Transcript Ch. 13

Chapter 13
Periodic Motion
Special Case:
Simple Harmonic Motion (SHM)
Simple Harmonic Motion (SHM)
• Only valid for small oscillation amplitude
• But SHM approximates a wide class of
periodic motion, from vibrating atoms to
vibrating tuning forks...
Starting Model for SHM:
mass m attached to a spring
Demonstration
Simple Harmonic Motion (SHM)
• x = displacement of mass m
from equilibrium
• Choose coordinate x so that
x = 0 is the equilibrium position
• If we displace the mass m, a restoring force
F acts on m to return it to equilibrium (x=0)
Simple Harmonic Motion (SHM)
• By ‘SHM’ we mean Hooke’s Law holds:
for small displacement x (from equilibrium),
F=–kx
ma = – k x
• negative sign: F is a ‘restoring’ force
(a and x have opposite directions)
Demonstration: spring with force meter
What is x(t) for SHM?
• We’ll explore this using two methods
• The ‘reference circle’:
x(t) = projection of certain circular motion
• A little math:
Solve Hooke’s Law
d 2x
F  m 2  kx
dt
The ‘Reference Circle’
P = mass on spring: x(t)
Q = point on reference circle
P = projection of Q
onto the screen
The ‘Reference Circle’
P = mass on spring: x(t)
Q = point on reference circle
A = amplitude of x(t)
(motion of P)
A = radius of reference circle
(motion of Q)
The ‘Reference Circle’
P = mass on spring: x(t)
Q = point on reference circle
f = oscillation frequency of P
= 1/T (cycles/sec)
w = angular speed of Q
= 2p /T (radians/sec)
w = 2p f
What is x(t) for SHM?
See additional notes or Fig. 13-4 for q
P = projection of Q onto screen.
We conclude the motion of P is:
x(t )  A cos q
q (t )  wt  
Alternative: A Little Math
2
• Solve Hooke’s Law:
d x
F  m 2  kx
dt
• Find a basic solution:
x(t )  A cos(wt   )
k
w
m
Solve for x(t)
See notes on x(t), v(t), a(t)
x(t )  A cos(wt  p / 3)
w  k / m  2p / T
• v = dx/dt
v=0
at x = A
|v| = max at x = 0
• a = dv/dt
|a| = max at x = A
a =0
at x = 0
Show expression for 
x(t )  A cos(wt   )
x0  A cos 
 v0 
  arctan 

 wx0 
x(t )  A cos(wt  0)
w  k / m  2p / T
• going from 1 to 3,
increase one of A, m, k
• (a) change A : same T
• (b) larger m : larger T
• (c) larger k : shorter T
Do demonstrations illustrating (a), (b), (c)
Summary of SHM
for an oscillator of mass m
F  kx
x(t )  A cos(wt   )
k
w
m
• A = amplitude of motion,  = ‘phase angle’
• A,  can be found from the values
of x and dx/dt at (say) t = 0
Energy in SHM
E  K U
1 2 1 2
 mv  kx
2
2
• As the body oscillates,
E is continuously
transformed from K to
U and back again
See notes on vmax
E = K + U = constant
Do Exercise 13-17
Summary of SHM
x(t )  A cos(wt   )
2
d x
2


w
x
2
dt
2p
w  2pf 
T
• x = displacement from equilibrium (x = 0)
• T = period of oscillation
• definitions of x and w depend on the SHM
Different Types of SHM
• horizontal (have been discussing so far)
• vertical (will see: acts like horizontal)
• swinging (pendulum)
• twisting (torsion pendulum)
• radial (example: atomic vibrations)
Horizontal SHM
Horizontal SHM
d 2x
2
 w x
2
dt
k
w block-spring 
m
• Now show: a vertical spring acts the same,
if we define x properly.
Vertical SHM
Show SHM occurs with x defined as shown
Do Exercise 13-25
‘Swinging’ SHM:
Simple Pendulum
2
d x
2


w
x
2
dt
w simple pendulum 
Derive w for small x
g
L
Do Pendulum Demonstrations
‘Swinging’ SHM:
Physical Pendulum
d 2q
2
 w q
2
dt
mgd
w physicalpendulum 
I
Derive w for small q
Do Exercises 13-39, 13-38
Angular SHM:
Torsion Pendulum (fiber-disk)
Application: Cavendish experiment (measures
gravitational constant G). The fiber twists
when blue masses gravitate toward red masses
Angular SHM:
Torsion Pendulum (coil-wheel)
d 2q
2
 w q
2
dt
w torsion pendulum 
Derive w for small q

