43
(4.0 10 N/m)(2.0 10 m)=80 N.
Fk
x
−
== ×
×
(e) At half of the maximum displacement,
1.0 mm
x
=
, and the force is
(4.0 10 N/m)(1.0 10 m)=40 N.
x
−
×
33. The textbook notes (in the discussion immediately after Eq. 157) that the
acceleration amplitude is
a
m
=
ω
2
x
m
, where
is the angular frequency and
x
m
= 0.0020 m
is the amplitude. Thus,
a
m
= 8000 m/s
2
leads to
= 2000 rad/s. Using Newton’s second
law with
m
= 0.010 kg, we have
=
=
+
=
80
2000
3
Fm
ama
t
t
m
−−
−
F
H
I
K
cos
N cos
ωφ
a
f
ch
a
f
π
where
t
is understood to be in seconds.
(a) Eq. 155 gives
T
= 2
π
/
= 3.1 × 10
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

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