Oscillatory
Motion
Cameron Rho
Background:
Oscillatory
motion,
or
Periodic
motion,
is
motion
of
an
object
that
repeats
itself
regularly—that
is,
the
object
returns
to
a
given
position
after
a
fixed
time
interval.
Major
Topics:
Motion
of
an
Object
Attached
to
a
Spring
Mathematical
Representation
of
Simple
Harmonic
Motion
Energy
of
the
Simple
Harmonic
Oscillator
Simple
Harmonic
Motion
and
Uniform
Circular
Motion
The
Pendulum
Dampened
and
Forced
Oscillations
Key
Phrases:
equilibrium
position:
the
position
where
Fs
=
0.
restoring
force:
a
force
directed
toward
the
equilibrium
position
and
opposite
displacement.
simple
harmonic
motion:
motion
in
which
acceleration
is
proportional
to
position
and
oppositely
directed
to
displacement.
amplitude:
the
maximum
value
of
position
in
either
the
positive
or
the
negative
direction.
period:
the
time
interval
required
for
one
full
cycle
of
motion.
frequency:
the
number
of
oscillations
per
time
interval.
dampening:
a
decrease
in
amplitude
and
motion
over
time
due
to
a
retarding
force.
resonance:
the
dramatic
increase
in
amplitude
near
the
natural
frequency.
Useful
Formulae:
Fs = −kx Hooke’s
Law
with
spring
constant
k
and
displacement
x.
1 2
U s = kx
2
Equation
for
finding
the
potential
energy
of
a
spring.
2π 1
T=
=
ω
f Equation
relating
period,
angular
frequency,
and
frequency.
m
Ts = 2π
k Equation
for
the
period
of
a
spring.
l
Tp = 2π
g Equation
for
the
period
of
a
pendulum.
€
€
€
€
€
Pendulum
Motion
Spring
Motion
F
θ
x
€
L
T
F =0
m
x
=
0
s
€
Fs €
mgsin θ
θ mgcosθ
s
s
x
€
€
€Fg = mg
€
Practice
Problems:
1.
[Easy]
In
an
engine,
a
piston
oscillates
with
simple
harmonic
motion
so
that
its
position
varies
according
to
the
expression
π
x
=
(5.00
cm)cos(2t
+
)
6
where
x
is
in
centimeters
and
t
is
in
seconds.
At
t
=
0,
find
(a)
the
position
of
the
piston,
(b)
its
velocity,
and
(c)
its
acceleration.
(d)
Find
the
period
and
amplitude
of
the
motion.
€
2.
[Medium]
A
cart
attached
to
a
spring
with
constant
3.24
N/m
vibrates
with
position
given
by
x
=
(5.00
cm)cos(3.60t
rad/s).
(a)
During
the
first
cycle,
for
0
<
t
<
1.75
s,
just
when
is
the
system’s
potential
energy
changing
most
rapidly
into
kinetic
energy?
(b)
What
is
the
maximum
rate
of
energy
transformation?
3.
[Hard]
A
solid
sphere
(radius
=
R)
rolls
without
slipping
in
a
cylindrical
trough
(radius
=
5R)
as
shown
in
the
Figure
below.
Show
that,
for
small
displacements
from
equilibrium
perpendicular
to
the
length
of
the
trough,
the
sphere
executes
simple
harmonic
motion
with
a
period
T = 2π 28R /5g .
€
5R
R
Solutions:
(final
solutions
are
underlined)
π
π
1.
(a)
x
=
(5.00
cm)cos(2t
+
);
so
at
t
=
0,
x
=
(5.00
cm)cos( )
=
4.33
cm
6
6
dx
π
π
(b)
v = =
–(10.0
cm/s)sin(2t
+
);
so
at
t
=
0,
v
=
–(10.0
cm/s)sin( )
dt
6
6
=
–5.00
cm/s
€
€
dv
π
π
(c)
a = =
–(20.0
cm/s2)cos(2t
+
);
so
at
t
=
0,
a
=
–(20.0
cm/s2)cos( )
dt
6
6
€
€
€
2
=
–17.3
cm/s 2π 2π
=
= 3.14 s
(d)
A
=
(5.00
cm)
and
T =
ω
2
€
€
€
€
2.
(a)
The
potential
energy
of
the
cart
is
1
1
U s = kx 2 = kA 2 cos2 (ωt ) 2
2
and
the
rate
of
change
of
the
same
is
dU s 1 2
1
= kA 2cos(ωt )[−ω sin(ωt )] = − kA 2ω sin2ωt .
dt
2
2
€
The
potential
energy
will
be
changing
most
rapidly
into
kinetic
energy
when
this
rate
of
change
is
at
a
maximum
negative
value,
when
π
2ωt = 2nπ + where
n
is
an
integer.
€
2
π ( 4n + 1)
π
Therefore,
t =
( 4n + 1) =
4ω
4 ( 3.60s−1 )
For
n
=
0,
this
gives
t
=
0.218
s
while
n
=
1
gives
t
=
1.09
s
all
other
n
values
give
times
outside
the
specified
range.
2
dU
1
1
(b)
€ s
= kA 2ω = ( 3.24 N m)(5.00 ×10−2 m) ( 3.60s−1 ) = 14.6
mW
dt max 2
2
1
1
3.
The
kinetic
energy
of
the
ball
is
K = mv 2 + Iω 2 ,
where
omega
is
the
rotation
2
2
rate
of
the
ball
about
its
center
of
mass.
Since
the
center
of
the
ball
moves
along
a
€
circle
of
radius
4R,
its
displacement
from
equilibrium
is
s = ( 4R)θ and
its
speed
is
 dθ 
ds
ds
€
v = = Rω and
thus
v = = 4R  .
Also,
since
the
ball
rolls
without
slipping,
 dt 
dt
dt
 dθ 
v
€
ω = = 4  .


R
dt
2
2
2
1  dθ  1  2 € 2  dθ  112mR 2  dθ 
€
The
kinetic
energy
is
then
K = m 4R  +  mR  4  =
   dt 
2 
dt  2  5
10  dt 
When
the
ball
has
an
angular
displacement
theta,
its
center
is
distance
€
h = 4R(1− cos θ ) higher
than
when
at
the
equilibrium
position.
Therefore,
the
θ2
€
potential
energy
is
U g = mgh = 4mgR(1− cosθ ) .
For
small
angles,
(1− cos θ ) ≈ .
2
2
2


112mR dθ
2
€
So,
U g ≈ 2mgRθ 2 ,
and
total
energy
is
E = K + U g =
  + 2mgRθ .
10  dt 
2
2
€
 dθ 
dE
112mR  dθ  d θ €
Since
E
=
constant
in
time,
=0=
  2 + 4mgRθ  .
 dt 
dt
5  dt  dt
2
2
€
 5g 
28R d θ €
dθ
+ gθ = 0 ,
or
2 = −
This
reduces
to
θ .
With
the
angular
2
 28R 
5 dt
dt
acceleration
equal
to
a
negative
constant
times
the
angular
position,
this
is
in
the
€
5g
defining
form
of
simple
harmonic
motion
with
ω =
.
The
period
of
the
28R
€
€
2π
28R
= 2π
simple
harmonic
motion
is
then
T =
.
ω
5g
€
€
€