which the wave moves.
These are also called compression waves.
– Example: Sound
– http://einstein.byu.edu/~masong/HTMstuff/WaveTrans
.html
What are two types of wave
shapes?
• Transverse
• Longitudinal
•
http://www.school-for-champions.com/science/sound.htm
Transverse Wave Type
Longitudinal Wave Type
Longitudinal vs Transverse
Other Waves Types Occurring in
Nature
•
•
•
•
•
Light: electromagnetic
Ocean waves: surface
Earthquakes: combination
Wave demos:
http://www.kettering.edu/~drussell/Demos/
waves/wavemotion.html
• http://www.kettering.edu/~drussell/Demos/
doppler/mach1.html
Properties of Waves
• Reflection occurs when a wave strikes a
medium boundary and “bounces back” into
the original medium.
• Those waves completely reflected have
the same energy and speed as the original
wave.
Types of Reflection
Fixed-end ReflectionThe wave reflects with
inverted phase.
Open-end ReflectionThe wave reflects with
The same phase.
www.iop.org/activity/education/Teaching_Resources
Refraction of Waves
 Wave is transmitted
from one medium to
another.
 Refracted waves may
change speed and
Wavelength
 Almost always is accompanied
by some reflection.
 Refracted waves do not
change frequency.
Sound - a longitudinal wave
• Sound travels through air about 340 m/s.
• Sound travels through other media as well,
often much faster than 340 m/s.
• Sound waves are started by vibration of
some other material, which starts the air
vibrating.
•
www.silcom.com/~aludwig/musicand.htm
Hearing Sounds
• We hear a sound as “high” or “low” pitch depending on
the frequency or wavelength. High-pitched sounds have
short wavelengths and high frequencies. Low-pitched
sounds have long wavelengths and low frequencies.
Humans hear from about 20 Hz to about 20,000 Hz.
• The amplitude of a sound’s vibration is interpreted as its
loudness. We measure loudness
(also known as sound intensity)
on the decibel scale, which is
logarithmic.
http://www.allegropianoworks.com/assets/rare_co
mpress.jpg
Doppler Effect
• The Doppler Effect is the apparent change in pitch of a
sound as a result of the relative motion of an observer
and the source of a sound. Coming toward you a car
horn appears higher pitched because the wavelength
has been effectively decreased by the motion of the car
relative to you. The opposite occurs when you are
behind the car.
http://people.finearts.uvic.ca/~aschloss/course_mat/MU207/images/Image2.gif
Pure Sound
• Sounds are longitudinal waves, but they
can be shown to look like transverse
waves.
• When air motion is graphed in a pure
sound tone versus position, we get what
looks like a sine or cosine function.
• A tuning fork produces a relatively pure
tone as does a human whistle.
Graphing a Sound Wave
Complex Sounds
• Because of superposition and
interference, real world waveforms may
not appear to be pure sine or cosine
functions.
• This is because most real world sounds
are composed of multiple frequencies.
• The human voice and most musical
instruments are examples.
The Oscilloscope
• With an Oscilloscope we can view waveforms. Pure tones will
resemble sine or cosine functions, and complex tones will show
other repeating patterns that are formed from multiple sine and
cosine functions added together. (Amplitude vs time.)
The Fourier Transform
•
The Fourier transform has long been used for
characterizing linear systems and for identifying the
frequency components making up a continuous
waveform. This mathematical technique separates a
complex waveform into its component frequencies.
•
The Fourier Transform´s ability to represent timedomain data in the frequency domain and viceversa has many applications. One of the most
frequent applications is analysing the spectral
(frequency) energy contained in data that has been
sampled at evenly-spaced time intervals. Other
applications include fast computation of
convolution (linear systems responses, digital
filtering, correlation (time-delay estimation,
similarity measurements) and time-frequency
analysis.
Fourier Transform - showing “time
domain” and “frequency domain”.
Superposition Principle
• When two or more waves pass a particular
point in a medium simultaneously, the
resulting displacement of the medium at
that point is the sum of the displacements
due to each individual wave.
• The waves are said to interfere with each
other.
Superposition of Waves
• When two or more waves meet, the
displacement at any point of the medium is
equal to the algebraic sum of the
displacements due to the individual waves.
Types of Interference
• If the waves are in phase, when crests and
troughs are aligned, the amplitude in
increased and this is called constructive
interference.
• If the waves are “out of phase”, when
crests and troughs are completely
misaligned, the amplitude is decreased
and can even be zero. This is called
destructive interference.