I
Radial SHM:
Atomic Vibrations
Show SHM results for small x (where r = R0+x)
Announcements
• Homework Sets 1 and 2 (Ch. 10 and 11):
returned at front
• Homework Set 5 (Ch. 14):
available at front, or on course webpages
• Recent changes to classweb access:
see HW 5 sheet at front, or course webpages
Damped
Simple Harmonic Motion
See transparency on damped block-spring
SHM: Ideal vs. Damped
• Ideal SHM:
• We have only treated the restoring force:
• Frestoring = – kx
• More realistic SHM:
• We should add some ‘damping’ force:
• Fdamping = – bv
Demonstration of damped block-spring
Damping Force
Fdamping
dx
 bv  b
dt
• this is the simplest model:
• damping force proportional to velocity
• b = ‘damping constant’
(characterizes strength of damping)
SHM: Ideal vs. Damped
• In ideal SHM, oscillator energy is constant:
E=K+U
,
dE/dt = 0
• In damped SHM, the oscillator’s energy
decreases with time:
E(t) = K + U
,
dE/dt < 0
Energy Dissipation
in Damped SHM
• Rate of energy loss due to damping:
dE
 Fdamping v
dt
 (bv)v
 bv
0
2
What is x(t) for damped SHM?
• We get a new equation of motion for x(t):
ma  Frestoring  Fdamping
2
d x
dx
m 2  kx  b
dt
dt
• We won’t solve it, just present the solutions.
Three Classes of Damping, b
d 2x
dx
m 2  kx  b
dt
dt
• small (‘underdamping’)
b  2 mk
• intermediate (‘critical’ damping)
b  2 mk
• large (‘overdamping’)
b  2 mk
‘underdamped’ SHM
‘underdamped’ SHM:
damped oscillation, frequency w´
b  2 mk
x(t )  Ae(b / 2 m )t cos(w t   )
2
k
b
w 

m 4m 2
‘underdamping’ vs. no damping
• underdamping:
• no damping (b=0):
2
w 
k
b

m 4m 2
w
k
m
w  w
‘critical damping’:
decay to x = 0, no oscillation
b  2 mk
x(t )  ( A  Bt )e (b / 2 m )t
• can also view this ‘critical’ value of b as
resulting from oscillation ‘disappearing’:
2
k
b
0

 w
2
m 4m
See sketch of x(t) for critical damping
‘overdamping’:
slower decay to x = 0, no oscillation
b  2 mk
x(t )  Ae(b / 2 m )t cosh(w overt   )
w over
2
b
k


2
4m m
See sketch of x(t) for overdamping
( a frequency! )
Application
• Shock absorbers:
• want critically damped
(no oscillations)
• not overdamped
(would have a
slow response time)
Forced Oscillations
(Forced SHM)
Forced SHM
• We have considered the presence of a
‘damping’ force acting on an oscillator:
Fdamping = – bv
• Now consider applying an external force:
Fdriving = Fmax coswdt
Forced SHM
• Every simple harmonic oscillator has a
natural oscillation frequency
• (w if undamped, w´ if underdamped)
• By appling Fdriving = Fmax coswdt we force
the oscillator to oscillate at the frequency wd
(can be anything, not necessarily w or w´)
What is x(t) for forced SHM?
• We get a new equation of motion for x(t):
ma  Frestoring  Fdamping  Fdriving
2
d x
dx
m 2  kx  b  Fmax cos w d t
dt
dt
• We won’t solve it, just present the solution.
x(t) for Forced SHM
• If you solve the differential equation, you
find the solution (at late times, t >> 2m/b)
x(t )  A cos(w d t   )
A
Fmax
(mw d2  k ) 2  (bw d ) 2
Amplitude A(wd)
• Shown (for  = 0):
A(wd) for different b
• larger b: smaller Amax
• Resonance:
Amax occurs at wR, near
the natural frequency,
w = (k/m)1/2
Do Resonance Demonstrations
Resonance Frequency (wR)
A
Fmax
(mw d2  k ) 2  (bw d ) 2
• Amax occurs at wd=wR (where dA/dwd=0):
k
b2
wR 

m 2m 2
natural, underdamped, forced:
w > w´ > wR
• natural frequency:
• underdamped frequency:
• resonance frequency:
w
k
m
w 
k
b2

m 4m 2
wR 
k
b2

m 2m 2
Introduction to
LRC Circuits
(Electromagnetic Oscillations)
See transparency on LRC circuit
Electric Quantity
Counterpart
• charge Q(t)
x(t)
• current I = dQ/dt
(moving charge)
(generates a magnetic field, B)
v = dx/dt
Electrical Concepts
• electric charge:
Q
• current (moving charge):
I = dQ/dt
• resistance (Q collides with atoms): R
• voltage (pushes Q through wire):
V = RI
Voltage (moves charges)
• resistance R causes charge Q to lose energy:
V = RI
• (voltage = potential energy per unit charge)
• C and L also cause energy (voltage) changes
Circuit Element
D(Voltage)
• R = resistance
(Q collides with atoms)
VR = RI
• C = capacitance
(capacity to store Q on plate)
VC = Q/C
• L = inductance
VL = L(dI/dt)
(inertia towards changes in I)
Change in Voltage = Change in Energy
• voltage = potential energy per unit charge
• recall, around a closed loop:
D( PE )  0
VL  VC  VR  0
VL  VC  VR  0
dI 1
L  Q  RI  0
dt C
d 2Q 1
dQ
L 2  QR
0
dt
C
dt
• Which looks like:
d 2x
dx
m 2  kx  b  0
dt
dt
Circuit Element
• 1/C = 1/capacitance
• L = inductance
• R = resistance
Counterpart
k
m
b
• (Extra Credit: Exercise 31-35)
• Use this table to write our damped SHM as
damped electromagnetic oscillations
In the LRC circuit, Q(t) acts just like x(t)!
underdamped, critically damped, overdamped
Driven (and resonance): Vdriving = Vmax coswdt