PHYS 1443 – Section 002
Lecture #25
Wednesday, Dec. 5, 2007
Dr. Jae Yu
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Wednesday, Dec. 5, 2007
Simple Harmonic Motion
Equation of the SHM
Simple Block Spring System
Energy of the SHO
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
1
Announcements
• Final exam
– Date and time: 11am – 12:30 pm, Monday, Dec. 10
– Location: SH103
– Covers: CH9.1 – CH14.3
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2
Vibration or Oscillation
What are the things
that vibrate/oscillate?
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So what is a vibration or oscillation?
Tuning fork
A pendulum
A car going over a bump
Buildings and bridges
The spider web with a prey
A periodic motion that repeats over the same path.
A simplest case is a block attached at the end of a coil spring.
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
3
Simple Harmonic Motion
Motion that occurs by the force that depends on displacement, and the
force is always directed toward the system’s equilibrium position.
What is a system that has such characteristics?
A system consists of a mass and a spring
When a spring is stretched from its equilibrium position
by a length x, the force acting on the mass is
F  kx
It’s negative, because the force resists against the change of
length, directed toward the equilibrium position.
From Newton’s second law
F  ma  kx
we obtain
a

k
x
m
k
This is a second order differential equation that can d 2 x
Condition for simple


x
harmonic motion
be solved but it is beyond the scope of this class.
m
dt 2
Acceleration is proportional to displacement from the equilibrium.
What do you observe
Acceleration is opposite direction to displacement.
from this equation?
This system
is doing
a Fall
simple
Wednesday, Dec. 5, 2007
PHYS
1443-002,
2007 harmonic motion (SHM).
4
Dr. Jaehoon Yu
Equation of Simple Harmonic Motion
d 2x
k


x
2
dt
m
The solution for the 2nd order differential equation
x  A cos t   
Amplitude
Phase
Angular
Frequency
Phase
constant
Generalized
expression of a simple
harmonic motion
Let’s think about the meaning of this equation of motion
What happens when t=0 and =0?
What is  if x is not A at t=0?
x  A cos0  0  A
x  A cos   x'
  cos 1 x'
What are the maximum/minimum possible values of x?
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
A/-A
An oscillation is fully
characterized by its:
•Amplitude
•Period or frequency
•Phase constant
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Vibration or Oscillation Properties
The maximum displacement from
the equilibrium is
Amplitude
One cycle of the oscillation
The complete to-and-fro motion from an initial point
Period of the motion, T
The time it takes to complete one full cycle
Unit?
sec
Frequency of the motion, f
The number of complete cycles per second
Unit?
s-1
Relationship between
period and frequency?
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
f

1
T
or
T

1
f
6
More on Equation of Simple Harmonic Motion
What is the time for full
cycle of oscillation?
The period
Since after a full cycle the position must be the same
x  Acos t  T      A cost  2p   
T

How many full cycles of oscillation
does this undergo per unit time?
2p

One of the properties of an oscillatory motion
f
1 
 
T 2p
Frequency
Let’s now think about the object’s speed and acceleration.
What is the unit?
1/s=Hz
x  Acost   
dx
 Asin t    Max speed v
max  A
dt
Max acceleration
dv
2
2
Acceleration at any given time a    A cos  t      x a   2 A
dt
max
What do we learn
Acceleration is reverse direction to displacement
about acceleration?
Acceleration and speed are p/2 off phase of each other:
When v is maximum, a is at its minimum
Speed at any given time
Wednesday, Dec. 5, 2007
v

PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
7
Simple Harmonic Motion continued
Phase constant determines the starting position of a simple harmonic motion.
x  Acost   
x t 0  A cos 
At t=0
This constant is important when there are more than one harmonic oscillation
involved in the motion and to determine the overall effect of the composite motion
Let’s determine phase constant and amplitude
xi  A cos 
At t=0
vi  A sin 
 vi 

