End of Thermodynamics
Waves
Review of Oscillations and SHM
Introducing Waves
Lana Sheridan
De Anza College
May 15, 2017
Last time
• entropy
• heat engines
Overview
• wrap up thermodynamics
• oscillations
• simple harmonic motion (SHM)
• spring systems
• energy in SHM
• pendula
• introducing waves
• kinds of waves
Carnot’s Theorem
Carnot’s Theorem
No real heat engine operating between two energy reservoirs can
be more efficient than a Carnot engine operating between the same
two reservoirs.
But how efficient is a Carnot engine?
and the PV dia
of two adiabati
The Carnot Cycle
P
A
The work done
during the cycle
equals the area
enclosed by the path
on the PV diagram.
Qh
B
Weng
D
Figure 22.11
Th
Qc
C
Tc
V
PV diagram for the
1. Process
The gas
ture Th .
ervoir th
piston.
2. In proce
thermal
energy e
peratur
raising t
3. In proce
energy r
peratur
and the
Efficiency of a Carnot Engine
First, we can relate the volumes at different parts of the cycle.
In the first adiabatic process:
Th VBγ−1 = Tc VCγ−1
In the second adiabatic process:
Th VAγ−1 = Tc VDγ−1
Taking a ratio, then the γ − 1 root:
VB
VC
=
VA
VD
Efficiency of a Carnot Engine
First law: ∆Eint = Q + W = 0 gives for the first isothermal process
VB
|Qh | = nRTh ln
VA
Second isothermal process:
|Qc | = nRTc ln
VC
VD
We will take a ratio of these to find the efficiency. Noting that
VB
VC
VA = VD :
|Qc |
|Tc |
=
|Qh |
|Th |
Efficiency of a Carnot Engine
Recall, efficiency of a heat engine:
e =1−
|Qc |
|Qh |
Efficiency of a Carnot engine:
e=
Th − Tc
Tc
=1−
Th
Th
(T is measured in Kelvin!)
This is the most efficient that any heat engine operating between
two reservoirs at constant temperatures can be.
Third Law of Thermodynamics
3rd Law
As the temperature of a material approaches zero, the entropy
approaches a constant value.
The constant value the entropy takes is very small. It is actually
zero if the lowest energy state of the material is unique.
Another way to express the third law:
3rd Law - alternate
It is impossible to reach absolute zero using any procedure and
only a finite number of steps.
Entropy in the Carnot Cycle
Since the working fluid returns to its initial state along reversible
paths, the change in the entropy for the whole cycle is
Entropy in the Carnot Cycle
Since the working fluid returns to its initial state along reversible
paths, the change in the entropy for the whole cycle is ∆S = 0.
Entropy in the Carnot Cycle
Since the working fluid returns to its initial state along reversible
paths, the change in the entropy for the whole cycle is ∆S = 0.
We can see this from an analysis also:
Z
1
∆S =
dQr
T
In the reversible adiabatic processes ∆S = 0.
In the reversible isothermal portions, T is constant so ∆S =
For the cycle
∆S =
|Qh | |Qc |
−
Th
Tc
Q
T.
Entropy in the Carnot Cycle
For the cycle
∆S =
We just found that
So
|Qh | |Qc |
−
Th
Tc
Th
|Qh |
=
|Qc |
Tc
|Qh |
|Qc |
=
Th
Tc
And for the cycle
∆S = 0
Entropy in the Carnot Cycle
We can represent the Carnot Cycle on a TS diagram:
QH
Temperature T
a
b
d
c
QL
Entropy S
Fig. 20-10
The Carnot cycle of
TH
TL
it is
is do
area
Figs.
quan
perf
invo
Carn
tere
gram
isoth
cycle
stan
ing t
Heat Engine question
Consider and ocean thermal energy conversion (OTEC) power
plant that operates on a temperature difference between deep 4◦ C
water and 25◦ C surface water. Show that the Carnot (ideal)
efficiency of this plant would be about 7%.
0
Hewitt, page 331, problem 2.
In a gasoline engine, six processes occur in each cycle; they are illustrated in Figure
In this discussion, let’s consider the interior of the cylinder above the piston
Car22.12.
Engines
to be the system that is taken through repeated cycles in the engine’s operation. For
a given cycle, the piston moves up and down twice, which represents a four-stroke
cycle consisting of two upstrokes and two downstrokes. The processes in the cycle
can be approximated by the Otto cycle shown in the PV diagram in Figure 22.13
(page 666). In the following discussion, refer to Figure 22.12 for the pictorial representation of the strokes and Figure 22.13 for the significance on the PV diagram of
the letter designations below:
Car engines work by burning fuel in cylinders with pistons.
