Chapter 13
Vibrations and Waves
1
•
When x is positive
,
F is negative
;
•
When at equilibrium (x=0),
F = 0 ;
•
When x is negative
,
F is positive
;
Hooke’s Law Reviewed
F
=
!
kx
2
Sinusoidal Oscillation
Pen traces a sine wave
3
Graphing x vs. t
A : amplitude
(length, m)
T : period
(time, s)
A
T
4
Some Vocabulary
f = Frequency
!
= Angular Frequency
T = Period
A = Amplitude
"
= phase
x
=
A
cos(
!
t
"
#
)
=
A
cos(2
$
ft
"
#
)
=
A
cos
2
$
t
T
"
#
%
&
’
(
)
*
f
=
1
T
!
=
2
"
f
=
2
"
T
5
Phases
Phase is related to starting time
90degrees changes cosine to sine
x
=
A
cos
2
!
t
T
"
#
$
%
&
’
(
)
=
A
cos
2
!
(
t
"
t
0
)
T
$
%
&
’
(
)
if
#
=
2
!
t
0
T
cos
!
t
"
#
2
$
%
&
’
(
)
=
sin
!
t
(
)
6
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a
x
v
•
Velocity is 90
°
“out of phase” with
x: When x is at max,
v is at min
....
•
Acceleration is 180°
“out of phase” with x
a = F/m =  (k/m) x
Velocity and
Acceleration vs. time
T
T
T
7
v and a vs. t
Find v
max
with E conservation
Find a
max
using F=ma
x
=
A
cos
!
t
v
=
"
v
max
sin
!
t
a
=
"
a
max
cos
!
t
1
2
kA
2
=
1
2
mv
max
2
v
max
=
A
k
m
!
kx
=
ma
!
kA
cos
"
t
=
!
ma
max
cos
"
t
a
max
=
A
k
m
8
What is
!
?
Requires calculus. Since
d
dt
A
cos
!
t
=
"
!
A
sin
!
t
v
max
=
!
A
=
A
k
m
!
=
k
m
9
Formula Summary
f
=
1
T
!
=
2
"
f
=
2
"
T
x
=
A
cos(
!
t
"
#
)
v
=
"
!
A
sin(
!
t
"
#
)
a
=
"
!
2
A
(cos
!
t
"
#
)
!
=
k
m
10
Example13.1
An blockspring system oscillates with an amplitude
of 3.5 cm. If the spring constant is 250 N/m and
the block has a mass of 0.50 kg, determine
(a) the mechanical energy of the system
(b) the maximum speed of the block
(c) the maximum acceleration.
a) 0.153 J
b) 0.783 m/s
c) 17.5 m/s
2
11
Example 13.2
A 36kg block is attached to a spring of constant
k=600 N/m. The block is pulled 3.5 cm away from its
equilibrium positions and released from rest at t=0.
At t=0.75 seconds,
a) what is the position of the block?
b) what is the velocity of the block?
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 Spring '06
 Pratt
 Physics, Cos, Transverse wave, Longitudinal wave, Sine wave

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