Transcript Simple Harmonic Motion - AdvancedPlacementPhysicsC

Oscillations and Simple
Harmonic Motion:
Mechanics C
Oscillatory Motion is repetitive back and forth
motion about an equilibrium position
Oscillatory Motion is periodic.
Swinging motion and vibrations are forms of
Oscillatory Motion.
Objects that undergo Oscillatory Motion are
called Oscillators.
Oscillatory Motion
The time to complete
one full cycle of
oscillation is a Period.
1
T
f

1
f
T
The amount of
oscillations per second
is called frequency and
is measured in Hertz.

Simple Harmonic Motion
What is the oscillation period for the
broadcast of a 100MHz FM radio station?
1
1
8
T 
110 s 10ns
8
f 110 Hz
Heinrich Hertz produced the
first artificial radio waves back
in 1887!
The most basic of all types
of oscillation is depicted on
the bottom sinusoidal
graph. Motion that follows
this pattern is called simple
harmonic motion or SHM.
Simple Harmonic Motion
An objects maximum
displacement from its
equilibrium position is
called the Amplitude (A) of
the motion.
Simple Harmonic Motion
Everywhere the slope (first
derivative) of the position
graph is zero, the velocity
graph crosses through zero.
What shape will a
velocity-time graph
have for SHM?
We need a position function
to describe the motion
above.
 2 t 
x  t   A cos 

 T 
 2 t 
x  t   A cos 

 T 
T
1
f
x  t   A cos  2 ft 
2

T
x(t) to symbolize
position as a function of
time
A=xmax=xmin
When t=T,
cos(2π)=cos(0)
x(t)=A
x  t   A cos t 
Mathematical Models of SHM
v t  
x  t   A cos t 
In this context we will
d  x t 
call omega Angular
dt
Frequency
v  t    A sin t 
What is the physical meaning of the product (Aω)?
The maximum speed of an
oscillation!
vmax  A
Mathematical Models of SHM
Recall: Hooke’s Law
Here is what we want to do: DERIVE AN
EXPRESSION THAT DEFINES THE
DISPLACEMENT FROM EQUILIBRIUM OF
THE SPRING IN TERMS OF TIME.
Fspring  kx
FNet  ma
d 2x
 kx  ma a 
dt
d 2x
 kx  m
dt
d 2x
k
 ( )x  0
dt
m
WHAT DOES THIS MEAN? THE SECOND DERIVATIVE OF A FUNCTION
THAT IS ADDED TO A CONSTANT TIMES ITSELF IS EQUAL TO ZERO.
What kind of function will ALWAYS do this?
An airtrack glider is attached to a
spring, pulled 20cm to the right, and
released at t=0s. It makes 15
oscillations in 10 seconds.
What is the period of oscillation?
f 
15 oscilations
10sec
1
 1.5Hz 
T
1
1
T 
 0.67 s
f 1.5Hz
Example:
An airtrack glider is attached to a
spring, pulled 20cm to the right, and
released at t=0s. It makes 15
oscillations in 10 seconds.
What is the object’s maximum speed?
vmax
A2
 A 
T
Example:
0.2m  2

vmax 
 0.67 s 
 1.88m / s
An airtrack glider is attached to a
spring, pulled 20cm to the right, and
released at t=0s. It makes 15
oscillations in 10 seconds.
What are the position and velocity at
t=0.8s?
x  t   A cos t    0.2m cos   0.8s    0.0625m
v  t    A sin t     0.2m  sin   0.8s    1.79m / s
Example:
A mass oscillating in SHM starts at
x=A and has period T. At what time,
as a fraction of T, does the object first
pass through 0.5A?
 2 t 
x  t   A cos 

 T 
x(t )  0.5 A
Example:
 2 t 
0.5 A  A cos 

 T 
T
cos 1  0.5  t
2
T
2
 
 t
3
T
t
6
x(t)  Acost 
When collecting and modeling data of SHM your
mathematical model had a value as shown below:
x(t)  Acost  C
 if your clock didn’t start at x=A or x=-A?
What
This value represents our initial conditions.
We call it the phase angle:

x(t)  Acost  
Model of SHM
Uniform circular
motion projected onto
one dimension is
simple harmonic
motion.
SHM and Circular Motion
Start with the x-component of
position of the particle in UCM
x(t)  Acos
Notice it started at angle zero


d
  t

dt
x(t)  Acost 
End
with the same result as
the spring in SHM!

