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d
a
y
d
t
=
d
(
–
ω
²
y
m
sin
(−ω
t+
φ))
d
t
=
y
m
ω
3
cos
(
−ω
t +
φ
)
< 0
at
t
= 0
.
This implies that
cos
φ
< 0 and consequently that
is in either the second or third
quadrant. The graph shows (at t = 0)
a
y
=
−
100 m/s², and (at another
t
)
a
max
= 400 m/s².
Therefore,
a
y
=
−
a
max
sin
(
−ω
t +
)

t
= 0
¡
=
sin
−
1
(
1
4
) =
0.25 rad
or
2.9 rad
(
bear in mind that sin
θ
= sin(
π −
)
)
, and we must choose
= 2.9 rad
because this is
about 166° and is in the second quadrant.
Of course, this answer added to 2n
π
is still a
valid answer (where n is any integer), so that, for example,
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics

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