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1
Unit 8
Simple Harmonic Motions
8.1
Periodic motion
8.2
Simple harmonic motion
8.3
Position, velocity and acceleration in S.H.M.
8.4
Simple pendulum
8.5
Energy of simple harmonic motion
8.6
Damped and forced oscillations
8.1
Periodic motion
A motion that repeats itself over and over is referred to as periodic motion.
Some useful quantities:
•
Period,
T
T
= Time required for one cycle of a periodic motion
Unit: second
•
Frequency,
f
f
= The number of oscillation per second
Unit: cycle s
−
1
= s
−
1
or Hz
Note that
T
= 1/
f.
•
Angular velocity,
ω
= Angular displacement per unit time
Unit: radian s
−
1
= s
−
1
Note that 1 period
→
2
π
, hence
= 2
f
(In words, the angular displacement is 2
f
radian per second)
or
T
2
=
•
Angular displacement,
θ
=
Angular displacement in time
t
=
t
•
Amplitude = The maximum displacement (angular displacement) of the motion.
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8.2
Simple harmonic motion
F =
0
x =
0
x =
0
v
=
0
F
max
x = A
F =
0
x =
0
v
max
F =
0
x =
0
v
max
x
=
0
v
=
0
F
max
x = A
Equilibrium position
t = 0
t = T
t = 3T/4
F
max
x =
0
x =
−
A
v
=
0
t = T/2
t = T/4
3
Hooke’s law states that the restoring force
F
is proportional to the displacement from its
equilibrium position:
F =
−
kx.
The negative sign is appeared in the force equation, which shows that the restoring force
and the displacement vector from the equilibrium position are in opposite direction. The
equilibrium position is defined as the point where the net force acting on the vibrating
mass is zero. For the above example,
x = 0
is the equilibrium point.
8.3
Position, velocity and acceleration in S.H.M.
Let the projection velocity of the particle on
AB
as
v
which is a component of the
tangential velocity
v
t
of the particle. We label the radius of the circle, that is distance
OA
as
A
.
)
(
cos
t
A
x
ω
=
(1)
)
(
sin
t
A
v
−
=
(2)
)
(
cos
2
t
A
a
−
=
(3)
Hence, we have
)
(
2
2
2
2
x
A
v
−
=
and
x
a
2
−
=
. Note that the latter equation is a
second order differential equation in the form
2
2
2
0
dx
x
dt
+=
or
0
2
=
+
x
x
±
±
.
Note also that
•
x
max
= A
O
A
N
P
x
B
θ
v
t
=
A
v
t
sin
v
t
P
2
A
2
A
cos
P
O
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•
ω
A
v
=
max
and
2
max
A
a
=
.
•
From (1) and (2), we know that
x
and
v
differs by
π
/2.
•
From (1) and (3), we know that
x
and
a
are out of phase, that is if
x
is at its
maximum, then
a
must be at its minimum and vice versa.
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This note was uploaded on 09/06/2010 for the course BSC PHY1417 taught by Professor Prof during the Spring '08 term at HKU.
 Spring '08
 Prof

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