By taking the ratio, one can obtain the phase constant   tan  
 xi 
1
By squaring the two equations and adding
them together, one can obtain the amplitude

A cos   sin 
2
2
2
Wednesday, Dec. 5, 2007

2  x 2   vi 
A i   
xi2  A2 cos 2 
vi2   2 A2 sin 2 
2
A
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
 vi 
2
xi   
 
2
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Sinusoidal Behavior of SHM
What do you think the trajectory will look if the
oscillation was plotted against time?
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
9
Sinusoidal Behavior of SHM
x  A cos  2p ft 
v  v0 sin  2p ft 
a  a0 cos  2p ft 
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
10
Example for Simple Harmonic Motion
An object oscillates with simple harmonic motion along the x-axis. Its displacement from
the origin varies with time according to the equation; x  4.00mcospt  p  where t is in seconds
and the angles in the parentheses are in radians. a) Determine the amplitude, frequency,
and period of the motion.
p




4
.
00
m
cos
p
t




Acos

t




x


A  4.00m The angular frequency, , is   p
From the equation of motion:
The amplitude, A, is
Therefore, frequency
and period are
T

2p


2p
p
 2s
f

1

p
1


 s 1
T
p
p 2
b)Calculate the velocity and acceleration of the object at any time t.
Taking the first derivative on the
equation of motion, the velocity is
By the same token, taking the
second derivative of equation of
motion, the acceleration, a, is
Wednesday, Dec. 5, 2007
p
dx





4
.
00

p
sin
p
t

v

m / s

dt

d 2 x  4.00  p 2 cos pt  p m / s 2


a 2


dt
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
11
Simple Block-Spring System
A block attached at the end of a spring on a frictionless surface experiences
acceleration when the spring is displaced from an equilibrium position.
This becomes a second
order differential equation
If we    k
d 2x
k
  x denote
2
m
dt
m
2
d
x

The resulting differential equation becomes



x
2
dt
Fspring  ma
 kx
k
a x
m
x  A cost   
Since this satisfies condition for simple
harmonic motion, we can take the solution
Does this solution satisfy the differential equation?
Let’s take derivatives with respect to time dx  A d cost      sin t   
Now the second order derivative becomes
dt
dt
d
d 2x
sin t      2  cost      2 x




2
dt
dt
Whenever the force acting on an object is linearly proportional to the displacement from some
equilibrium position and is in the opposite direction, the particle moves in simple harmonic motion.
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
12
More Simple Block-Spring System
How do the period and frequency of this harmonic motion look?
Since the angular frequency  is
The period, T, becomes
So the frequency is
Special case #1
x   cost
f
T



2p


 2p
1

1


T
2p
2p
k
m
m
k
k
m
What can we learn from these?
•Frequency and period do not
depend on amplitude
•Period is inversely proportional
to spring constant and
proportional to mass
Let’s consider that the spring is stretched to a distance A, and the block is
let go from rest, giving 0 initial speed; xi=A, vi=0,
2
dx
d
x
v
  sin t a  2   2  cos t ai   2   kA/ m
dt
dt
This equation of motion satisfies all the conditions. So it is the solution for this motion.
Suppose a block is given non-zero initial velocity vi to positive x at the
instant it is at the equilibrium, xi=0
Is this a good
p
vi 

p
1 
1
  tan     
 tan  
x   cos t    A sin t  solution?
x 


Special case #2


i

Wednesday, Dec. 5, 2007


PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
13
Example for Spring Block System
A car with a mass of 1300kg is constructed so that its frame is supported by four
springs. Each spring has a force constant of 20,000N/m. If two people riding in the
car have a combined mass of 160kg, find the frequency of vibration of the car after it is
driven over a pothole in the road.
Let’s assume that mass is evenly distributed to all four springs.
The total mass of the system is 1460kg.
Therefore each spring supports 365kg each.
From the frequency relationship
based on Hook’s law
f
Thus the frequency for
1
f 
vibration of each spring is
2p

