Chapter 13: Oscillatory Motions
Simple harmonic motion
 Spring
and Hooke’s law
• When a mass hanging from a spring and
in equilibrium, the Newton’s 2nd law says:
F
y
y
 ma  Fs  Fg  0  Fs  Fg
This means the force due to the spring is
equal to the force by gravity and opposite
in direction when the spring is stretched.
• Hooke’s law states that increasing the
weight by equal amounts increases the
stretch of the spring by equal amount.
Therefore, the force due to the spring must be proportional to the
stretch of the spring.
x is deviation from the spring w/o weight


Fs  kx where k is the spring constant.
This is also true when the spring shrinks
Simple harmonic motion
 Simple
harmonic motion (SHM)
• Let’s study a motion of the
mass m. When the mass is
attached to the spring, the
spring stretches by x0. Then
lift the mass by A and release
it.
F
y
y
y
y
 ma  Fs  Fg  ma
• The initial stretch is x0-x and from Hooke’s law:
k ( x0  x)  mg  ma
• Since in equilibrium Fs  Fg  0  kx0  mg ,
dv d  dx  d 2 x
 kx  ma  a  (k / m) x and a 
   2
dt dt  dt  dt
d 2x
k
2


x



x Equation for SHM
2
dt
m
Simple harmonic motion
 Simple
harmonic motion (SHM) (cont’d)
d 2x
k
2


x



x 
2
dt
m
k
: angular frequency (rad/s)
m
phase constant
• Solution:
x(t )  A cos(t   )
As x(0)  A cos   x0 ,   arccos
x0
A
Frequency : f   / 2 , period : T  1 / f , amplitude : A
Hz
s
dx(t )
v(t ) 
 A sin( t   ), v( x)   A2  x 2
dt
dv(t )
a (t ) 
  2 A cos(t   ), a ( x)   2 x
dt
velocity
acceleration
Simple harmonic motion
 Simple
harmonic motion (SHM) (cont’d)
f=/(2)
• Solution:
x(t )  A cos(t   )
Acos
• What is SHM/SHO?
t=-/ t=0
A simple harmonic motion is the motion
0; 20 ; 30
of an oscillating system which satisfies
the following condition:
1. Motion is about an equilibrium position
at which point no net force acts on the
system.
2. The restoring force is proportional to
and oppositely directed to the
displacement.
3. Motion is periodic.
By Dr. Dan Russell, Kettering University
Simple harmonic motion
 Connection
between SHM and circular motion
• For an object in circular motion, the angular
velocity is defined as,

d
   t  
dt
• The tangential velocity is related to the
angular velocity : v  r
• The centripetal acceleration is also related
to the angular velocity:
v 2 (r ) 2
a

 r 2
r
r

Vector r is called a phasor
• The position, velocity and acceleration
of the object as a function of time are:
SHM! x  r cos   x(t )  r cos(t   )
v  r sin   v(t )  r sin( t   )
a   2 r cos   a(t )   2 r cos(t   )   2 x(t )
Simple harmonic motion
 Displacement,
velocity and acceleration in SHM
• Displacement
x(t )  A cos(t )
• Velocity
v(t ) 
dx(t )
 A sin( t )
dt
• Acceleration
a (t ) 
Note:
dv(t )
  2 A cos(t )
dt
A  x 2 (0)  v 2 (0) /  2
 0
Energy in SHM
 Energy
conservation
dU
Fs   kx  
dx
Ch.7
E  K U
x
1 2
U  U    Fs dx  kx
 Energy conservation in a SHM
0
2
No friction
1 2
1 2
K  mv ; U  kx
2
2
1 2 1 2
E  mv  kx  const.
2
2
BTW:
1 2 1 2 1 2
kA  mv  kx  E
2
2
2
2
(
k 2
v
A  x2
m
)
Energy in SHM
 Energy
conservation in a SHM (cont’d)
1 2 1 2
E  mv  kx  const.
2
2
kinetic energy
energy
energy
E
distance from equilibrium point
potential energy
Time
Applications of SHM
 Simple
pendulum
• The forces on the mass at the end are
gravity and the tension. The tension, however,
exerts no torque about the top of the string.
g

g

  I  mg sin   m 2     sin     for small angle
d d 2
g

 2  
dt
dt

Angular frequency of a simple pendulum

g

1
,f 


2 2
g
1

, T   2

f
g
mg
Applications of SHM
 Physical
pendulum
• A simple pendulum has all its mass concentrated
at a point and oscillates due to gravitational torques.
Objects that do not have their mass concentrated at
a point also oscillate due to gravitational torques.
mgr
  I  mgr sin   I     I sin 
Angular frequency of a physical pendulum
mgr