SHM and Circular
Motion
We will not always start our
clocks at one amplitude.
  t  0
x(t)  Acost   0 
v x (t)  Asin t   0 
v x (t)  v max sin t   0 
Initial conditions:
Phi is called the phase of the
oscillation
  t  0
Phi naught is called the phase
constant or phase shift. This
value specifies the initial
conditions.
Different values of the phase constant correspond
to different starting points on the circle and thus
to different initial conditions
The Phase Constant:
Phase Shifts:
An object on a spring oscillates with a period of 0.8s and
an amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its position
and direction of motion at t=2s?
x(t)  Acost   0 
Initial conditions:
x 0  5cm  Acos 0 
2
1x 0 
1 5cm 
 0  cos   cos 
 120   rads
A 
10cm
3
From the period we get:
2 2


 7.85rad /s
T 0.8s
An object on a spring oscillates with a period of 0.8s and
an amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its position
and direction of motion at t=2s?
x(t)  Acost   0 

2 
x(t)  0.1cos7.852  

3 
2
 0   rads
3
  7.85rad/s
x(t)  0.05m
A  0.1m
t  2s 

Total mechanical energy is conserved
for our SHM example of a spring with
constant k, mass m, and on a
frictionless surface.
1 2 1 2
E  K  U  mv  kx
2
2

The particle has all potential energy
at x=A and x=–A, and the particle
has purely kinetic energy at x=0.
We have modeled SHM mathematically.
Now comes the physics.
At turning points:
1 2
E  U  kA
2
At x=0:



1
2
E  k  mv max
2
From conservation:
1 2 1 2
kA  mv max
2
2
Maximum speed as related to
amplitude:
vmax
k

A
m
From energy considerations:
vmax
k

A
m
From kinematics:
v max  A


Combine these:
k

m
1 k
f
2 m

m
T  2
k
a 500g block on a spring is pulled a distance of 20cm and
released. The subsequent oscillations are measured to
have a period of 0.8s. at what position or positions is the
block’s speed 1.0m/s?
The motion is SHM and energy is conserved.
1 2 1 2 1 2
mv  kx  kA
2
2
2
kx  kA  mv
2
2
m 2
x A  v
k
2

2
x A 
2
v
2

2
2 2


 7.85rad /s
T 0.8s

x  0.15m
Acceleration is at a maximum when the particle is at
maximum and minimum displacement from x=0.
dv x (t) dAsin t 
2
ax 

  Acost 
dt
dt
Dynamics of SHM
ax   Acost 
2
x  Acost 
a x   x
2



Acceleration is
proportional to the
negative of the
displacement.
Dynamics of SHM
a x   x
2
According to Newton’s 2nd
Law:
F  ma x  kx
ma x  kx
As we found with energy

considerations:
k
ax 
x
m

Dynamics of SHM
Acceleration is not
constant:
2
dx
ax  2
dt
2
dx
k
 x
2
dt
m
This is the equation of
motion for a mass on a
spring. It is of a general
form called a second order
differential equation.
Unlike algebraic equations, their solutions are not
numbers, but functions.
In SHM we are only interested in one form so we can
use our solution for many objects undergoing SHM.
Solutions to these diff. eqns. are unique (there is only
one). One common method of solving is guessing the
solution that the equation should have…
2
dx
k
 x
2
dt
m
From
evidence, we
expect the
solution:
x  Acost   0 
2nd-Order Differential Equations:

