Oscillations: Simple Harmonic
Motion and Energy
1
Oscillations: Simple Harmonic
Motion and Energy
T
g
Ball and massless spring
Ball experiencing
gravity and string tension...
2
Motion of a particle with mass m
attached to a massless spring
Two common types of forces:
Gravitational force on earth : mg
Constant force!
The restoring force of a spring or a rubber band
Force is proportional to the extension!
3
Newton's Second Law and the
Gravitational Force Fg
2
d x
dv
F=m a=m dt =m 2
dt
Apply it to the case where the applied force is the
gravitational force
F=m a=F
g =mg
2
d x
a= 2 = g
dt
4
Newton's Second Law and the
Spring Force Fs
2
d x
dv
F=m a=m dt =m dt 2
Apply it to the case where the applied force is the
force of a spring
F=m a=F s =k x
Restoring force, increases with extension
k is called the spring constant
5
The mathematical reprentation is
called the Equation of Motion
2
d x
m 2 =kx
dt
We write can write the equation as:
2
k
d x
2
=
x
x=
2
m
dt
k
2
where: =
m
6
Visualize what this equation means!
So far as this equation is concerned,
how tough the spring is,
and how massive the particle is,
do not individually matter.
All that matters is the ratio of the two.
2
d x
2
=
x
2
dt
where: =
k
m
7
The solution to this differential
equation is:
x = A cos t
Where from A , ?
OK, lets just assume its correct and work out
the derivatives. Derivative number 1:
dx /dt = A sin t
8
Almost all the Greek
that physicists know:
: little omega
:big OMEGA
: little phi
:big PHI
:lambda
:mu
:epsilon
: pi
English Equivalents
o
O
f
F
l
m
e
p
... (excepting of course physicists from Greece)
!
9
The solution to this differential
equation is:
dx /dt = A sin t
Derivative number 2:
2
2
2
d x /dt = A cos t
2
d x
2
=
x
2
dt
10
Lecture 2
Today
Understand the physical meaning of
amplitude and phase
Discuss the meaning (and the units)
of angular frequency, frequency and
period of oscillation
11
Lecture 2
The solution to this differential
equation
2
d x
2
=
x
2
dt
is
12
The solution to this differential
equation is:
x t=
x= A cos
cost
t
Where from A , ?
OK, lets just assume its correct and work out
the derivatives. Derivative number 1:
dx /dt = A sin t
13
The solution to this differential
equation is:
x t=
x= A cos
cost
t
A is the “amplitude” of the wave:
It determines how high or low it goes
φ is the “phase” of the wave:
It determines where the wave starts
14
x(t), v(t), a(t)
Displacement: x(t)
x t= A cos t
Velocity: v(t) = dx/dt
v t = A sin t
Acceleration: a(t)=dv/dt
2
a t = A cos t
15
x(t), v(t), a(t)
Displacement: x(t)
x t= A cos t
Velocity: v(t) = dx/dt
v t = A sin t
Acceleration: a(t)=dv/dt
2
a t = A cos t
16
Simple harmonic Motion
Displacement: x(t)
x t= A cos t
T
Period of oscillation T:
k
But: =
m
T =2 2
m
T=
=2
k
17
Simple harmonic Motion
Displacement: x(t)
x t= A cos t
T
1
Frequency of oscillation f: f =
T
k
1
=
=
f=
m
2 2
k
m
18
Simple harmonic Motion:
Angular
Frequency
k
=
m
1
=
Frequency f =
2 2
Units
rad/s
k
m
2
m
Period T =
=2
k
1/s
or Hertz (Hz)
s
19
Simple harmonic Motion:
Angular
units
Units
2 radians = 360 But angle is really unitless. Here's why
(sort of).
The perimeter l of a circle of radius R is:
l=2 R
or
l
2 =
R
Therefore angle is a ratio of two lengths
and thus unitless.
20
Oscillations Example: A BlockSpring System
But first: what is a system ?
System is “science-speak” for
a collection of interacting or interdependent
entities.
A block with a mass of m=200 g is displaced 5.0
cm from equilibrium and released from rest.
Find the solution for its equation of motion.
21
Oscillations Example: A BlockSpring System
A block with a mass of m=200 g is displaced 5.0
cm from equilibrium and released from rest.
Find the solution for its equation of motion.
k =5.0 N /m
m=0.2 kg
1 N =1 kg m/ s
2
N /m
k
1
=
= 5.0
=5 s
m
0.2kg
Note: we sometimes explicitly say rad/s for angular frequency but we dont have to.
22
Oscillations Example: A BlockSpring System
A block with a mass of m=200 g is displaced 5.0
cm from equilibrium and released from rest.
Find the solution for its equation of motion.
1
k =5.0 N /m
=5 rad s
m=0.2 kg
Initial position represented by
x i=0.05 m subscript i.
v i=0
Initial velocity is `zero (“at rest”)
23
Oscillations Example: A BlockSpring System
Lets assume we know k, m, xi and vi,
and solve the problem in this general form,
without plugging in the numbers yet.
x t= A cos t
v t = A sin t
At initial time (t=0), we have:
x 0= A cos . 0
v 0 = A sin . 0
= A cos=x i
= A sin =v i
We know ω, xi and vi , we do not know A and φ
24
Oscillations Example: A BlockSpring System
We know ω, xi and vi , we do not know A and φ
x 0= A cos . 0
= A cos=x i
v i A sin =
= tan xi
A cos v 0 = A sin . 0
= A sin =v i
v i
=arctan
xi
A=x i /cos
25
Oscillations Example: A BlockSpring System
Solutions for A and φ
v i
=arctan
xi
when vi = 0, =0
A =x i /cos =x i when =0
when φ = 0, A=x i =0.05 m
1
x t= 0.05m cos5 s t m
1
1
v t =0.05m 5s sin 5 s t
=0.25 sin 5t
m/s
26
Energy in Simple Harmonic Motion
Kinetic Energy
1
2
K t = m v t 2
1
2
Potential Energy
U t = k x t 2
x t= A cos t
We have our
equations:
v t = A sin t
= k /m
27
Energy in Simple Harmonic Motion
1
2 2
2
Kinetic Energy
K t = m A sin t
2
1
2
2
= k A sin t
2
1
2
2
Potential Energy U t = k A cos t
2