P16_003.PDF
(
64 KB
)
Pobierz
Chapter 16 - 16.3
3. (a) The motion repeats every 0
.
500s so the period must be
T
=0
.
500s.
(b) The frequency is the reciprocal of the period:
f
=1
/T
=1
/
(0
.
500s) = 2
.
00Hz.
(c) The angular frequency
ω
is
ω
=2
πf
=2
π
(2
.
00Hz) = 12
.
57rad
/
s.
(d) The angular frequency is related to the spring constant
k
and the mass
m
by
ω
=
k/m
.Wesolve
for
k
:
k
=
mω
2
=(0
.
500kg)(12
.
57rad
/
s)
2
=79
.
0N
/
m.
(e) Let
x
m
be the amplitude. The maximum speed is
v
m
=
ωx
m
=(12
.
57rad
/
s)(0
.
350m) = 4
.
40m
/
s.
(f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by
F
m
=
kx
m
=(79
.
0N
/
m)(0
.
350m) = 27
.
6N.
Plik z chomika:
kf.mtsw
Inne pliki z tego folderu:
P16_096.PDF
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P16_013.PDF
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P16_009.PDF
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P16_002.PDF
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Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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