Let’s put this possible solution into our equation and
see if we guessed right!
x  Acost   0 
dx
 Asin t 
dt
2
dx
k
 x
2
dt
m
k
 Acost   Acost 
2
dx
m
2
  Acost

2
dt
k
2
 
m
IT WORKS. Sinusoidal oscillation of
SHM 
is a result of Newton’s laws!
2
2nd-Order Differential Equations:
The equilibrium
position gets
shifted
downward
Vertical
springs
oscillate differently than horizontal springs
because there are 2 forces acting.
Hanging at rest:
Fnet  kL  mg  0
kL  mg
m
L  g
k


this is the equilibrium
position of the system.

What about vertical oscillations of
a spring-mass system??
Now we let the system
oscillate. At maximum:
Fnet  k L  y  mg
Fnet  kL  mg  ky
But:
kL  mg  0

So:

Fnet  ky
 we have learned about
Everything that
horizontal oscillations is equally valid for
vertical oscillations!

You need to show how to derive
the Period of a Pendulum equation
T = 2∏√l/g
s  L
Fnet t  mg sin   ma t

The Pendulum
2
ds
 gsin 
2
dt
Equation of motion
for a pendulum


When θ is about
0.1rad or less, h
and s are about the
same.
d 2s
 gsin 
2
dt
sin   
2
s
ds



g
cos 1
L
dt 2
2
tan  sin  1 
ds
mgs
Fnet t  m 2  
dt
L

Small Angle
Approximation:

Equation of
motion for a
pendulum
2
ds
gs

2
dt
L

g

L
x(t)  Acost   0 
(t)  max cost   0 

The Pendulum
What length pendulum will have a period of exactly 1s?
L
T  2
g
g

L
2

 T 
g   L
2 
2
1s 
L  9.8m/s    0.248m
2 
2
A Pendulum Clock
Notice that all objects that
we look at are described
the same mathematically.
Any system with a linear restoring
force will undergo simple
harmonic motion around the
equilibrium position.
Conditions for SHM
   I  mgd  mgl sin 
Small Angle Approx.
d
mgl


2
dt
I
2


mgl

I
A Physical Pendulum

when there is
mass in the
entire pendulum,
not just the bob.
All real oscillators are damped
oscillators. these are any that slow
down and eventually stop.
a model of drag force for
slow objects:
Fdrag  bv
b is the damping
constant (sort of like a
coefficient of friction).
Damped Oscillations
Solution to 2ndorder diff eq:
x(t)  Aebt / 2m cost  0 
F  F F
s
drag
 kx  bv  ma

2
2
dx
dx
kx  b  m 2  0
dt
dt
k
b


2
m 4m
2
b
Another
diff eq.
  
2
4m
Damped Oscillations

2nd-order
2
0
A slowly changing line
that provides a border to
a rapid oscillation is
called the envelope of
the oscillations.
x(t)  Ae
bt / 2m
cost  0 
Damped Oscillations
Not all oscillating objects are disturbed from rest
then allowed to move undisturbed.
Some objects may be subjected to a periodic
external force.
Driven
Oscillations
All objects have a natural frequency at which
they tend to vibrate when disturbed.
Objects may be exposed to a periodic force with
a particular driving frequency.
If the driven
frequency matches
the natural
frequency of an
object, RESONANCE
occurs
Driven
Oscillations
Spring
Best example of simple harmonic oscillator.
T = 2m/k
m
Simple Pendulum
Acts as simple harmonic oscillator only
when angle of swing is small.
T = 2L/g

Conical Pendulum
Not really a simple harmonic oscillator,
but equation is similar to simple
pendulum.

