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
C
After Two Exams
Average = 74.1
High = 100
20%
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%
2.8%
1.5%
2.9% 3.3%
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
4πr
Sphere with
radius r
Source PS
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).
v=
elastic
inertial
R. Field 11/12/2013
University of Florida
(wave speed)
v=
FT
μ
PHY 2053
Transverse wave on a string:
FT = string tension
μ = M/L = linear mass density
Page 3
Constructing Traveling Waves
y = f(x) at time 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 + φ) at time t = 0,
where k is the "wave number“, k = 2π/λ, λ 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 ) + φ ) = A sin( kx − ωt + φ ) where ω = kv.
y = y ( x, t ) = A sin( kx − ωt + φ )
y = y ( x, t ) = A sin( kx + ωt + φ )
λ
y=Asin(kx)
1.0
A
0.5
0.0
-0.5
-1.0
kx (radians)
Speed of propagation!
Harmonic wave traveling to the right
Harmonic wave traveling to the left
The phase angle φ determines y at x = t = 0, y(x=t=0) = Asinφ. If y(x=t=0) = 0 then φ = 0.
R. Field 11/12/2013
PHY 2053
University of Florida
v=
Page 4
ω
k
Waves: Mathematical Description
y = y ( x, t ) = A sin(Φ ) = A sin( kx − ωt )
y
y-axis
y(x,t)
wave traveling to the right
A
A
Φ = kx-ωt
φ
Φ
Vector A with length A undergoing uniform circular motion with phase Φ = kx – ωt. The projection
onto the y-axis gives y = Asin(kx – ωt).
y = y ( x, t = 0) = A sin( kx)
If t = 0 then
y
y-axis
y(x)
λ
A
Φ = kx
x
One circular revolution corresponds to Φ = 2π = kλ, and hence k = 2π/λ (“wave number”).
R. Field 11/12/2013
University of Florida
PHY 2053
Page 5
Waves: Mathematical Description
y = y ( x, t ) = A sin(Φ ) = A sin( kx − ωt )
y
y-axis
y(x,t)
wave traveling to the right
A
A
Φ = kx-ωt
φ
Φ
Vector A with length A undergoing uniform circular motion with phase Φ = kx – ωt. The projection
onto the y-axis gives y = Asin(kx – ωt).
If x = 0 then
y
y-axis
y(t)
y = y ( x = 0, t ) = A sin( −ωt )
T
Φ = -ωt
A
t
One circular revolution corresponds to Φ = 2π = ωT and hence T = 2π/ω (“period”).
R. Field 11/12/2013
University of Florida
PHY 2053
Page 6
Waves: Mathematical Description
y = y ( x, t ) = A sin(Φ ) = A sin( kx − ωt + φ )
In General
y
y-axis
y(x,t)
A
wave traveling to the right
A
Φ = kx-ωt+φ
φ
Φ
Overall phase Φ = kx – ωt + φ.
yy = y ( x, t ) = A sin(Φ ) = A sin( kx + ωt + φ )
y-axis
y(x,t)
A
wave traveling to the left
Φ = kx+ωt+φ
A
Φφ
Period (in s)
T=
2π
ω
Frequency (in Hz)
1
f =
T
R. Field 11/12/2013
University of Florida
Wave Number (in rad/m)
Angular Frequency (in rad/s)
ω = 2πf
PHY 2053
k=
2π
λ
Wave Speed (in m/s)
v=
ω
k
= λf
Page 7
Waves: Mathematical Description
y
y = y ( x, t ) = A sin(Φ ) = A sin( kx − ωt + φ )
wave traveling to the right
A
vnode
φ
x
nth node
point on string
• Waves Propagation: A node is a point on the wave where y(x,t) vanishes:
Φ = kx − ωt + φ = nπ n = 0,1,2, L kx1 − ωt1 + φ = nπ k ( x2 − x2 ) = ω (t 2 − t1 )
dx ω
kx2 − ωt 2 + φ = nπ
kΔx = ωΔt
=
vnode =
dt k (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
= −ωA cos(kx − ωt + φ )
dt
umax = ωA
R. Field 11/12/2013
University of Florida
transverse acceleration of a point on the string
a=
du
= ω 2 A sin( kx − ωt + φ )
dt
a = −ω 2 y
amax = ω2 A
SHM
PHY 2053
Page 8
Waves: Example Problems
• A transverse wave on a taught string has amplitude A, wavelength λ
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?
ω
umax
2πA
ωA
v
=
u
=
ω
A
kA
=
=
=
Answer: 2πA/λ
max
v
ω/k
λ
k
• The function y(x,t) = Acos(kx - ωt) describes a wave on a taut string
with the x-axis parallel to the string. The wavelength is λ = 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
umax = ωA
R. Field 11/12/2013
University of Florida
v=
ω ωλ
=
k 2π
λumax (3.14cm)(10m / s) _
v=
=
= 50m / s
2πA
2π (0.1cm)
PHY 2053
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=
λ = 4Δx = 4(10cm) = 0.4m
d 12cm
=
= 40m / s
Δt 3ms
f =
R. Field 11/12/2013
University of Florida
v
λ
=
40m / s
= 100 Hz
0.4m
PHY 2053
Page 10