Constructive Interference
Crests are
Aligned 
the waves are
“in phase”
Destructive Interference
Crests are
aligned with
troughs 
Waves are
“out of
phase”
Constructive & Destructive
Interference
Interference Problem: Draw the waveform
from the two components shown below.
Standing Waves
• A standing wave is one which is reflected
back and forth between fixes ends of a
string or pipe.
• Reflection may be fixed or open-ended.
• Superposition of the wave upon itself
results in a pattern of constructive and
destructive interference and an enhanced
wave. Let’s see a simulation
• http://www.5min.com/Video/The-Rubens-TubeFrequency-of-Fire-1858291
Fixed-end standing waves - guitar
or violin string
•
Fundamental
•
•
1st harmonic
l = 2L
•
First overtone
•
•
2nd harmonic
l=L
•
Second Overtone
•
•
3rd harmonic
l = 2L/3
http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html
Problem
• A string of length 12 m that’s fixed at both ends
supports a standing wave with a total of 5 nodes.
What are the harmonic number and wavelength
of this standing wave?
• L = 4(1/2 l )  l = 2L/4 4th harmonic since it
matches ln = 2L/n for n = 4
• wavelength: l4 = 2(12m) / 4 = 6 m
Open-ended standing waves flute & clarinet
l= 4L
l = 2L
l = (4/3)L
l=L
l = (4/5L
l = (2/3)L
physics.indiana.edu/~p105_f02/standing_waves_...
http://upload.wikimedia.org/wikiboo
ks/en/3/32/Fhsst_waves40.png
•
open ends one end
–
closed
both ends
closed
Sample Problem
• 12-30. a) Determine the length of an organ pipe that
emits middle C (262Hz). The air temp. is 21oC.
• A) v = 331m/s + 0.6 m/soC(21oC) = 344m/s
• A) l = 2L v = fl = 2lf
L = v/2f = 344m/s/{2(262/s)]
•
L = 0.656m
• B) What are the wavelength and frequency of the 1st
harmonic?
• Frequency is 262 Hz
• Wavelength is twice the length of the pipe, 1.31 m.
• C) What is the wavelength and frequency in the traveling
sound wave produced in the outside air?
• They are the same because it is air that is resonating in
the organ pipe: 262Hz and 1.31 m
Superposition of 2 sound waves
http://www.ece.utexas.edu/~nodog/me379m/superposition.html
Resonance and Beats
• Resonance occurs when a vibration from
one oscillator occurs at a natural
frequency for another oscillator.
• The first oscillator will cause the second to
vibrate.
• See next slide.
Resonance
•
http://www.isd-dc.org/ISDWash/GIFS%20Pictures%20&%20Whatnots/tuningforkresonance.jpg
Beats
• The word physicists use to describe the
characteristic loud/soft pattern that
characterizes two nearly matched
frequencies.
• Musicians call this “being out of tune”.
Beats
• When two sound waves whose
frequencies are close but not exactly the
same, the resulting sound modulates in
amplitude changing from loud to soft to
loud. This is called beat frequency and is
shown by:
•
fbeat =
f1-f2
Diffraction
• Bending of a wave around a barrier
• Diffraction of waves combined with
interference of the diffracte waves causes
“diffraction patterns”.
• Here is an example using a “ripple tank”.
• http://www.falstad.com/ripple/
Double-slit or multi-slit diffraction
•
•
micro.magnet.fsu.edu/.../doubleslit/
Remove frame
Single Slit Diffraction
• n l = s sin q
 n -- dark band number
 l -- wavelength (m)
 s -- slit width (m)
 q -- angle defined by central band, slit, and
dark bank
Sample Problem
• Light of wavelength 360 nm is passed through a
diffraction grating that has 10,000 slits per cm.
If the screen is 2.0 m from the grating, how far
from the central bright band is the first order
bright band?
Sample Problem
• Light of wavelength 560 nm is passed through two
slits. It is found that, on a screen 1.0 m from the slits,
a bright spot is formed at x = 0, and another is
formed at x = 0.03m. What is the spacing between
the slits?
Sample Problem
• Light is passed through a single slit of width 2.1 x 10-6
m. How far from the central bright band do the 1st and
2nd order dark bands appear if the screen is 3.0 m away
from the slit?
