PHYSICS 231
INTRODUCTORY PHYSICS I
Lecture 20
Last Lecture
Heat Engine
Refrigerator, Heat Pump
Qhot
engine
Qhot
W
fridge
Qcold
W
Qcold
W
Tcold
e
 eCarnot 1
Qhot
Thot
Qc,h
Tc,h
COP 
 COPCarnot 
W
Thot  Tcold
W  Qhot  Qcold
Entropy
• Measure of Disorder of the system
(randomness, ignorance)
• Entropy: S = kBlog(N)
N = # of possible arrangements for fixed E and Q
1000
900
800
700
600
500
400
300
200
100
0
(0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0)
Number of ways for 12 molecules to arrange
themselves in two halves of container.
S is greater if molecules spread evenly in both halves.
2nd Law of Thermodynamics
(version 2)
The Total Entropy of the Universe can never
decrease.
(but entropy of system can increase or decrease)
On a macroscopic level, one finds that adding heat
raises entropy:
S  Q / T
Temperature in Kelvin!
Why does Q flow from hot to cold?
• Consider two systems, one with TA and one with TB
• Allow Q > 0 to flow from TA to TB
• Entropy changes by:
S = Q/TB - Q/TA
• This can only occur if S > 0, requiring TA > TB.
• System will achieve more randomness by exchanging
heat until TB = TA

Carnot Engine
Carnot cycle is most efficient possible, because
the total entropy change is zero.
It is a “reversible process”.
For real engines:
Qcold Qhot
S  Senvironment 

0 
Tcold Thot
W
Qcold
Tcold
e
1
1
 eCarnot
Qhot
Qhot
Thot
Chapter 13
Vibrations and Waves
Hooke’s Law Reviewed
F  kx
• When x is positive
F is negative
;
,
• When at equilibrium (x=0),
F = 0 ;
QuickTime™ and a
Animation decompressor
are needed to see this picture.
• When x is negative
F is positive
;
,
Sinusoidal Oscillation
If we extend the mass,
and let go, the pen
traces a sine wave.
Graphing x vs. t
A
T
A : amplitude (length, m)
T : period (time, s)
Period and Frequency
A
T
x  Acost
T  2
Amplitude: A

Period: T

Frequency: f = 1/T
Angular frequency: 
T

2

,

f 
2
Phases
Often a phase
the peak:
is included to shift the timing of
x  Acost   
 Acos (t  t 0 )
for peak at
t  t0
Phase of 90-degrees changes cosine to sine



cos   t    sin  t 

2
Velocity and
Acceleration vs. time
• Velocity is 90
“out of phase” with x:
When x is at max,
v is at min ....
• Acceleration is 180°
“out of phase” with x
a = F/m = - (k/m) x
x
T
v
T
a
T
v and a vs. t
x  A cos  t
v  vmax sin  t
a  amax cos  t
Find vmax with E conservation
1 2 1 2
kA  mvmax
2
2
k
vmax  A
m
Find amax using F=ma
kx  ma
kA cos  t  mamax cos  t
k
amax  A
m
Connection to Circular Motion
Projection on axis
circular motion
with constant
angular velocity 
Simple Harmonic
Motion
QuickTime™ and a
Animation decompressor
are needed to see this picture.
What is ?
Circular motion
Angular speed: 
Radius: A
Simple Harmonic Motion
Cons. of E: v max
=> Speed: v=A
k
A
m


QuickTime™ and a
Animation decompressor
are needed to see this picture.

k
m
k

m
Formula Summary
1
f 
T
2
  2 f 
T
x  A cos( t   )
v   Asin( t   )
a   2 A(cos  t   )

k
m
Example13.1
An block-spring system oscillates with an amplitude of 3.5 cm.
If the spring constant is 250 N/m and the block has a mass of
0.50 kg, determine
(a) the mechanical energy of the system
(b) the maximum speed of the block
(c) the maximum acceleration.
a) 0.153 J
b) 0.783 m/s
c) 17.5 m/s2
Example 13.2
A 36-kg block is attached to a spring of constant
k=600 N/m. The block is pulled 3.5 cm away from its
equilibrium positions and released from rest at t=0.
At t=0.75 seconds,
a) what is the position of the block?
b) what is the velocity of the block?
a) -3.489 cm
b) -1.138 cm/s
Example 13.3
A 36-kg block is attached to a spring of constant k=600
N/m. The block is pulled 3.5 cm away from its
equilibrium position and is pushed so that is has an initial
velocity of 5.0 cm/s at t=0.
a) What is the position of the block at t=0.75 seconds?
a) -3.39 cm
Example 13.4a
An object undergoing simple harmonic motion follows the
expression,
x(t)  4  2cos[ (t  3)]
Where x will be in cm if t is in seconds
The amplitude of the motion is:
a) 1 cm
b) 2 cm
c) 3 cm
d) 4 cm
e) -4 cm
Example 13.4b
An object undergoing simple harmonic motion follows the
expression,
x(t)  4  2cos[ (t  3)]
Here, x will be in cm if t is in seconds
The period of the motion is:
a) 1/3 s
b) 1/2 s
c) 1 s
d) 2 s
e) 2/ s
Example 13.4c
An object undergoing simple harmonic motion follows the
expression,
x(t)  4  2cos[ (t  3)]
Here, x will be in cm if t is in seconds
The frequency of the motion is:
a) 1/3 Hz
b) 1/2 Hz
c) 1 Hz
d) 2 Hz
e)  Hz
Example 13.4d
An object undergoing simple harmonic motion follows the
expression,
x(t)  4  2cos[ (t  3)]
Here, x will be in cm if t is in seconds
The angular frequency of the motion is:
a) 1/3 rad/s
b) 1/2 rad/s
c) 1 rad/s
d) 2 rad/s
e)  rad/s
Example 13.4e
An object undergoing simple harmonic motion follows the
expression,
x(t)  4  2cos[ (t  3)]
Here, x will be in cm if t is in seconds
The object will pass through the equilibrium position
at the times, t = _____ seconds
a)
b)
c)
d)
e)
…,
…,
…,
…,
…,
-2, -1, 0, 1, 2 …
-1.5, -0.5, 0.5, 1.5, 2.5, …
-1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, …
-4, -2, 0, 2, 4, …
-2.5, -0.5, 1.5, 3.5,