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22–1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s.
SOLUTION $ mg - k(y + yst) = my
+ T ΣFy = may;
where kyst = mg
k $ y + y = 0 m
Hence
p =
=
k Bm
B
Where k =
8(9.81) 0.175
448.46 = 7.487 8 $ y + (7.487)2y = 0
6
= 448.46 N>m
$ y + 56.1y = 0
Ans.
The solution of the above differential equation is of the form: y = A sin pt + B cos pt
(1)
# v = y = Ap cos pt - Bp sin pt
(2)
At t = 0, y = 0.1 m and v = v0 = 1.50 m>s From Eq. (1)
0.1 = A sin 0 + B cos 0
B = 0.1 m v0 1.50 = = 0.2003 m p 7.487
From Eq. (2)
v0 = Ap cos 0 - 0
Hence
y = 0.2003 sin 7.487t + 0.1 cos 7.487t
At t = 0.22 s,
y = 0.2003 sin [7.487(0.22)] + 0.1 cos [7.487(0.22)]
A =
Ans.
= 0.192 m
1190
Ans: $ y + 56.1 y = 0 y 0 t = 0.22 s = 0.192 m