Mechanical Energy and
Simple Harmonic Oscillator
8.01
Week 09D1 2006
Change in Mechanical Energy
Ftotal  Fc total  Fnc total
Total force:
final
Total work:
W total 

initial
final
F total  dr 
 F
total
c

 Fnctotal  dr
initial
W total  U total  Wnc
Total work done is change in kinetic energy:
W total  K
Mechanical Energy Change:
K  U total  Wnc
E mechanical  K  U total  Wnc
Modeling the Motion
Choose initial and final states.
Draw all relevant free body force diagrams
Identify non-conservative forces.
Calculate non-conservative work
final
Wnc   Fnc  dr .
initial
Choose zero point P for potential energy for each interaction
in which potential energy difference is well-defined.
Mechanical Energy Accounting
Initial state:
 Total initial kinetic energy
 Total initial potential energy
 Total initial Mechanical energy
Kinitial  K1,initial  K2,initial 
Uinitial  U1,initial  U2,initial 
mechanical
Einitial
 Kinitial  Uinitial
Final state:
 Total final kinetic energy
Kfinal  K1,final  K2,final 
 Total final potential energy
Ufinal  U1,final  U2,final 
 Total final mechanical energy
 Apply Energy Law:
mechanical
Efinal
 Kfinal  Ufinal
mechanical
mechanical
Wnc  Efinal
 Einitial
Example: Energy Changes
A small point like object of mass m rests on top of a sphere of radius
R. The object is released from the top of the sphere with a negligible
speed and it slowly starts to slide. Find an expression for the angle 
with respect to the vertical at which the object just loses contact with
the sphere.
Kinitial
K final 
0
Uinitial  0
1 2
mv f
2
Ufinal  mgR(1 cos f )
mechanical
Einitial
0
mechanical
Efinal

1 2
mv f  mgR(1  cos f )
2
1 2

Wnc  0  0  0   mv f  mgR(1  cos f ) 
2
1 2
mv f  mgR(1  cos f )
2
Recall Modeling the Motion:
Newton ‘s Second Law
 Define system, choose coordinate system.
 Draw force diagram.
 Newton’s Second Law for each direction.
 Example: x-direction
ˆi : F total
x
 Example: Circular motion
rˆ : F
d 2x
m 2 .
dt
total
r
v2
 m .
R
Example (con’t): Free Body
Force Diagram
Newton’s Second Law
v2
rˆ : N  mg cos   m
R
2
d

θˆ : mg cos  mR 2
dt
Constraint condition:
N  0 at    f
Radial Equation becomes
mg cos f  m
v 2f
R
Energy Condition:
Conclusion:

1 2 R
mv f  mg cos f
2
2
1 2
mv f  mgR(1  cos f )
2
R
mgR(1  cos f )  mg cos f 
2
2
1  2 
cos f 
  f  cos  
3
 3
Group Problem: Block-Spring
System with Friction
A block of mass m slides along a horizontal surface with
speed v 0. At t =0 it hits a spring with spring constant k
and begins to experience a friction force. The coefficient
of friction is variable and is given by µk= bv where b is a
constant. Find the loss in mechanical energy when the
block has first come momentarily to rest.
Simple Harmonic Motion
Hooke’s Law
Define system, choose coordinate system.
Draw free-body diagram.
Hooke’s Law
Fspring  kx ˆi
2
d x
kx  m 2
dt
Concept Question
Which of the following functions x(t) has a second
derivative which is proportional to the negative of the
function
2
d x
 x ?
2
dt
1.
2.
3.
4.
1 2
x(t)  at
2
x(t)  Aet /T
x(t)  Aet /T
 2 
x(t)  Acos  t 
 T 
Concept Question
The first derivative vx  dx / dt of the sinusoidal
function
 2 
is:
x  Acos  t 
 T 
1.
 2 
vx (t)  Acos  t 
 T 
2.
 2 
vx (t)   Asin  t 
 T 
3.
 2 
2
vx (t)  
Asin  t 
T
 T 
4.
 2 
2
vx (t) 
Acos  t 
T
 T 
Simple Harmonic Motion
Equation of Motion:
d 2x
kx  m 2
dt
Solution: Oscillatory with Period T
Position:
 2 
 2 
x  Acos  t   Bsin  t 
 T 
 T 
 2  2
 2 
dx
2
vx 

