Physics 2111
Unit 21
Today’s Concepts:
Periodic Motion
Simple Harmonic Motion
Mass on a Spring
Damped Harmonic Motion
Forced Harmonic Motion - Resonance
Unit 21, Slide 1
Periodic Motion
Any motion that repeats itself over and
over again.
Period, T: time it takes to complete one cycle
Frequency, f: number of cycles completed
every second
f = 1/T
Unit 21, Slide 2
Simple Harmonic Motion
k
m
-A
0
A
x
2x
d
F=-kx = ma = m 2
dt
Unit 21, Slide 3
Most general solution:
x(t) = Acos(wt-f)
Unit 21, Slide 4
What is w?
Tells you how fast the object oscillates.
wT = 2p
w = 2p/T w = 2pf
What is A?
Amplitude – maximum stretch of the spring.
Unit 21, Slide 5
What do they mean graphically?
T
-p
-p
A
p
p

Change A
Change w
and T
Change f
Unit 21, Slide 6
Example 21.1 (mass on spring)
k
m
-A
0
A
x
A 2kg mass is attached to a spring with a
spring constant of k=5N/m. The spring is
stretched 20cm and then released at t=0.
a) What is the period of the mass?
b) What is the equation of motion of the
mass?
Unit 21, Slide 7
Unit 21, Slide 8
Example 21.2 (mass on spring - velocity)
k
m
-A
0
A
x
A 2kg mass is attached to a spring with a
spring constant of k=5N/m. The spring is
stretched 20cm and then released at t=0.
a) What is it’s maximum velocity?
b) What is it’s maximum acceleration?
Unit 21, Slide 9
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) = -w2 y(t)
a<0
v<0
a<0
v>0
y(t)
(A)
(D)
t
(B)
a>0
v<0
(C)
a>0
v>0
The answer is (D).
Mechanics Lecture 21, Slide 10
Energy
k
m
-A
0
A
x
For mass on a spring we know PE = ½ kx2
For any object we know KE = ½ mv2
What does that tell us in the case of a SHO?
Mechanics Lecture 21, Slide 11
Example 21.3 (mass on spring - energy)
k
m
-A
0
x
A
A 2kg mass is attached to a spring with a
spring constant of k=5N/m. The spring is
stretched 20cm and then released at t=0.
a) What is its total mechanical energy?
b) What is its maximum kinetic energy?
Mechanics Lecture 21, Slide 13
SHM Dynamics
What does angular frequency w have to do with moving back and
forth in a straight line?
y = R cos = R cos(wt)
y
3
1
4
1
2
3
1
x
6
8
7
2
8

5
1
0
-1
p
p
2
4
7
2p

6
5
Mechanics Lecture 21, Slide 14
Damped Harmonic Motion
Physical oscillator come to rest due
“damping force”
k
FD = -bv
m
SF = ma
-kx2 -b dx/dt = m d2x/dt2
Solution: x(t) = xmax e-bt/2m cos(w’t +f)
′
𝜔 =
𝑘
2
𝑏
𝑚 −
4𝑚2
Mechanics Lecture 21, Slide 15
Damped Harmonic Motion
Notice we’re taking the square root of a
difference. What if it’s negative?
k
′
𝜔 =
m
𝑘
2
𝑏
𝑚 −
4𝑚2
k/m > (b/2m)2
under damped
k/m < (b/2m)2
over damped
k/m = (b/2m)2
critically damped
(w’ = 0) no oscillations
Mechanics Lecture 21, Slide 16
Damped Harmonic Motion
Overdamped
Critically damped
Under damped
Which one do you want for your shocks on
your car?
Mechanics Lecture 21, Slide 17
Underdamped Harmonic Motion
x(t) = xmax e-bt/2m cos(w’t +f)
k
m
Like a SHO with a time dependent amplitude
Mechanics Lecture 21, Slide 18
Underdamped Harmonic Motion
x(t) = xmax e-bt/2m cos(w’t +f)
k
m
Mechanics Lecture 21, Slide 19
Example 21.5 (Damped Harmonic Oscillator)
k
m
A 10kg mass is hung from a spring
with a constant of 50N/m. The air
drag on the mass causes a damping
constant of b=10Nsec/m. If the
spring is stretched 20cm and
released, how far will is oscillate at
the end of the first cycle?
This is 20cm
What is this?
Mechanics Lecture 21, Slide 20
Forced Harmonic Motion
wd
Add third force
• restoring force (e.g. spring)
-kx
• damping force (e.g. air drag) -bv
• driving force (e.g. motor)
k
Fmax cos(wd + f)
m
Amplitude
of mass
=
Fmax
(m2(wo2 - wd2) + b2wd2))0.5
Natural Frequency of
mass (w2 = k/m)
Frequency of driving
force
Mechanics Lecture 21, Slide 21