Transcript 2053_Lecture_11-12-13

PHY2053 Exam 2
PHY 2053 Fall13 Exam 2
80
Number of Students = 621
Average = 11.6
Median = 12
High = 20
Low = 3
Number
60
40
20
0
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15 16 17 18 19 20
Grade
Average = 11.6
High = 20 (2 students)
Low = 3
R. Field 11/12/2013
University of Florida
PHY 2053
After adding 2 points
to everyone’s score.
Page 1
Estimated Course Grades
18 students have
100 points!
PHY 2053 Fall13 Estimated Course Grades
≥ 40 D≥ 45 D
≥ 50 D+
≥ 55 C≥ 60 C
≥ 65 C+
≥ 70 B≥ 75 B
≥ 82 B+
≥ 87 A≥ 92 A
30
Number
25
20
15
10
5
20%
After Two Exams
C
Average = 74.1
High = 100
B
A
Percent of Students
35
PHY 2053 Fall13 Estimated Course Grades
15%
After Two Exams
Number = 646
A or A- = 22.9%
>=B = 50.2%
>=C = 84.4%
16.3%
13.2% 12.8%
12.7%
11.0%
10%
10.2%
8.2%
5.1%
5%
2.9% 3.3%
2.8%
1.5%
0%
0
0
5
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
Points (100 max)
E
D-
D
D+
C-
C
C+
B-
B
B+
A-
A
Grade
1. Assumes that you get the same grade on the Final Exam that you averaged on Exam 1
plus Exam 2.
2. Include the first 8 quizzes and assumes that you get the same average on all your
remaining quizzes that you have for the first 8 quizzes.
3. Includes the first 9 WebAssign HW assignments and assumes that you get the same
average on all your remaining homework assignments that you have for the first 9
assignments.
4. Includes your HITT scores through 10/31/13 and assumes you maintain the same average
on the remaining HITT questions.
5. Includes your first 4 Sakai HW assignments and assumes you maintain the same average
on the remaining assignments.
R. Field 11/12/2013
University of Florida
PHY 2053
Page 2
Traveling Waves: Energy Transport
A “wave” is a traveling disturbance that transports
energy but not matter.
• Intensity:
I 
Ppower
v
Intensity I = power per unit area
(measured in Watts/m2)
Area
Intensity is proportional
to the square of the
amplitude A!
• Variation with Distance: If sound is emitted isotropically
(i.e. equal intensity in all directions) from a point source with
power Psource and if the mechanical energy of the wave is
conserved then
P
I 
source
2
4 r
S p h ere w ith
rad iu s r
S ou rce P S
r
(intensity from isotropic point source)
• Speed of Propagation: The speed of any mechanical wave depends on both
the inertial property of the medium (stores kinetic energy) and the elastic
property (stores potential energy).
Transverse wave on a string:
v
elastic
inertial
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University of Florida
(wave speed)
v
FT
m
PHY 2053
FT = string tension
m = M/L = linear mass density
Page 3
Constructing Traveling Waves
y = f(x) at tim e t= 0
y = f(x-vt)
v
x=0
x = vt
• Constructing Traveling Waves: To construct a wave with shape y = f(x) at
time t = 0 traveling to the right with speed v replace x by x-vt.
Traveling Harmonic Waves: Harmonic waves
have the form y = A sin(kx + f) at time t = 0,
where k is the "wave number“, k = 2/,  is the
"wave length". and A is the "amplitude". To
construct a harmonic wave traveling to the right
with speed v, replace x by x-vt as follows:
y  A sin( k ( x  vt )  f )  A sin( kx   t  f ) where  = kv.

y= A sin (k x)
1 .0
A
0 .5
0 .0
-0 .5
-1 .0
k x (rad ian s)
Speed of propagation!
y  y ( x , t )  A sin( kx   t  f ) Harmonic wave traveling to the right
y  y ( x , t )  A sin( kx   t  f ) Harmonic wave traveling to the left
The phase angle f determines y at x = t = 0, y(x=t=0) = Asinf. If y(x=t=0) = 0 then f = 0.
R. Field 11/12/2013
PHY 2053
University of Florida
v

k
Page 4
Waves: Mathematical Description
y  y ( x , t )  A sin( F )  A sin( kx   t )
y
y-axis
y(x,t)
wave traveling to the right
A
A
F = kx-t
f
F
Vector A with length A undergoing uniform circular motion with phase F = kx – t. The projection onto
the y-axis gives y = Asin(kx – t).
y  y ( x , t  0 )  A sin( kx )
If t = 0 then
y
y-axis
y(x)