T = 2L(cos )/g
The small angle approximation
for a simple pendulum
Fr sin     I
 mg sin  ( L)  (mL2 )

mgcos
mg
mgsin
 g sin   L if  , sin   
g
  ( )  0
If the angle is small,
L
the “radian” value for
g
2
theta and the sine of

, 
the theta in degrees
L
T
Tpendulum  2
A simple pendulum is one where a
mass is located at the end of string.
The string’s length represents the
radius of a circle and has negligible
mass.
l
g
will be equal.
Once again, using our sine function
model we can derive using circular
motion equations the formula for the
period of a pendulum.
Torsional Pendulum
Twists back and forth
through equilibrium
position.
T = 2I/
Physical Pendulum
Anything that doesn’t fall into any of the
other categories of pendulums.
T = 2I/
 = Mgd
Energy Conservation in
Pendulums
 K1
+ U 1 = K 2 + U2
◦K = 1/2mv2
◦U = mgh
 1/2mv12+mgh1
h
=1/2mv22+
mgh2
 v12 + 2gh1 = v22+ 2gh2
A physical pendulum is an oscillating body that
rotates according to the location of its center of
mass rather than a simple pendulum where all the
mass is located at the end of a light string.
Fr sin     I
 mg sin d  I , d  L
2
if  , sin   
It is important to understand
that “d” is the lever arm
distance or the distance from
the COM position to the point of
rotation. It is also the same “d” in
the Parallel Axes theorem.
 mgd  I
mgd
 (
)  0
I
mgd
2

, 
I
T
T physical pendulum  2
The Physical Pendulum
I
mgd
Example
A spring is hanging from the ceiling. You know that
if you elongate the spring by 3.0 meters, it will
take 330 N of force to hold it at that position: The
spring is then hung and a 5.0-kg mass is
attached. The system is allowed to reach
equilibrium; then displaced an additional 1.5
meters and released. Calculate the:
Spring Constant
Fs  kx 330  (k )(3)
k
Angular frequency
110 N/m
k
k
110
 



m
m
5
2
4.7 rad/s
Example
A spring is hanging from the ceiling. You know
that if you elongate the spring by 3.0 meters,
it will take 330 N of force to hold it at that
position: The spring is then hung and a 5.0-kg
mass is attached. The system is allowed to
reach equilibrium; then displaced an
additional 1.5 meters and released. Calculate
the:
Amplitude
Stated in the question as 1.5 m
Frequency and Period
2
  2f 
T
 4.7
f 


2 2
2 2
T


 4.7
0.75 Hz
1.34 s
Example
A spring is hanging from the ceiling. You know
that if you elongate the spring by 3.0 meters, it
will take 330 N of force to hold it at that
position: The spring is then hung and a 5.0-kg
mass is attached. The system is allowed to
reach equilibrium; then displaced an additional
1.5 meters and released. Calculate the:
Total Energy
Maximum velocity
U s  1 kx2  1 kA2
2
2
U  1 (110)(1.5) 2  123.75 J
2
v  A  (1.5)( 4.7) 
7.05 m/s
A spring is hanging from the ceiling. You know that if you
elongate the spring by 3.0 meters, it will take 330 N of
force to hold it at that position: The spring is then hung
and a 5.0-kg mass is attached. The system is allowed to
reach equilibrium; then displaced an additional 1.5
meters and released. Calculate the:
Position of mass at maximum velocity
At the equilibrium position
Maximum acceleration of the mass
a   2 A  (4.7) 2 (1.5) 
33.135 m/s/s
Position of mass at maximum acceleration
At maximum amplitude, 1.5 m
A “seconds
pendulum”
beats seconds;
that is, it takes 1
s for half a cycle.
a) Find the length of a simple
seconds pendulum.
b) What assumption have
you made in this
calculation?
A thin uniform rod of mass 0.112 kg and length 0.096 m is
suspended by a wire through its center and perpendicular to its
length. The wire is twisted and the rod set to oscillating. The
period is found to be 2.14 s. When a flat body in the shape of
an equilateral triangle is suspended similarly through its center
of mass, the period is 5.83. Find the rotational inertia of the
triangle.
Problem #2
A uniform disk is pivoted at its rim.
Find the period for small oscillations
and the length of the equivalent
simple pendulum.
Problem #3
THE
END