*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **ine or arccosine.
Since x(t) is expressed in terms of sine, we choose to express φ = π − arcsin 3 . Hence,
5
x(t) = 5 sin 2t + π − arcsin 3 . Since the amplitude of x(t) is 5, the object will travel
5
at most 5 feet above the equilibrium position. To ﬁnd when this happens, we solve the
3
= −5, the negative once again signifying that
equation x(t) = 5 sin 2t + π − arcsin 5
the object is above the equilibrium position. Going through the usual machinations, we get
t = 1 arcsin 3 + π + πk for integers k . The smallest of these values occurs when k = 0,
2
5
4
1
that is, t = 2 arcsin 3 + π ≈ 1.107 seconds after the start of the motion. To check our
5
4
answer using the calculator, we graph y = 5 sin 2x + π − arcsin 3
on a graphing utility
5
and conﬁrm the coordinates of the ﬁrst relative minimum to be approximately (1.107, −5).
x
3
2
1 −1 π
4 π
2 3π
4 π t −2
−3 x(t) = 3 sin 2t + 14 π
2 y = 5 sin 2x + π − arcsin For conﬁrmation, we note that Aω cos(φ) = v0 , which in...

View
Full
Document