Physics 101: Lecture 22
Simple Harmonic Motion
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Today’s lecture will cover Textbook Sections 10.4 - 10.6
Physics 101: Lecture 22, Pg 1
Review: Ideal Springs
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Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched or
compressed from its relaxed position.
ÎFX = -k x
Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
(often called “spring constant”)
relaxed position
FX = - kx < 0
x>0
x
x=0
Physics 101: Lecture 22, Pg 2
Review: Simple Harmonic Motion
Uniform circular motion <-> Motion of an object attached to an ideal spring
Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
Angular frequency =
ω = 2πf = 2π/T
Physics 101: Lecture 22, Pg 3
Simple Harmonic Motion:
Quick Review
x(t) = [A]cos(ωt)
v(t) = -[Aω]sin(ωt)
a(t) = -[Aω2]cos(ωt)
x(t) = [A]sin(ωt)
OR
v(t) = [Aω]cos(ωt)
a(t) = -[Aω2]sin(ωt)
xmax = A
Period = T (seconds per cycle)
vmax = Aω
Frequency = f = 1/T (cycles per second)
amax = Aω2
Angular frequency =
ω = 2πf = 2π/T
Physics 101: Lecture 22, Pg 4
Review: Period of a Spring
For simple harmonic oscillator
ω = 2πf = 2π/T
For mass M on spring with spring constant k
ω=
k
m
T = 2π
m
k
Physics 101: Lecture 22, Pg 5
Concept Question
If the amplitude of the oscillation (same block and same spring) was
doubled, how would the period of the oscillation change? (The period is
the time it takes to make one complete oscillation)
1. The period of the oscillation would double.
2. The period of the oscillation would be halved
3. The period of the oscillation would stay the same
CORRECT
x
+2A
t
-2A
Physics 101: Lecture 22, Pg 6
Potential Energy of a Spring
1 2
PES = kx
2
Where x is measured from
the equilibrium position
m
x=0
x
PES
0
x
Physics 101: Lecture 22, Pg 7
Same thing for a vertical spring:
y
PES =
1 2
ky
2
Where y is measured from
the equilibrium position
y=0
m
PES
0
y
Physics 101: Lecture 22, Pg 8
In either case...
y
m
x
x=0
m
y=0
Etotal = 1/2 Mv2 + 1/2 kx2 = constant
KE
PE
KEmax = 1/2Mv2max=1/2Mω2A2 =1/2kA2
PEmax = 1/2kA2
Etotal = 1/2kA2
Physics 101: Lecture 22, Pg 9
Concept Question
In Case 1 a mass on a spring oscillates back and forth. In Case 2, the
mass is doubled but the spring and the amplitude of the oscillation is
the same as in Case 1.
In which case is the maximum kinetic energy of the mass the biggest?
1. Case 1
2. Case 2
3. Same
CORRECT
Physics 101: Lecture 22, Pg 10
Concept Question
1/ kx2
2
PE =
KE = 0
x=-A
x=0
x=+A
PE = 0
KE = KEMAX
same
for both
same
for both
x=-A
x=0
x=+A
Physics 101: Lecture 22, Pg 11
Concept Question
The same would be true for vertical springs...
PE = 1/2k y2
Y=0
PE = 1/2k y2
Y=0
Physics 101: Lecture 22, Pg 12
Pendulum
ω=
g
L
2π
L
T=
= 2π
ω
g
For “small oscillation”, period does not depend on
•mass
•amplitude
Physics 101: Lecture 22, Pg 13
Concept Question
Suppose a grandfather clock (a simple pendulum) runs slow. In order to
make it run on time you should:
1. Make the pendulum shorter
CORRECT
2. Make the pendulum longer
g
ω=
L
2π
L
T=
= 2π
ω
g
Physics 101: Lecture 22, Pg 14
Concept Question
A pendulum is hanging vertically from the ceiling of an elevator. Initially
the elevator is at rest and the period of the pendulum is T. Now the
pendulum accelerates upward. The period of the pendulum will now be
1. greater than T
2. equal to T
3. less than T
CORRECT
g
ω=
L
2π
L
T=
= 2π
ω
g
“Effective g” is larger when accelerating upward
(you feel heavier)
Physics 101: Lecture 22, Pg 15