A
F = kx
x
One circular revolution corresponds to F = 2 = k, and hence k = 2/ (“wave number”).
R. Field 11/12/2013
University of Florida
PHY 2053
Page 5
Waves: Mathematical Description
y  y ( x , t )  A sin( F )  A sin( kx   t )
y
y-axis
y(x,t)
wave traveling to the right
A
A
F = kx-t
f
F
Vector A with length A undergoing uniform circular motion with phase F = kx – t. The projection onto
the y-axis gives y = Asin(kx – t).
If x = 0 then
y
y-axis
y(t)
y  y ( x  0 , t )  A sin(   t )
T
F = -t
A
t
One circular revolution corresponds to F = 2 = T and hence T = 2/ (“period”).
R. Field 11/12/2013
University of Florida
PHY 2053
Page 6
Waves: Mathematical Description
yy
In General
y-axis
 y ( x , t )  A sin( F )  A sin( kx   t  f )
wave traveling to the right
y(x,t)
A
A
F = kx-t+f
f
F
Overall phase F = kx – t + f.
y  y ( x , t )  A sin( F )  A sin( kx   t  f )
y-axis
wave traveling to the left
y(x,t)
A
F = kx+t+f
F
Period (in s)
T 
2

Frequency (in Hz)
f 
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University of Florida
1
T
Wave Number (in rad/m)
Angular Frequency (in rad/s)
  2 f
PHY 2053
k 
2

Wave Speed (in m/s)
v

 f
k
Page 7
Waves: Mathematical Description
y  y ( x , t )  A sin( F )  A sin( kx   t  f )
y
wave traveling to the right
A
vnode
f
x
nth node
point on string
• Waves Propagation: A node is a point on the wave where y(x,t) vanishes:
F  kx   t  f  n  n  0 ,1, 2 ,  kx1   t1  f  n 
k ( x 2  x 2 )   ( t 2  t1 )
v node 
dx
dt


k
kx 2   t 2  f  n 
k x    t
(wave speed)
• Transverse Speed & Acceleration: For “transverse” waves the points on
the string move up and down while the wave moves to the right.
transverse speed of a point on the string
u 
dy
dt
   A cos( kx   t  f )
u max   A
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transverse acceleration of a point on the string
a
du
dt
  A sin( kx   t  f )
2
a   y
2
a max   A
2
SHM
PHY 2053
Page 8
Waves: Example Problems
• A transverse wave on a taught string has amplitude A, wavelength 
and speed v. A point on the string only moves in the transverse
direction. If its maximum transverse speed is umax, what is the ratio
umax/v?

u max
A
2 A
v



kA

u


A
Answer: 2A/
max
v
k
 /k

• The function y(x,t) = Acos(kx - t) describes a wave on a taut string
with the x-axis parallel to the string. The wavelength is  = 3.14 cm
and the amplitude is A = 0.1 cm. If the maximum transverse speed of
any point on the string is 10 m/s, what is the speed of propagation of
the travelling wave in the x-direction?
Answer: 50 m/s
u max   A
R. Field 11/12/2013
University of Florida
v

k


2
v
 u max
2 A

PHY 2053
( 3 . 14 cm )( 10 m / s ) _
2 ( 0 . 1cm )
 50 m / s
Page 9
Waves: Example Problems
• A sinusoidal wave moving along a string is
shown twice in the figure. Crest A travels in the
positive direction along the x-axis and moves a
distance d = 12 cm in 3 ms. If the tick marks
along the x-axis are 10 cm apart, what is the
frequency of the traveling wave?
Answer: 100 Hz
v
d
t

12 cm
  4  x  4 (10 cm )  0 . 4 m
 40 m / s
3 ms
f 
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University of Florida
v


40 m / s
 100 Hz
0 .4 m
PHY 2053
Page 10