**Unformatted text preview: **ow the equilibrium position with
an upward velocity of 8 feet per second, ﬁnd the equation of motion of the object, x(t). What
is the longest distance the object travels above the equilibrium position? When does this ﬁrst
happen? Conﬁrm your result using a graphing utility.
Solution. In order to use the formulas in Theorem 11.1, we ﬁrst need to determine the spring
constant k and the mass of the object m. To ﬁnd k , we use Hooke’s Law F = kd. We know the
object weighs 64 lbs. and stretches the spring 8 ft.. Using F = 64 and d = 8, we get 64 = k · 8, or
k = 8 lbs. . To ﬁnd m, we use w = mg with w = 64 lbs. and g = 32 ft. . We get m = 2 slugs. We can
ft.
s2
now proceed to apply Theorem 11.1.
k
1. With k = 8 and m = 2, we get ω = m = 8 = 2. We are told that the object is released
2
3 feet below the equilibrium position ‘from rest.’ This means x0 = 3 and v0 = 0. Therefore,
√
2
A=
x 2 + v0
= 32 + 02 = 3. To determine the phase φ, we have A sin(φ) = x0 ,
0
ω
13
The sign convention...

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