The four stroke cycle:
1. During the intake stroke (Fig. 22.12a and O S A in Figure 22.13), the piston
moves downward and a gaseous mixture of air and fuel is drawn into the
The intake valve
opens, and the air–
fuel mixture enters
as the piston moves
down.
The piston moves
up and compresses
the mixture.
The spark plug
fires and ignites
the mixture.
The hot gas
pushes the piston
downward.
The exhaust valve
opens, and the
residual gas escapes.
The piston moves
up and pushes the
remaining gas out.
Spark plug
Air
and
fuel
Exhaust
Piston
Intake
Compression
Spark
Power
Release
Exhaust
a
b
c
d
e
f
Figure 22.12
The four-stroke cycle of a conventional gasoline engine. The arrows on the piston
indicate the direction of its motion during each process.
Car Engines
The intake valve
opens, and the air–
fuel mixture enters
as the piston moves
down.
Spark plug
Air
and
fuel
Piston
Intake
Car Engines
The piston moves
up and compresses
the mixture.
Spark plug
Piston
Compression
b
es downward and a gaseous mixture of air and fuel is drawn into the
Car Engines
valve
the air–
e enters
n moves
The piston moves
up and compresses
the mixture.
The spark plug
fires and ignites
the mixture.
The hot gas
pushes the piston
downward.
The exha
opens, an
residual g
Spark plug
ke
Piston
Compression
Spark
Power
Relea
a gaseous mixture of air and fuel is drawn into the
Car Engines
on moves
compresses
ture.
ession
The spark plug
fires and ignites
the mixture.
Spark
The hot gas
pushes the piston
downward.
Power
The exhaust valve
opens, and the
residual gas escapes.
Release
The pis
up and
remain
Exha
re of air and fuel is drawn into the
Car Engines
spark plug
s and ignites
mixture.
The hot gas
pushes the piston
downward.
The exhaust valve
opens, and the
residual gas escapes.
The piston moves
up and pushes the
remaining gas out.
Exhaust
Spark
Power
Release
Exhaust
c
d
e
f
22.13), the piston
drawn into the
Car Engines
ot gas
s the piston
ward.
The exhaust valve
opens, and the
residual gas escapes.
The piston moves
up and pushes the
remaining gas out.
Exhaust
Power
Release
Exhaust
d
e
f
Car Engines and the Otto Cycle
666
Chapter 22 Heat Engines, Entro
The Otto cycle approximates the real 4-stroke cycle we just
discussed.
P
T A TC
C
Adiabatic
processes
Qh
B
D
O
A
V2
V1
Qc
V
Figure 22.13 PV diagram for
cylind
energ
as pot
from
sion s
from
2. Durin
ton m
volum
work
area u
3. Comb
22.13)
Car Engines and the Otto Cycle
The efficiency of the Otto cycle is
e =1−
1
(V1 /V2 )(γ−1)
(See example 22.5 for a proof of this expression.)
A typical value for the volume compression is V1 /V2 = 8, which
would give an efficiency of 56%.
Real efficiencies of car engines are much less than this, ∼ 20%.
There is heat loss, work lost overcoming friction, and imperfect
combustion.
22.12. In this discussion, let’s consider the interior of the cylinder above the piston
to be the system that is taken through repeated cycles in the engine’s operation. For
a given cycle, the piston moves up and down twice, which represents a four-stroke
cycle consisting of two upstrokes and two downstrokes. The processes in the cycle
can be approximated by the Otto cycle shown in the PV diagram in Figure 22.13
(page 666). In the following discussion, refer to Figure 22.12 for the pictorial representation of the strokes and Figure 22.13 for the significance on the PV diagram of
the letter designations below:
Car Engines
The four stroke cycle:
1. During the intake stroke (Fig. 22.12a and O S A in Figure 22.13), the piston
moves downward and a gaseous mixture of air and fuel is drawn into the
The intake valve
opens, and the air–
fuel mixture enters
as the piston moves
down.
The piston moves
up and compresses
the mixture.
The spark plug
fires and ignites
the mixture.
The hot gas
pushes the piston
downward.
The exhaust valve
opens, and the
residual gas escapes.
The piston moves
up and pushes the
remaining gas out.
Spark plug
Air
and
fuel
Exhaust
Piston
Intake
Compression
Spark
Power
Release
Exhaust
a
b
c
d
e
f
Figure 22.12
The four-stroke cycle of a conventional gasoline engine. The arrows on the piston
indicate the direction of its motion during each process.
