Transcript Simple Harmonic Motion

Oscillations and Simple
Harmonic Motion:
AP Physics C: Mechanics
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
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

dt
  t
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:
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!

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
THE
END