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105. (a) From the graph, it is clear that
x
m
= 0.30 m.
(b) With
F
= –
kx
, we see
k
is the (negative) slope of the graph — which is 75/0.30 = 250
N/m. Plugging this into Eq. 1513 yields
=2
028
T
m
k
π=
..
s
(c) As discussed in
§
152, the maximum acceleration is
22
2
1.5 10 m/s .
mmm
k
axx
m
ω
===
×
Alternatively, we could arrive at this result using
a
m
= (2
π
/
T
)
2
x
m
.
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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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