Transcript Chapter 15 Lecture Notes

Chapter 15
Oscillatory Motion
Recall the Spring
Fspring  kx
Negative because it is a restoring force.
In other words, if x is positive, the force and acceleration is negative and vice versa.
This makes the object oscillate and therefore, it undergoes simple harmonic motion.
Since F=ma, this can be rewritten as:
max  kx
ax  
k
x
m
Now pay attention
Remember that:
So
ax  
k
x
m
dx
v
dt
dv
a
dt
d 2x
a 2
dt
can be rewritten as:
d 2x
k
 x
2
dt
m
2
d x
k
 x
2
dt
m
To make this differential equation easier to solve, we make
then
k
 2
m
d 2x
2



x
2
dt
After solving, we get:
x(t )  A cos(t   )
Note: This requires knowledge on how to solve differential equations.
It is more important to know what the solution is.
Simple Harmonic Motion
x(t )  A cos(t   )
Equation for distance (x) as time (t) changes.
General formula for simple harmonic motion.
A, ω, ϕ are constants
A is Amplitude
- For springs: max value of distance (x) (positive or negative)
- Maximum value the wave alternates back and forth between
ω is Angular frequency
→
- How rapidly oscillations occur
- Units are rad/s
Remember:
k
 
m
k

m
2
x(t )  A cos(t   )
(t   )

is the phase
Essentially is the shifts of the wave
is the phase constant
These determine the starting position of the wave
General Concepts
F  kx
d 2x
k
 x
2
dt
m
x(t )  A cos(t   )
Anything with behaviors which have formulas that look like
these are undergoing simple harmonic motion and can be
measured using the same method as the spring.
Period and frequency
T
For springs
For springs
2
Period: time for 1 full oscillation

m
 2
k
1
f 
T
1 k

2 m
frequency: number of oscillations per second
Measured in cycles per second
- Hertz (Hz)
Note: frequency (f) and angular frequency (ω) measure the same thing but
with different units. They differ by a factor of 2 pi.
Velocity and Acceleration
Velocity of oscillation
Acceleration of oscillation
dx
v
 A sin( t   )
dt
d 2x
a  2   2 A cos(t   )
dt
Note: magnitude of maximum values are when the sin and cosine arguments equal 1
Energy of Simple Harmonic Oscillators
Remember that:
1 2
K  mv
2
1 2
U  kx
2
E  K U
After substituting the equations of velocity(v) and distance(x)
for simple harmonic oscillations, we get:
1 2
E  kA
2
Applications: Simple Pendulum

The restoring force for a pendulum is
T
 mg sin 
Ftension  matension
d 2s
 mg sin   m 2
dt
mg
thus
d 2 L
 mg sin   m
dt 2
d 2
 mg sin   mL 2
dt
g
d 2
 sin   2
L
dt
where s  L
which is the arclength
or the path the ball
travels along
Simple Pendulums continued
Notice that
d 2
g


sin 
2
dt
L
2
almost looks like d x   k x
2
dt
m
According to the small angle approximation, which states that sinθ ≈ θ if θ is
small (about less than 10°)
2
d

g
We can rewrite the equation to be
   for small angles
dt 2
L
which is exactly in the form for simple harmonic motions
where

g
L
so then
2
L
T
 2

g
we can now use all the other formulas for simple harmonic motions for
the case of a pendulum
Applications: Torsional Pendulum
When a torsion pendulum is twisted, there exists a restoring torque
which is equal to:
  
This looks just like
F  kx
but in rotational form
Thus, we can apply what we know about angular motion to get information about
this object’s simple harmonic oscillations
Torsion Pendulum continued
 I
Remember:
  
After substituting
we get
   I
d 2
   I 2
dt
d 2

 
2
dt
I
where


I
and
T
2

 2
I