Jet Engines
Jet engines are even simpler and more efficient than car engines.
However, they require more advanced materials...
1
Turbofan schematic from Wikipedia by K. Aainsqatsi.
Advanced Jets: Scramjet
...and higher speeds of operation.
1
Scramjet schematic from Wikipedia by User:Emoscopes.
Transition
End of Thermodynamics material.
The test will be on the topics in Chapters 19-22 of the textbook.
Beginning of Waves.
We start with a quick review of oscillation, Chapter 15.
Oscillations and Periodic Motion
Many physical systems exhibit cycles of repetitive behavior.
After some time, they return to their initial configuration.
Examples:
• clocks
• rolling wheels
• a pendulum
• bobs on springs
Oscillations
oscillation
motion that repeats over a period of time
amplitude
the magnitude of the vibration; how far does the object move from
its average (equilibrium) position.
period, T
the time for one complete oscillation.
After 1 period, the motion repeats itself.
Oscillations
frequency
The number of complete oscillations in some amount of time.
Usually, oscillations per second.
f =
1
T
Units of frequency: Hertz. 1 Hz = 1 s−1
If one oscillation takes a quarter of a second (0.25 s), then there
are 4 oscillations per second. The frequency is 4 s−1 = 4 Hz.
Period and frequency question
What is the period, in seconds, that corresponds to each of the
following frequencies?
1
10 Hz
2
0.2 Hz
3
60 Hz
1
Hewitt, page 350, Ch 18, problem 2.
Simple Harmonic Motion
The oscillations of bobs on springs and pendula are very regular
and simple to describe.
It is called simple harmonic motion.
simple harmonic motion (SHM)
any motion in which the acceleration is proportional to the
displacement from equilibrium, but opposite in direction
The force causing the acceleration is called the “restoring force”.
SHM and Springs
If a mass is attached to a spring, the force on the mass depends on
its displacement from the spring’s natural length.
Hooke’s Law:
F = −kx
where k is the spring constant and x is the displacement (position)
of the mass.
Hooke’s law gives the force on the bob ⇒ SHM.
The spring force is the restoring force.
SHM and Springs
How can we find an equation of motion for the block?
Newton’s second law:
Fnet = Fs = ma
SHM and Springs
How can we find an equation of motion for the block?
Newton’s second law:
Fnet = Fs = ma
Using the definition of acceleration: a =
d2 x
k
2 = −mx
dt
Define
r
ω=
k
m
and we can write this equation as:
d2 x
= −ω2 x
dt2
d2 x
dt2
SHM and Springs
To solve:
d2 x
2
2 = −ω x
dt
notice that it is a second order linear differential equation.
We can actually find the solutions just be inspection.
A solution x(t) to this equation has the property that if we take its
derivative twice, we get the same form of the function back again,
but with an additional factor of −ω2 .
SHM and Springs
To solve:
d2 x
2
2 = −ω x
dt
notice that it is a second order linear differential equation.
We can actually find the solutions just be inspection.
A solution x(t) to this equation has the property that if we take its
derivative twice, we get the same form of the function back again,
but with an additional factor of −ω2 .
Candidate: x(t) = A cos(ωt), where A is a constant.
d2 x
2
X
= −ω A cos(ωt) = −ω2 x
dt2
SHM and Springs
d2 x
= −ω2 x
dt2
In fact, any solutions of the form:
x = B1 cos(ωt + φ1 ) + B2 sin(ωt + φ2 )
where B1 , B2 , φ1 , and φ2 are constants are correct.
However, since sin(θ) = cos(θ + π/2), in general any solution can
be written in the form:
x = A cos(ωt + φ)
Waveform
x = A cos(ωt + φ)
x
T
xi
A
ø
a
f =
1
1
T
Figure from Serway & Jewett, 9th ed.
t
Oscillations and Waveforms
Any oscillation can be plotted against time. eg. the position of a
vibrating object against time.
The result is a waveform.
Oscillations and Waveforms
Any oscillation can be plotted against time. eg. the position of a
vibrating object against time.
The result is a waveform.
From this wave description of the motion, a lot of parameters can
be specified.
This allows us to quantitatively compare one oscillation to another.
Examples of quantities: period, amplitude, frequency.
Oscillating Solutions
y = A sin(ωt)
x = A cos(ωt)
Phasor diagram
y plotted against t
Here, ωt is the phase at time t.
