Download Mechanical Energy and Simple Harmonic Oscillator

Simple Harmonic Oscillator
8.01
Week 13D1
Today’s Reading Assignment Young
and Freedman: 14.1-14.6
1
Announcements
Math Review Tuesday Nov 29 from 9-11 pm
Pset 11 due Thursday Dec 1 at 9 pm
W013D2 Reading Assignment Young and Freedman:
14.1-14.6
2
Mass on a Spring C2:
Simple Harmonic Motion
3
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
Mass on a Spring
(1)
(3)
F  kx
(2)
(4)
5
Concept Question: Simple
Harmonic Oscillator
Which of the following functions x(t) has a second derivative
which is proportional to the negative of the function
d 2x
 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 Answer:
Simple Harmonic Oscillator
Answer 4. By direct calculation, when
 2 
x(t)  Acos 
t
 T 
 2 
 2 
dx(t)
    Asin 
t
dt
 T 
 T 
2
2
 2 
 2 
 2 
d x(t)
    Acos 
t      x(t)
2
 T 
 T 
 T 
dt
2
Demo slide: spray paint oscillator C4
Illustrating choice of alternative
representations for position as a
function of time (amplitude and phase
or sum of sin and cos)
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Simple Harmonic Motion
d 2x
x(t)  C cos( 0t)  D sin( 0t)
kx  m 2 
dt
 0  k / m  2 f  2 / T
x(t)  Acos( 0t   ) 
C  Acos( )
D   Asin( )
A  C 2  D2
tan    D / C9
Simple Harmonic Motion
Equation of Motion:
d 2x
kx  m 2
dt
Solution: Oscillatory with
Period
T  2 /  0  2 m / k
x  C cos( 0t)  D sin( 0t)
Position:
Velocity:
dx
vx 
  0C sin( 0t)   0 Dcos( 0t)
dt
Initial Position at t = 0:
x0  x(t  0)  C
Initial Velocity at t = 0:
vx,0  vx (t  0)   0 D
General Solution:
x  x0 cos( 0t)   0vx,0 sin( 0t)
Worked Example: Simple Harmonic
Motion Block-Spring
A block of mass m , attached
to a spring with spring
constant k, is free to slide
along a horizontal frictionless
surface. At t = 0 the blockspring system is stretched an
amount x0 > 0 from the
equilibrium position and
released from rest. What is the
x -component of the velocity of
the block when it first comes
back to the equilibrium?
Worked Example: Simple Harmonic
Motion Block-Spring
At t = 0, the block is at x(t = 0) = x0, and it is released from rest
so vx(t = 0) = 0. Therefore the initial conditions are
x(t  0)  C  x0
vx (t  0)   0 D  0
So the position and x-component of the velocity of the block
are given by of the block as a function of time is
x(t)  x0 cos( 0t)
vx (t)   0 x0 sin( 0t)
The particle first returns to equilibrium at t = T/4, therefore
sin( 0T / 4)  1  vx   0 x0  
 k / m x
0
Mass on a Spring: Energy
x(t)  Acos( 0t   )
vx (t)   0 Asin( 0t   )
0  k / m
A  x02  vx2 /  02
Constant energy oscillates between kinetic and potential energies
K(t)  (1 / 2)m(vx (t))2  (1 / 2)kA2 sin 2 ( 0t   )
U (t)  (1 / 2)kx  (1 / 2)kA cos ( 0t   )
2
2
2
E  K(t)  U (t)  (1 / 2)kA2  (1 / 2)kx02  (1 / 2)mvx2  constant
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Energy and Simple Harmonic Motion
E  K(t)  U(t)  (1/ 2)kx(t)2  (1/ 2)mvx (t)2
dE / dt  0 
dv x
dx
0  kx
 mv x
dt
dt
d 2x
0  kx  mx 2
dt
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Concept Question: Simple
Harmonic Motion Velocity
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 compressed 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
CQ Answer: Simple Harmonic
Motion Velocity
Answer 3. The particle starts with potential energy. When it
first returns to equilibrium it now has only kinetic energy. Since
the energy of the block-spring system is constant:
(1/ 2)mvx2  (1/ 2)kx02
Take positive root because object is moving in positive xdirection when it first comes back to equilibrium, and xo < 0, we
require that vx,o > 0, therefore
k
vx  
x0
m
Table Problem: 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?
Kinetic Energy vs. Potential Energy
State
Kinetic
energy
Potential
energy
Mechanical
energy
1 2
K0  mvx,0
2
1 2
U 0  kx0
2
E0  K 0  U 0
1 2
K f  mvx, f
2
1 2
U f  kx f
2
Initial
x0  0
vx,0  0
Final
xf  0
vx, f  0
E f  K f  U f  E0
Source
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Energy Diagram
Choose zero point for potential energy:
U (x  0)  0
Potential energy function:
1 2
U (x)  kx , U (x  0)  0
2
Mechanical energy is represented by a
horizontal line
E
mechanical
1 2 1 2
 K(x)  U(x)  mvx  kx
2
2
K  E mechanical  U
Graph of Potential energy function
U(x) vs. x
Concept Question: Harmonic
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
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Concept Question: Harmonic
Energy Diagram Answer
The position and x-component of velocity of a particle is given by
x(t )  D cos(t )  D sin t  , D  0
vx (t)   0 D sin( 0t)   0 D cos( 0t)
Hence
vx,0   0 D  0
Thus the particle starts out with a positive x- coordinate and the initial
x-component of the velocity is negative therefore it is moving toward
the origin. It has both non-zero potential and kinetic energies and
hence it is not at it’s maximal position so at it is located at position 4.
22
Oscillating Systems:
Simple Harmonic Motion
23
Demo: Air Spring with
Oscillating steel ball C12
24
Worked Example: fluid oscillations
in a U-tube
A U-tube open at both ends to
atmospheric pressure is filled with an
incompressible fluid of density r. The
cross-sectional area A of the tube is
uniform and the total length of the
column of fluid is L. A piston is used to
depress the height of the liquid column
on one side by a distance x, and then
is quickly removed. What is the
frequency of the ensuing simple
harmonic motion? Assume streamline
flow and no drag at the walls of the Utube. The gravitational constant is g.
25
Demonstration:
U-tube oscillations
26
Table Problem: Earthquake
In 1993, to try to protect its historic courthouse from damage due to earthquakes,
the city of San Francisco designed a special foundation. The building rests on N
identical supports. Each support has a small diameter spherical cap anchored to
the ground resting in an inverted spherical cup of radius R attached to the
building. As a result, if the building undergoes a small horizontal displacement x,
it rises by an amount h  x 2 / 2R as shown in the diagram. Building has
mass M.
a) Assume that there is no friction. Find the differential equation which describes
the dynamics of small displacements x(t) of the center of the cups from the
centers of the caps.
b) A sudden shock sets the building moving with the initial conditions x(0) = 0
and vx (0)  v0 . Find an expression for x(t) for all subsequent times.
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