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**Unformatted text preview: **t should be traveling upwards when it ﬁrst passes through it.
To check this answer, we graph one cycle of x(t). Since our applied domain in this situation
π
is t ≥ 0, and the period of x(t) is T = 2π = 22 = π , we graph x(t) over the interval [0, π ].
ω
Remembering that x(t) > 0 means the object is below the equilibrium position and x(t) < 0
means the object is above the equilibrium position, the fact our graph is crossing through the
t-axis from positive x to negative x at t = π conﬁrms our answer.
4 2. The only diﬀerence between this problem and the previous problem is that we now release
the object with an upward velocity of 8 ft . We still have ω = 2 and x0 = 3, but now
s
we have v0 = −8, the negative indicating the velocity is directed upwards. Here, we get
2 A = x2 + v0 = 32 + (−4)2 = 5. From A sin(φ) = x0 , we get 5 sin(φ) = 3 which gives
0
ω
3
sin(φ) = 5 . From Aω cos(φ) = v0 , we get 10 cos(φ) = −8, or cos(φ) = − 4 . This means
5
that φ is a Quadrant II angle which we can describe in terms of either arcs...

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