Announcements
§ smartPhysics homework deadlines have been reset to 3:30 PM on December 15 (beginning of final exam). You can get 100% credit if you go back and correct ANY problem on the HW from the beginning of the semester!
§ Last year’s final exam has been posted
§ Final exam is worth 200 points and is 2 hours:
Ø Tuesday December 15, 3:30 – 5:30 pm
Ø Room: JFB 101 & 103 (both)
Mechanics Lecture 21, Slide 1
Exam 3 Statistics
Mechanics Lecture 21, Slide 2
Lecture 21: Simple Harmonic Motion
Today’s Concept:
Simple Harmonic Motion: Mass on a Spring
Mechanics Lecture 21, Slide 3
Fspring
a
Most general solution:
x(t) = Acos(ω t−ϕ)
Mechanics Lecture 21, Slide 4
demo
Mechanics Lecture 21, Slide 5
SHM Dynamics
What does angular frequency ω have to do with moving back and forth in a straight line?
y = R cosθ = R cos(ω t)
y
3
4
1
2
θ
5
1
x
6
7
8
0
-­1
1
1
2
8
3
π
π
2
4
7
2π
θ
6
5
Mechanics Lecture 21, Slide 6
SHM Solution
Drawing of A cos(ω t)
✓
T = 2π /ω
A
θ
−2π
−π
A
π
2π
Mechanics Lecture 21, Slide 7
SHM Solution
Drawing of A cos(ω t − ϕ)
✓
ϕ
−2π
−π
π
2π
θ
Mechanics Lecture 21, Slide 8
SHM Solution
Drawing of Acos(ω t − π /2) = Asin(ω t)
π /2
ωt
−2π
−π
π
2π
Mechanics Lecture 21, Slide 9
In the slide titled, "example", the decision to choose the equation "A*cos(w*t+'phi') seemed like an arbitrary decision. Please explain why that equation was chosen over the other general solution equations. The description of the phase angle is a little confusing too.
ϕ
−2π
−π
π
2π
θ
Drawing of Acos(ω t − ϕ)
Mechanics Lecture 21, Slide 10
In the slide titled, "example", the decision to choose the equation "A*cos(w*t+'phi') seemed like an arbitrary decision. Please explain why that equation was chosen over the other general solution equations. The description of the phase angle is a little confusing too.
Is this a sine or a cosine?
Mechanics Lecture 21, Slide 11
Any linear combination of sines and cosines having the same frequency will result in a sinusoidal curve with the same frequency.
Mechanics Lecture 21, Slide 12
CheckPoint: Superposition
1.00
0.50
1000
900
800
700
600
500
400
300
200
100
0
0.00
-­0.50
-­1.00
Suppose the two sinusoidal curves shown above are added together. Which of the plots shown below best represents the result?
0.60
0.40
0.20
1000
900
800
700
600
500
400
300
200
A)
100
0
0.00
-­0.20
-­0.40
-­0.60
2.00
1.50
1.00
0.50
500
600
700
800
900
1000
500
600
700
800
900
1000
400
300
200
100
0.00
-­0.50
0
B)
-­1.00
-­1.50
-­2.00
0.6
0.4
0.2
400
300
200
-­0.2
100
0
0
C)
-­0.4
-­0.6
Mechanics Lecture 21, Slide 13
Can we talk more about the trigonometrical functions? Like in question #2.
⎛ a+b ⎞
⎛ a −b ⎞
sin(a) + sin(b) = 2sin ⎜
⎟ cos ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎠
Suppose:
a =ωt +α
b =ωt +β
⎛
⎛α − β⎞
α +β⎞
sin(ω t + α ) + sin(ω t + β ) = 2sin ⎜ ω t +
cos ⎜
⎟
2 ⎠
⎝
⎝ 2 ⎟⎠
⎛
⎛ −φ ⎞
φ⎞
⇒ sin(ω t) + sin(ω t + φ ) = 2sin ⎜ ω t + ⎟ cos ⎜ ⎟
2⎠
⎝
⎝ 2⎠
⎛
φ⎞
= Asin ⎜ ω t + ⎟
2⎠
⎝
Mechanics Lecture 21, Slide 14
Mechanics Lecture 21, Slide 15
Mechanics Lecture 21, Slide 16
ACT
A
B
C
D
A mass oscillates up & down on a spring. Its position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration?
y(t)
(A)
(C)
t
(B)
Mechanics Lecture 21, Slide 17
dy
The slope of y(t) tells us the sign of the velocity since v y =
dt
y(t) and a(t) have the opposite sign since a(t) = −ω2 y(t)
a<0
v<0
a<0
v>0
y(t)
(A)
(C)
The answer is (C).
t
(B)
a>0
v>0
Mechanics Lecture 21, Slide 18
ACT
A
B
C
D
A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describes its velocity and acceleration as a function of time?
A) v(t) = −vmax sin(ωt)
B) v(t) = vmax sin(ωt)
C) v(t) = vmax cos(ωt)
a(t) = −amax cos(ωt)
a(t) = amax cos(ωt)
a(t) = −amax cos(ωt)
k
t=0
y
m
d
0
(both vmax and amax are positive numbers)
Mechanics Lecture 21, Slide 19
ACT
A
B
C
D
Since we start with the maximum possible displacement at t = 0
we know that:
y = d cos(ωt)
dy
vy =
= −ω d sin(ω t ) = −vmax sin(ω t )
dt
ay =
dv y
dt
= −ω 2 d cos(ωt ) = −amax cos(ωt )
k
t=0
y
m
d
0
Mechanics Lecture 21, Slide 20
Δy
kΔy
mg
mg = kΔy
mg
k=
Δy
k
ω=
m
Mechanics Lecture 21, Slide 21
At t = 0, y = 0, moving down
y(t ) = − Asin(ωt )
v(t ) = −ω Acos(ωt )
a(t ) = −ω 2 A sin (ωt )
Use energy conservation to find A
1 2
1 2
mvmax = kA
2
2
A = vmax
m
k
v(t ) = −ω Acos(ωt )
Mechanics Lecture 21, Slide 22
a(t ) = ω 2 A sin (ωt )
amax = ω 2 A
Or similarly
amax
Fmax kΔymax kA
=
=
=
m
m
m
Mechanics Lecture 21, Slide 23
F (t ) = ky (t ) = kA sin (ωt )
Mechanics Lecture 21, Slide 24
1 2
U = ky
2
y(t ) = − A sin (ωt )
Mechanics Lecture 21, Slide 25