I
Applications of SHM
 Angular
SHM
• An angular version of SHM is called torsion
oscillation and shown on the right.
• A disk suspended by a wire experiences a
restoring torsion when rotated by a small
angle  :
  
c.f. F  kx
: torsion constant
d 2

   I  2   
dt
I
Angular frequency of an angular SHM :


I
k
c.f.  
m
Damped oscillations
 Oscillation
with friction
• In real world dissipative forces such as friction between a block and
a table exist. Such a dissipative force will decrease the amplitude of an
oscillation – damped oscillation.
The friction reduces the mechanical energy of the system as time
passes, and the motion is said to be damped.
Damped oscillations
 A simple
example of damped oscillation
• Consider a simple harmonic oscillation with a frictional damping
force that is directly proportional to the velocity of the oscillating
object.
F  kx  bv  ma
2
dx
d x
 kx  b  m 2
dt
dt
If the damping force is relatively small, the motion is described by:
x(t )  Ae (b / 2 m )t cos( ' t   ) where
k
b2
' 

m 4m 2
Damped oscillations (cont’d)
 A simple
example of damped oscillation
x(t )  Ae (b / 2 m )t cos( ' t   ) where
k
b2
' 

2
m 4m
Ae
 (b / 2 m )t
By Dr. Dan Russell, Kettering University
Forced oscillations and resonance
 Driving
force
An example of resonantly driven damped harmonic oscillator
Push
Wait 1
period
Forced oscillations and resonance
 Driving
force (cont’d)
• The additional force that pushed by the person in the animation
on the previous page is called a driving force.
• When a periodically varying driving force with angular frequency
d is applied to a damped harmonic oscillator, the resulting motion
is called a forced oscillation.
driving force : cos(dt )
d=0.4
d=1.01
By Dr. Dan Russell, Kettering University
d=1.6
Forced oscillations and resonance
 Forced
oscillation and resonance
Damped SHM
Fixed
natural frequency
k
b2
' 

m 4m 2
Damped
Forced damped SHM
Moving/driving force F (t )  Fmax cos d t
Forced oscillations and resonance
 Forced
oscillation and resonance (cont’d)
Amplitude for a forced damped oscillation:
2
Fmax
When
k

m

, A has a maximum
d
A
(k  md2 ) 2  b 2d2
near d  k / m resonance:
The fact that there is
an amplitude peak at
driving frequencies close
to the natural frequency
of the system is called
resonance
A
natural frequency
angular freq. of driving force
Exercises
 Problem
1
a) The speed of the pan and the steak
immediately after the collision (total
inelastic collision):
k=400 N/m
Initial speed of the meat just before the
collision: v  2 gh
i
Final speed of the meat-pan just after the
M=2.2 kg
collision:
M
v f  vi
mM
h=0.40 m
2 .2
2
 2(9.80 m/s )(0.40 m)
 2.6 m/s
2 .4
m=0.20 kg
Exercises
 Problem
1(cont’d)
b) The amplitude of the subsequent motion:
When the steak hits the pan, the pan is
Mg/k above the new equilibrium position.
k=400 N/m
v f   A2  x 2f  A2  [ x 2f  v 2f /  2 ]2
where x f  Mg / k ,   k /( m  M )
So the amplitude is:
2
2 ghM 2
 Mg 
A 
 0.21 m.
 
k
k
(
m

M
)


c) The period:
(m  M )
T  2
 0.49 s.
k
M=2.2 kg
h=0.40 m
m=0.20 kg
Exercises
 Problem
2
k
Ma   f  kx ( f : friction)
fR  I cm , I cm  (1 / 2) MR 2
a  R
Each:M/2, R
cylinders
rolls w/o
slipping
kx
k
2
a


x



x.
2
M  I cm / R
(3 / 2) M
stretched by x
and then released
kx

f
T
2

 2
3M
.
2k
Exercises
 Problem
3
Two identical, thin rods, each with mass m
and length L, are joined at right angles to
form an L-shaped object. This object is
balanced on top of a sharp edge. If the Lshaped object is deflected slightly, it oscillates.
Find the frequency of the oscillation.
L
L
Solution:
The moment of inertia about the pivot:
2(1 / 3)mL2  (2 / 3)mL2
The center of gravity is located when balanced at a distance
d  L /( 2 2 )
d  L /( 2 2 ) below the pivot.
Think the L-shaped object as a physical
pendulum and is represented by the center
of gravity. The period T is:
T  2
I
mgd
L
L
Exercises
 Problem
4
Find the effective spring constant.
F1  k1 x1  F2  k2 x2  F
F1=-k1x1
F2=-k2x2
F  keff x, x  x1  x2
x
F
F F
 x1  x2   
keff
k1 k2
keff 
k1k 2
k1  k 2
F  F1  F2
F1=-k1x
F  keff x, F1  k1 x, F2  k 2 x
keff  k1  k 2
F2=-k2x