Asin  t  
Bcos  t 
dt
T
 T  T
 T 
Initial Position at t = 0: x0  x(t  0)  A
2
Initial Velocity at t = 0: vx,0  vx (t  0)  T B
Velocity:
General Solution:
 2  T
 2 
x  x0 cos  t  
vx,0 sin  t 
 T  2
 T 
Period and Angular Frequency
d 2x
kx  m 2
dt
 2 
 2 
x  Acos  t   Bsin  t 
 T 
 T 
Equation of Motion:
Solution: Oscillatory with Period T
x -component of velocity:
vx 
 2  2
 2 
dx
2

Asin  t  
cos  t 
dt
T
 T  T
 T 
x -component of acceleration:
ax 
Period:
2
2
2
 2 
 2   2 
 2 
 2 
d x


Acos
t

Bsin
t


x
2










 T 
 T   T 
 T 
 T 
dt
2
2
2
 2 
 2 
d x
1
kx  m 2  m   x  k  m    T 
2
 T 
 T 
dt
2
Angular frequency
2
k


T
m
m
k
Example: Block-Spring System with
No Friction
A block of mass m slides along a frictionless horizontal
surface with speed v x,0. At t = 0 it hits a spring with spring
constant k and begins to slow down. How far is the spring
compressed when the block has first come momentarily
to rest?
Modeling the Motion: Energy
Choose initial and final states:
Change in potential energy:
1
U ( x f )  U ( x0 )  k  x 2f  x02 
2
Choose zero point for potential energy:
Potential energy function:
U(x  0)  0
1 2
U (x)  kx , U (x  0)  0
2
Mechanical energy is constant (Wnc  0)
mechanical
mechanical
Efinal
 Einitial
Initial and Final Conditions
2
B
Initial state: x0  A  0 and vx,0 
T
 2 
T


x(t) 
vx,0 sin  t  vx (t)  vx,0 cos 2 t
 T 
2
 T 
First comes to rest when v (t )  0  2 t  
x f
T f 2
Since at time t f  T / 4
Final position
 2 

sin 
t f   sin    1 
 T 
 2
T
m
x(t f ) 
vx,0 
vx,0
2
k
Kinetic Energy vs. Potential Energy
State
Kinetic
energy
Potential
energy
Mechanical
energy
Initial
x0  0
vx,0  0
1 2
K 0  mvx,0
2
U0  0
1 2
E0  mvx,0
2
Kf  0
1 2
U f  kx f
2
1 2
E f  kx f
2
Final
xf  0
vx, f  0
Conservation of Mechanical Energy
E f  E0
1 2 1 2

kx f  mvx ,0
2
2
The amount the spring has compresses when
the object first comes to rest is
xf 
m
vx,0
k
Concept Question: Simple
Harmonic Motion
A block of mass m is attached to a spring with spring constant
k is free to slide along a horizontal frictionless surface.
At t = 0 the block-spring system is stretched an amount x0 > 0
from the equilibrium position and is released from rest. What is
the x -component of the velocity of the block when it first
comes back to the equilibrium?
1.
T
vx  x0
4
2.
3.
k
vx  
x0
m
4.
T
vx  x0
4
vx 
k
x0
m
Energy Diagram
Choose zero point for potential energy:
U(x  0)  0
Potential energy function:
U (x) 
1 2
kx , U (x  0)  0
2
Mechanical energy is represented by a
horizontal line since it is a constant
E
mechanical
1 2 1 2
 K(x)  U(x)  mvx  kx
2
2
Kinetic energy is difference between
mechanical energy and potential energy
(independent of choice of zero point)
K  E mechanical  U
Graph of Potential energy function
U(x) vs. x
Concept Question: Energy
Diagram
The position of a particle is given by
x(t )  D cos(t )  D sin t  , D  0
Where was the particle at t = 0?
1)
2)
3)
4)
5)
6)
7)
1
2
3
4
5
1 or 5
2 or 4
Concept Question: Energy
Diagram 1
A particle with total mechanical energy E has
position x > 0 at t = 0
1) escapes
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
Concept Question: Energy
Diagram 2
A particle with total mechanical energy E has
position x > 0 at t = 0
1) escapes
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
Concept Question: Energy
Diagram 3
A particle with total mechanical energy E has
position x > 0 at t = 0
1) escapes
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
Concept Question: Energy
Diagram 4
A particle with total mechanical energy E has
position x > 0 at t = 0
1) escapes
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information
Concept Question: Energy
Diagram 5
A particle with total mechanical energy E has
position x > 0 at t = 0
1) escapes
2) approximates simple harmonic motion
3) oscillates around a
4) oscillates around b
5) periodically revisits a and b
6) not enough information