Mathematical Description of a
Traveling Wave
• Y = A sin (v t + k x )
• Y dependent of x and t; y(x,t) or “y of x & t”
• If the - sign is used, wave is traveling in +x
direction 
• A is amplitude of the wave
 m (omega) is angular frequency (m = 2pf)
 k angular wave # (k = 2 pk, k = 1/l)
Other forms
• Important features of the wave: amplitude,
frequency f (through v), period T (which is
1/f = 2p/v), wavelength (l = 2p/k) and wave
speed v (which is lf = v/k)
• y = Asin2p[ft + (1/l)x] or
• y = Asin(2p/l)(vt + x)
Sample Problem:
The vertical position y of any point on a rope that supports a transverse wave traveling
horizontally is given by the equation
y = 0.1 sin (6 p t + 8 p x)
Find:
amplitude: 0.1
angular frequency: v = 6 p s -1
frequency: f = v / (2 p) = (6 p s -1 )/ (2 p ) = 3 Hz
angular wave number: k = k / (2 p ) = 8 p m -1) / ((2 p) =
4 m-1
wavelength: l = 1/ k = 1/ (4 m-1) = 0.25 m
period: T = 1/f = 1 / 3 Hz = 0.33s
wave speed: v = f l = 0.25m (3 Hz) = 0.75 m/s
Assignment
• P15/MC1-4
Sound Level
• Intensity: Rate at which sound waves
transmit energy is measured in energy per
unit area: watt/m2 or watt/cm 2
• Intensity level or loudness level, B
• B = 10log I/Io where Io = 1x10-12 w/m2
•
or 1x10-16 w/cm2
More math…
b = 10 log I
Io
10
-16
w/cm
2
Intensity level = 10 log Intensity / threshhold of
hearing
We all don’t hear the same, so this is a
comparative measurement in decibels
Flow chart for b problem
• If I = 4.7 x 10^-10 w/cm^2:
• 10xlog(4.7 2nd EE -10 / 1 2nd EE -16) =
66.7 dB
• If I = 2.9 x 10^-3 w/cm^2:
• 10xlog(2.9 2nd EE -3 / 1 2nd EE -16) =
135 dB
Problem
• Now we are going backwards from intensity level
(dB) to intensity (w/cm2)
• If the intensity level is 83 dB, convert that
to intensity in w/cm2.
• B = 10 log I / Io
get to a working eqtn:
• B /10 = log I / Io
• Log-1(B/10) = Log-1(log I/Io)
• Log-1(B/10) = I/Io
• Let’s say that the intensity level of a sound
is 25.3 dB. What is the intensity of the
sound in w/cm2?
• B = 10 log I / Io
• 25.3 dB = 10 log (I/10-16 w/cm2)
• 2.53 = log I – log 10-16
• 2.50 + log 10-16 = log I
• 2.50 – 16 = log I
• 25.3 dB = 10 log (I/10-16 w/cm2)
• 2.53 = log I – log 10-16
• 2.50 + log 10-16 = log I  what power do you
raise 10 to, to get 10-16?
• 2.50 – 16 = log I
• Adding on the left 
-13.5 = log I
• Raise 10 to the -13.5 power by this sequence:
• 2nd 10x (-13.5)  3.16 x 10-14 w/cm2 = I
Doppler Effect
• Doppler Effect is the apparent change in frequency as a
result of relative motion between the source of a sound
and an observer.
•
•
•
•
•
•
f’ frequency heard by observer
f frequency of source
v velocity of sound in air
vd velocity of detector
vs velocity of source
Sample Problem
• A source of 4 kHz sound waves travel at
1/9 the speed of sound toward a detector
that’s moving at 1/9 the speed of sound,
toward the source.
– a. what is the frequency of the waves as
they’re received by the detector?
– b. how does the wavelength of the detected
waves compare to the wavelength of the
emitted waves?
= v + 1/9 v x f = 5/4 f = 5/4 (4 kHz) = 5 kHz
v - 1/9 v
+ sign on top as detector moves toward source
-sign on bottom as source moves toward det.
-Frequency is shifted up by a factor of 5/4, the l will
shift down by the same factor.
ld = ls / 5/4 = 4/5 l s
Sample
• A person yells, emitting a constant
frequency of 200 Hz, as he runs at 5m/s
toward a stationary brick wall. When the
reflected waves reach the person, how
many beats per second will he hear? (Use
343 m/s for the speed of sound.)
