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constant
φ
in Eq. 152 is zero. Also,
f
= 0.25 Hz is given, so we have
ω
= 2
π
f
=
π
/2 rad/s.
The variable
t
is understood to take values in seconds.
(a) The period is
T
= 1/
f
= 4.0 s.
(b) As noted above,
=
π
/2 rad/s.
(c) The amplitude, as observed above, is 0.37 cm.
(d) Eq. 153 becomes
x
= (0.37 cm) cos(
π
t
/2).
(e) The derivative of
x
is
v
= –(0.37 cm/s)(
π
/2) sin(
π
t
/2)
≈
(–0.58 cm/s) sin(
π
t
/2).
(f) From the previous part, we conclude
v
m
= 0.58 cm/s.
(g) The accelerationamplitude is
a
m
=
2
x
m
= 0.91 cm/s
2
.
(h) Making sure our calculator is in radians mode, we find
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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

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