In general, if y = A sin(ωt + φ), the phase at time t is ωt + φ.
1
Figure from School of Physics webpage, University of New South Wales.
Oscillations
angular frequency
angular displacement per unit time in rotation, or the rate of
change of the phase of a sinusoidal waveform
ω=
2π
= 2πf
T
SHM and Springs
x = A cos(ωt + φ)
ω is the angular frequency of the oscillation.
When t = 2π
ω the block has returned to the position it had at
t = 0. That is one complete cycle.
SHM and Springs
x = A cos(ωt + φ)
ω is the angular frequency of the oscillation.
When t = 2π
ω the block has returned to the position it had at
t = 0. That is one complete cycle.
Recalling that ω =
p
k/m:
r
Period, T = 2π
m
k
Only depends on the mass of the bob and the spring constant.
Does not depend on the amplitude.
SHM and Springs Question
A mass-spring system has a period, T . If the mass of the bob is
quadrupled (and everything else is unchanged), what happens to
the period of the motion?
(A) halves, T /2
(B) remains unchanged, T
(C) doubles, 2T
(D) quadruples, 4T
SHM and Springs Question
A mass-spring system has a period, T . If the mass of the bob is
quadrupled (and everything else is unchanged), what happens to
the period of the motion?
(A) halves, T /2
(B) remains unchanged, T
(C) doubles, 2T
←
(D) quadruples, 4T
SHM and Springs Question
A mass-spring system has a period, T . If the spring constant is
halved (and everything else is unchanged), what happens to the
period of the motion?
(A) halves, T /2
(B) is reduced by a factor of
√
√
2, so, T / 2
(C) remains unchanged, T
(D) is increased by a factor of
√
√
2, so, 2T
SHM and Springs Question
A mass-spring system has a period, T . If the spring constant is
halved (and everything else is unchanged), what happens to the
period of the motion?
(A) halves, T /2
(B) is reduced by a factor of
√
√
2, so, T / 2
(C) remains unchanged, T
(D) is increased by a factor of
√
√
2, so, 2T
←
eration versus time. Notice that at
A Springs
of
7v2and
any specified time the velocity is
SHM
s extreme
908theout
position
The position of
bobofatphase
a givenwith
time the
is given
by:
and the acceleration is 1808 out of
on agrees
x =the
A cos(ωt
+ φ)
phase with
position.
A is the amplitude of the oscillation. We could also write xmax = A.
we define
he spring
x!0
and v(t)
A
m
t!0
xi ! A
vi ! 0
The speed of Figure
the particle
at anyApoint
in time is:
15.6
block–spring
for f, the
sitive and
ion is
system that
dx begins its motion from
−Aω sin(ωt + φ)
v=
rest with dt
the=block
at x 5 A at t 5 0.
460
Chapter 15
Oscillatory Motion
Energy in SHM
S
amax
%
100
50
0
%
100
50
0
%
100
50
0
a
S
vmax
b
S
amax
c
0
T
4
T
2
%
100
50
0
S
vmax
d
S
amax
e
S
v
f
x
–A
t
0
A
x
3T
4
%
100
50
0
T
%
100
50
0
t
Kinetic Potential Total
energy energy energy
Energy in SHM
Potential Energy:
1
1
U = kx 2 = kA2 cos2 (ωt + φ)
2
2
Kinetic Energy:
1
1
K = mv 2 = mA2 ω2 sin2 (ωt + φ)
2
2
Using ω2 = k/m
K +U =
=
1 2
kA cos2 (ωt + φ) + sin2 (ωt + φ)
2
1 2
kA
2
This does not depend on time! (Energy is conserved.)
Energy in SHM
15.3 Energy of the Simple H
In either plot, notice that
K " U ! constant.
U
1
K, U
K, U
1
2
2 kA
1
2
2 kA
T
2
a
K ! 12 mv 2
U ! 2 kx 2
K
T
t
–A
A
x
O
b
1
K + U = kA2
2 or zero. Because v2 5 k/m, we can
e see that K and U are always positive quantities
press the 1total
of9th
theed.
simple harmonic oscillator as
Figuremechanical
from Serway energy
& Jewett,
Pendula and SHM
A pendulum bob that is displaced to
one side by a small amount and released
follows SHM to a good approximation.
Gravity and the tension in the string
provide the restoring force.
modeled as simple harmonic
motion about the equilibrium
position u ! 0.
Pendula and SHM
u
S
L
T
m
s
m g sin u
u
The simple pendulum is anot
It consists of a particle-like b
that is fixed at the upper end
vertical plane and is driven b
the angle u is small (less than
harmonic oscillator.