Determine what f will be reflected and heard by the runner. person is source and
wall is detector
fwall = v
f
v - vrunner
Reflected sound wave (no change in f): wall is the source and runner is the
detector.
frunner
= v + v runner
f
v
Combine these two formulae: frunner = v + vrunner f
v - vrunner
f = (343+5)m/s (200Hz) = 206 Hz
(343-5)m/s
Beat frequency is fbeat = 206Hz - 200 Hz = 6 Hz.
p. 343 Princeton
Doppler Effect for Light
•
wps.prenhall.com/.../ch25_SWA/images/8.gif
•
•
c = 3 x 10 8 m/s
u = speed of the source and the detector.
www.sv.vt.edu/.../class95/physics/doppler.gif
Periodic Motion …
• Motion that repeats itself over a fixed and
reproducible period of time
• An example is a planet moving about its
sun. This is called the period (T) or year
of the planet
• Mechanical devices on earth that have
periodic motion are useful timers and are
called oscillators.
Simple Harmonic Motion
• Attach a weight to a spring, stretch the spring
past its equilibrium point and release. The
weight bobs up and down with a reproducible
period, T
• Plot position vs time to get a graph that
resembles a sine or cosine function. The graph
is “sinusoidal”, so the motion is referred to as
simple harmonic motion.
• Springs and pendulums undergo simple
harmonic motion and are referred to as simple
harmonic oscillators.
Graph Analysis
Graph Analysis
Oscillator Definitions
• Amplitude
– Maximum displacement from equilibrium
– Related to energy
• Period
– Length of time required for one oscillation
• Frequency
– How fast the oscillator is oscillating
– f = 1/T in Hz or s-1
Problem 11-5
• An elastic cord vibrates with a frequency
of 3.0 Hz when a mass of 0.60 kg is hung
from it. What is its frequency if only 0.38
kg hangs from it?

)
 2.62kgkg) 9.80m
)  9.80ms s )
F F mgmg  2.62
k k= = = = = =
= 81.5
N Nm m
= 81.5
xx xx
0.315
mm
0.315
2 2
Springs
• Springs are a common type of simple
harmonic oscillator.
• Our springs are “ideal springs”, which
means:
They are massless
They are both compressible and extensible.
• They will follow Hooke’s Law.
F = -kx
Hooke’s Law Review
The force constant
of a spring can be
determined by
attaching a weight
and seeing how far
it stretches.
Period of a Spring
T = period (s)
m = mass (kg)
k = force constant (N/m)
Sample Problem
• Calculate the period of a 350 g mass
attached to an ideal spr5ing with a force
constant of 35 N/m.
Sample Problem
• A 300 g mass attached to a spring undergoes
simple harmonic motion with a frequency of 25 Hz.
What is the force constant of the spring?
Sample Problem
• An 80 g mass attached to a spring hung vertically
causes it to stretch 30 cm from its unstretched
position. If the mass is set into oscillation on the
end of the spring, what is its period?
Combinations of Springs
• In parallel, springs work together
• In series, springs work independently
What do you think?
Does this combination of springs act
as parallel or series?
Answer: parallel !
Problem
• If you want to double the force constant of
a spring, you
• A. double its length by connecting it to
another one just like it.
• B. cut it in half.
• C. add twice as much mass.
• D. take half of the mass off.
• Answer: B
Conservation of Energy
• Springs and pendulums obey conservation
of energy
• The equilibrium position has high kinetic
energy and low potential energy.
• The positions of maximum displacement
have high potential energy and low kinetic
energy.
• Total energy of the oscillating system is
constant.
Problem 11-20
A block of mass m is supported
by 2 identical parallel vertical
springs, each with spring stiffness
constant k. What will be the
frequency of vibration?
 Fy = FA  FB  mg = 0  FA  FB = mg
FA
FB
FA
FB
x
mg
mg
F
y
= Fnet = FA  FB  mg = FA  kx  FB  kx  mg = 2kx   FA  FB  mg ) = 2kx
General form of a restoring force producing SHM with spring
constant of 2k
Sample Problem
Spring Problem
Another spring problem
Pendulums
• Pendulums can be thought of as simple
harmonic oscillators.
• The displacement needs to be small for it
to work properly.
Conservation of Energy
• Pendulums also obey conservation of
energy.
• The equilibrium position has high
kinetic energy and low potential
energy.
• The positions of maximum
displacement have high potential
energy and low kinetic energy.
• Total energy of the oscillating system
is constant.
Pendulum Forces
2005 Pearson Prentice Hall Fig 11-12
Period of a Pendulum
T -- period (s)
l -- length of string (m)
g -- gravitational acceleration (m/s2)
Problem:
• Predict the period of a pendulum consisting of a
500 g mass attached to a 2.5 m long string.
Problem:
• Suppose you notice that a 5 kg weight fixed to a
string swings back and forth 5 times in 20
seconds. How long is the string?
Last problem!
• The period of a pendulum is observed to be T.
Suppose you want to make the period 2T. What
do you do to the pendulum?