1

1


T
2p
2p
k
1

m
2p
k
m
20000
 1.18s 1  1.18Hz
365
How long does it take for the car to complete two full vibrations?
The period is
T 
Wednesday, Dec. 5, 2007
1
 2p
f
m
 0.849s
k
For two cycles
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
2T  1.70s
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Example for Spring Block System
A block with a mass of 200g is connected to a light spring for which the force constant is
5.00 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced
5.00 cm from equilibrium and released from reset. Find the period of its motion.
From the Hook’s law, we obtain

X=0
X=0.05
k

m

5.00
 5.00s 1
0.20
As we know, period does not depend on the
amplitude or phase constant of the oscillation,
therefore the period, T, is simply
T

2p


2p
 1.26 s
5.00
Determine the maximum speed of the block.
From the general expression of the
simple harmonic motion, the speed is
Wednesday, Dec. 5, 2007
dx
 Asin t   
dt
 A  5.00  0.05  0.25m/ s
vmax 
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
15
Energy of the Simple Harmonic Oscillator
What do you think the mechanical energy of the harmonic oscillator look without friction?
Kinetic energy of a
1
1
2

mv

m 2  2 sin 2 t   
KE
harmonic oscillator is
2
2
The elastic potential energy stored in the spring
PE

1 2 1
kx  k 2 cos 2 t   
2
2
Therefore the total
1
2 2
2
2
2
mechanical energy of the E  KE  PE  m  sin t     k cos t   
2
harmonic oscillator is




1
1 2
2
2
2
2
k







k

sin

t



k

cos

t



kA
Since
m E  KE  PE
2
2
Total mechanical energy of a simple harmonic oscillator is proportional to the square of the amplitude.
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
16
Energy of the Simple Harmonic Oscillator cont’d
1
1
2
KEmax  mvmax  k 2
2
2
vmax  k A
m
Maximum KE is
when PE=0
Maximum speed
The speed at any given
point of the oscillation
E
v
1
1 2 1
2
2

mv

kx

k

 KE  PE
2
2
2

 k m A x
2
2

 vmax
KE/PE
-A
Wednesday, Dec. 5, 2007
x
1  
 A
2
E=KE+PE=kA2/2
A
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
x
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Oscillation Properties
Amplitude?
•
•
•
•
•
Wednesday, Dec. 5, 2007
A
When is the force greatest?
When is the speed greatest?
When is the acceleration greatest?
When is the potential energy greatest?
When is the kinetic energy greatest?
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
18
Example for Energy of Simple Harmonic Oscillator
A 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a
horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of
the cube if the amplitude of the motion is 3.00 cm.
k  20.0N / m
A  3.00cm  0.03m
1 2 1
2
3
 KE  PE  kA  20.0 0.03  9.00 10 J
2
2
From the problem statement, A and k are
The total energy of
the cube is
E
E
1
1 2
 kA2
mvmax
2
2
Maximum speed occurs when kinetic
energy is the same as the total energy
vmax  A k  0.03 20.0  0.190m / s
m
0.500
KEmax 
b) What is the velocity of the cube when the displacement is 2.00 cm.
velocity at any given
displacement is
v




k m A2  x 2  20.0  0.032  0.02 2 / 0.500  0.141m / s
c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.
1 2 1
Potential
2
3
Kinetic
KE  mv  0.500  0.141  4.97 10 J
2
2
energy, PE
energy, KE
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
PE 
1 2 1
2
kx  20.0  0.02  4.00 10 3 J
2
2
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Congratulations!!!!
You all have done very well!!!
I certainly had a lot of fun with ya’ll
and am truly proud of you!
Good luck with your exam!!!
Have safe holidays!!
Wednesday, Dec. 5, 2007
PHYS 1443-002, Fall 2007
Dr. Jaehoon Yu
20