Transcript Simple Harmonic Motion

Periodic Motion
What is periodic motion?

When a vibration or oscillation repeats
itself over and over the motion is said to
be periodic.


Most objects vibrate briefly when given an
impulse
Electrical oscillations occur in TV’s and radio’s,
atoms vibrate around a fixed spot
Terminology






Oscillation or Vibration – a motion that repeats itself
with no net displacement.
Equilibrium Position – the point that the object
oscillates around. Also known as the rest position.
Displacement – how far the mass is from the
equilibrium point (x)
Maximum displacement – how far the mass moves
from the equilibrium position. (xmax occurs at A)
Amplitude (A) – the distance from the equilibrium
point to the maximum displacement.
Cycle – a complete to and fro motion.
Terminology



Period (T) – the time needed to complete
one cycle. (units – seconds)
Frequency (f) – the number of cycles
completed in one second. Units are Hertz.
(Hz = 1/s = s-1)
Formula


f =1/T
T=1/f
Simple Harmonic Motion

Simple Harmonic Motion is any motion in
which the restoring force is proportional to
displacement.

Examples:
An acrobat swinging on a trapeze
 Child on a playground swing
 Pendulum of a clock or metronome
 Mass on the end of a spring


Restoring force – the force that pushes or
pulls the mass back to equilibrium.
Hooke’s Law




In 1678 Robert Hooke found that most massspring systems obey a simple relationship
between force and displacement for small
displacements.
This is a restoring force.
FSpring = FElastic = – kx
The force, F, is negative because it is always
a restoring force, pulling or pushing the
opposite direction of the displacement.
The Applied Force is in the opposite direction
of the Spring Force.
Spring Constant “k”



The value of the constant measures the
‘stiffness’ of the spring.
The larger the value, the stiffer the spring
Unit for the spring constant, k, is N/m.
Period of a Spring



Period of a spring is given by the following
equation:
m = mass in kg
k = spring constant in N/m
m
T  2
k



1
f
2
k
m
What happens to the period as mass increases?
What happens to the period as the spring
constant increases?
What is the frequency equation?
Elastic Potential Energy




At maximum displacement the potential energy
is at its maximum and kinetic is at its minimum.
At the equilibrium position the potential is at its
minimum and kinetic is at its maximum.
US = USpring = ½ k x2
KE = ½ m v2
Mass on a Spring
HORIZONTAL MOTION


If the object is in MOTION then at the
equilibrium position (x=0) the velocity is at
the maximum.
At the maximum displacement, spring force
and acceleration reaches a maximum and
the velocity is zero.


Known as a turning point.
The acceleration is in the opposite direction of
the motion.
E=K+U
E = 1/2mv2 + 1/2kx2
Maximum displacement is A = x for the example below.
A
E = 1/2kA2
E = 1/2mv02
A
E = 1/2kA2
x0
Conservation of energy review


Remember that all energy is conserved
 The energy just changes from one form to an other.
Initial Energy = Final Energy
 Ki + Ui = Kf +PUf
The spring is release from "A";
what is the maximum speed ?
1
1
1
1
2
2
2
 m  0   k  A   m  vf   k  02
2
2
2
2
The example is for a horizontal spring
k 2
2
system.
vf   A
Solving for “vf” as for a spring released
m
at “A” and finding the maximum velocity
k 2
k as the spring passes through the
vf 
A  A
m
m equilibrium position.

Finding the velocity in general depends on the
original amplitude and the location in the cycle, x.
KE  U S  E
1 mv 2  1 / 2kx 2  1 2 kA2
2
2
2
1
2
kA

1
2
kx
k 2 k 2
2
v 
 A  x
1 2m
m
m
k 2
x  2
x 
v  A  1  2   v 0  1  2 
m 
A 
A 

2
2
2
v  v0

x 
 1  2 
A 

2
Potential Energy in a Spring
Example
A spring with a force constant of 5.2 N/m has a relaxed
length of 2.45 m. When a mass is attached to the end of
the spring and allowed to come to rest, the vertical length
of the spring is 3.57 m. Calculate the elastic potential
energy stored in the spring.
Spring Constant in a Spring
Example

