Physics 203 – College Physics I
Department of Physics – The Citadel
Physics 203
College Physics I
Fall 2012
S. A. Yost
Chapter 11
Simple Harmonic Motion
Physics 203 – College Physics I
Department of Physics – The Citadel
Announcements
Problem set 10B is due Thursday.
Read sections 1 – 4 and 7 – 9 if you haven’t
Topics: simple harmonic motion, intro to waves.
Next Tuesday: Ch. 11, sec. 7 – 9 and 11 – 13 with
some material from Ch. 12, sec. 1 – 4 & 7 mixed
in. (The topics are related.)
Physics 203 – College Physics I
Clicker Question
Are you here?
A = Yes
B = No
C = Both
D = Neither
E = Can’t be determined
Department of Physics – The Citadel
Physics 203 – College Physics I
Department of Physics – The Citadel
Hooke’s Law
From chapter 6:
Hooke’s Law describes a linear restoring force
when a spring is displaced from its equilibrium
position.
x
F = -k x
Elastic potential energy:
U = ½ kx2
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
When a mass oscillates under a linear restoring
force F = -kx, the acceleration is always opposite
the displacement from equilibrium, but proportional
to it.
a = F/m = -(k/m) x.
This is called simple harmonic motion.
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
y
v
Where else have we seen
a linear restoring force?
Look at uniform circular
motion in components.
m
r
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
The acceleration vector
always points toward the
center and has magnitude
→
y
v
→
a=
a
v2/r
→
r
v=rw
x
a= w2 r
In vectors:
→
m
→
a = - w2 r
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
The components of the
acceleration vector
→
a= -
→
2
w r
y
→
a
are
m
→
ax = - w2 x
ay = - w2 y
The acceleration in each
direction is proportional
to the displacement.
r
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
Just focus on the x
component:
a = - w2 x
This is simple harmonic
motion, if we match the
Hooke’s Law condition:
a = F/m =- (k/m) x.
Identify the angular
velocity for SHM as
___
w = √k/m
y
→
a
m
→
r
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
Circular motion
x = r cos q = r cos wt
The maximum value of x is
called the amplitude.
In SHM, it is usually
written as A:
x = A cos wt
y
→
a
m
q
→
r
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
General feature of simple harmonic
motion:
The frequency doesn’t depend on the
amplitude.
No matter how far you displace the object
from equilibrium, w = √ k/m
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
x = A cos wt with angular frequency w = √ k/m
T = 2p/w is the period of the motion.
A
x
0
t
T
-A
f = 1/T = w/2p is the frequency of the motion.
Units: 1/s = Hz.
Physics 203 – College Physics I
Department of Physics – The Citadel
Circular Motion Analogy
The velocity vector in
uniform circular motion
has magnitude v = rw
and is perpendicular to
the radius vector:
vx = - r w sin q
vy = -r w cos q, q = wt
For SHM, take the x
component, with r →A:
v = - A w sin wt
→
y v
q
m
→
q
r
x
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
v = - A w sin wt
x= A cos wt
Note that v = 0 at the turning points x = ± A.
The maximum speed is vmax = Aw, where x = 0.
A
x
v=0
0
t
vmax
-A
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
Acceleration:
a = F/m = - k x/m = - w 2 A cos wt
Note that a = 0 at the equilibrium points x = 0.
The maximum acceleration is amax = w2A
a = - w2A
A
x
0
t
a=0
-A
a = +w2A
Physics 203 – College Physics I
Department of Physics – The Citadel
Simple Harmonic Motion
Energy is conserved:
K = ½ mv2,
Total energy:
U = ½ kx2.
E = K + U = ½ kA2.
•The potential energy is maximum at the
turning points.
•The kinetic energy is maximum at the
equilibrium position.
Kmax = ½ mvmax2 = ½ kA2 → vmax = Aw.
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
A 3.0 kg object is attached to a spring with
k = 280 N/m and is oscillating in SHM. When it is
2.0 cm from equilibrium, it moves 0.55 m/s.
What is the frequency of the motion?
___
______________
w = √k/m = √(280 N/m) / 3.0 m = 9.66 s-1
f = w/2p = 1.5 Hz
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
A 3.0 kg object is attached to a spring with
k = 280 N/m and is oscillating in SHM. When it is
2.0 cm from equilibrium, it moves 0.55 m/s.
What is the amplitude of the motion?
K = ½ mv2,
Total energy:
U = ½ kx2.
E = K + U = ½ kA2 = ½ mv2 + ½ kx2
A2 = (m/k) v2 + x2 = (v/w)2 + x2
= [(0.55 m/s)/(9.66 s-1)]2 + (0.020 m)2 = 0.00364 m2
A = 0.060 m = 6.0 cm.
Physics 203 – College Physics I
Department of Physics – The Citadel
Example
A 3.0 kg object is attached to a spring with
k = 280 N/m and is oscillating in SHM. When it is
2.0 cm from equilibrium, it moves 0.55 m/s.
What is the maximum force on the object?
F = kA = (280 N/m) (0.060 m) = 17 N.
What is its maximum speed?
vmax = Aw = (0.060 m) (9.66 s-1) = 0.58 m/s