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Question #160919
A 200g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250s. The total energy of the system is 2.00J. Find
(a) the force constant of tge spring
(b) the amplitude of the motion.
Expert's answer
(a)
T
=
2
π
m
k
,
T=2\pi\sqrt{\dfrac{m}{k}},
T
=
2
π
k
m
,
k
=
4
π
2
m
T
2
,
k=\dfrac{4\pi^2m}{T^2},
k
=
T
2
4
π
2
m
,
k
=
4
π
2
⋅
0.2
k
g
(
0.25
s
)
2
=
126.3
N
m
.
k=\dfrac{4\pi^2\cdot0.2\ kg}{(0.25\ s)^2}=126.3\ \dfrac{N}{m}.
k
=
(
0.25
s
)
2
4
π
2
⋅
0.2
k
g
=
126.3
m
N
.
(b)
E
=
1
2
k
A
2
,
E=\dfrac{1}{2}kA^2,
E
=
2
1
k
A
2
,
A
=
2
E
k
,
A=\sqrt{\dfrac{2E}{k}},
A
=
k
2
E
,
A
=
2
⋅
2.0
J
126.3
N
m
=
0.177
m
.
A=\sqrt{\dfrac{2\cdot2.0\ J}{126.3\ \dfrac{N}{m}}}=0.177\ m.
A
=
126.3
m
N
2
⋅
2.0
J
=
0.177
m
.
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