The forces acting on the bo
tational force m S
g . The tange
always acts toward u 5 0, oppo
tion. Therefore, the tangenti
Newton’s second law for moti
Ft 5
m g cos u
S
mg
where the negative sign indic
(vertical)
position
The net force on theFigure
pendulum
is −mg sin θrium
along
the arc
s, and s
15.16 Abob
simple
when the string is atpendulum.
an angle θ to the vertical. expressed the tangential acc
Because s 5 Lu (Eq. 10.1a wit
Therefore, Newton’s second law for rotations (tangential
component only) gives:
I
d2 θ
= −Lmg sin θ
dt2
Pendula and SHM
Noting that moment of inertia, I = mL2
d2 θ
g
2 = − L sin θ
dt
Pendula and SHM
Noting that moment of inertia, I = mL2
d2 θ
g
2 = − L sin θ
dt
This is a non-linear second order differential equation, and the
exact solution is a bit ugly.
However, we can approximate this motion as SHM. (Remember,
for SHM acceleration is proportional to displacement but in the
opposite direction.)
For small values of θ (measured in radians!), sin θ ≈ θ.
Pendula and SHM
The equation of motion becomes:
d2 θ
= −ω2 θ
dt2
where ω =
q
g
L.
Solutions will be of the form
θ = θmax cos(ωt + φ)
where θmax is the amplitude and T =
2π
ω
s
Period, T = 2π
is the period.
L
g
Problem
An astronaut on the Moon attaches a small brass ball to a 1.00 m
length of string and makes a simple pendulum. She times 15
complete swings in a time of 75 seconds. From this measurement
she calculates the acceleration due to gravity on the Moon. What
is her result?1
1
Hewitt, “Conceptual Physics”, problem 8, page 350.
Problem
An astronaut on the Moon attaches a small brass ball to a 1.00 m
length of string and makes a simple pendulum. She times 15
complete swings in a time of 75 seconds. From this measurement
she calculates the acceleration due to gravity on the Moon. What
is her result?1
1.58 m/s2
1
Hewitt, “Conceptual Physics”, problem 8, page 350.
Waves
Very often an oscillation or one-time disturbance can be detected
far away.
Plucking one end of a stretched string will eventually result in the
far end of the string vibrating.
The string is a medium along which the vibration travels.
It carries energy from on part of the string to another.
Wave
a disturbance or oscillation that transfers energy through matter or
space.
Wave Pulses
484
Chapter 16
As the pulse moves along the
string, new elements of the
string are displaced from their
equilibrium positions.
Figure 16.1 A hand moves the
Wave Motion
which the pebble is dropped
motion: energy is transferred
16.1 Propagatio
The introduction to this
fer of energy through spac
of energy transfer mecha
and electromagnetic radi
nism, matter transfer, the
through space with no wa
All mechanical waves r
taining elements that can
which elements of the me
wave motion is to flick o
opposite end fixed as sho
a pulse) is formed and tr
represents four consecutiv
eling pulse. The hand is
Wave pulses
A point P in the middle of the string moves up and down, just as
the hand did.
Kinds of Waves
medium
a material substance that carries waves. The constituent particles
are temporarily displaced as the wave passes, but they return to
their original position.
Kinds of waves:
• mechanical waves – waves that travel on a medium, eg. sound
waves, waves on string, water waves
• electromagnetic waves – light, in all its various wavelengths,
eg. x-rays, uv, infrared, radio waves
• matter waves – wait for Phys4D!
Kinds of Waves
Kinds of waves:
• transverse – displacement perpendicular to direction of wave
travel
• longitudinal – displacement parallel to direction of wave travel
Transverse
Longitudinal
Transverse vs. Longitudinal
Examples of transverse waves:
• vibrations on a guitar string
• ripples in water
• light
• S-waves in an earthquake (more destructive)
Examples of longitudinal waves:
• sound
• P-waves in an earthquake (initial shockwave, faster moving)
Earthquakes
Earthquakes
Sound waves
Summary
• finished thermodynamics
• oscillations
• simple harmonic motion (SHM)
• spring and pendulum systems
• intro to waves
Homework Serway & Jewett:
• prev: Ch 22, OQs: 1, 3, 7; CQs: 1; Probs: 1, 3, 9, 15, 20, 23,
29, 37, 67, 73, 81
• new: Ch 15, onward from page 472. OQs: 13; CQs: 5, 7;
Probs: 1, 3, 9, 35, 41, 86