A mass of 0.30 kg is attached to a spring
and is set into vibration with a period of
0.24 s. What is the spring constant of the
spring?
Period and Frequency in a
Spring Example

A spring of spring constant 30.0 N/m is
attached to different masses, and the
system is set in motion. Find the period
and frequency of vibration for masses of
the following magnitudes:



2.3 kg
15 g
1.9 kg
Simple Graph of SHM
Cosine Graph
x  t   A cos  t 
This means the position is a
function of time.
Graph of Unit Circle
x0
1.2
1
q=0
Theta
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
Time
2
2.5
x0
1.30
q=/2
Theta
0.80
0.30
-0.20 0
1
2
3
4
-0.70
-1.20
Time
5
6
7
8
x0
1.50
1.00
Theta
0.50
0.00
0
1
2
3
4
-0.50
q=
-1.00
-1.50
Time
5
6
7
8
x0
1.50
1.00
Theta
0.50
q=3/2
0.00
0
1
2
3
4
-0.50
-1.00
-1.50
Time
5
6
7
8
Notice that the radius of the circle equals
the amplitude of the spring.
x0
Cosine Graph
1.50
1.00
Theta
q=2
0.50
0.00
-0.50 0
2
4
6
-1.00
-1.50
Time
8
10
Amplitude is independent of the period!!!

The maximum velocity is equal to the path
length of the circle (2r) divided by time.
2r 2A
v0 

 2Af
t
T
m
T  2
k
1 kA2  1 mv02
2
2
A
m

v0
k
Simple Pendulum
Is a pendulum simple harmonic motion?




Simple pendulum is a mass on the end of a
string.
The mass is called a “BOB”.
Assume the mass is concentrated at a point.
Neglect air resistance and friction.
The restoring force is a component of the
bob’s weight (-mg sin q).


If the restoring force is proportional to the
displacement the pendulum’s motion is
simple harmonic.
There are two forces acting on the
pendulum:


The tension in the string
The weight of the bob
Motion of a Pendulum
Ftension
Ftension
Weightx
Weight
Weighty
Weight
Restoring force



As the pendulum is pulled back the x component
of the weight gets larger and the y component
gets smaller.
Therefore the greater the displacement the
larger the restoring force
For small displacements the pendulums motion
is simple harmonic.
Energy of a Pendulum



Energy is conserved.
At maximum displacement: Velocity is
zero, acceleration is largest, Energy is all
potential.
At equilibrium: Velocity is the largest,
acceleration is zero, Energy is all kinetic.
Amplitude, Period, and Frequency


The time it takes for a pendulum to swing
from one side to the other and back again
is one period.
The number of complete cycles in one
second is the frequency.
Small Angle Notes
Extra Notes
Figure 14.19
Figure 14.19B
Figure 14.19A
What is this length here?
Answer is l cos q
What is this length
here?
Answer is l (1 - cos q
l sin q
Trig of a Pendulum
Restoring force = -mg sinq
For small angles sin q = q
Using x = L q gets:
F = -(mg/L)x
This is similar to Hooke’s
Law with k = mg/L
Using the equations
derived for a spring :
q
L
FT
mg sinq
m
m
  2
 2
mg
k
L
L
 2
g
Note that the period
of Pendulum does
NOT depend on the
mass of the bob!!
x
q
mg
mg cosq
Period of a Pendulum


Depends on the length and free fall
acceleration.
For small amplitudes the period DOES
NOT depend on the amplitude.
L
T  2
g
Damped Harmonic Motion


The amplitude of any real oscillating
spring slowly decreases. This is damped
harmonic motion
Damping is due to friction and air
Forced Vibrations/ Resonance



When a system is set in motion then left alone it
vibrates at its natural frequency (f0)
When an outside force is constantly applied it
creates forced vibrations
The amplitude of the forced vibration depends
on the difference between f and f0.
Now with x(t) find v(t) and a(t)
x  t   A cos  t 
take the derivative of x w.r.t. time
dx d
  A cos  t     A sin  t 
dt dt
v (t )   A sin  t 
take the derivative of v w.r.t. time
dv d
2
   A sin  t     A cos  t 
dt dt
a (t )   2 A cos  t 