When the block is moved from its equilibrium position, the spring cable and dashpot will exert forces on the block. The equilibrium equation can be written as,
T + T + R = ma
Due to the pulley, any displace of the block, x, will be twice the distance in the spring. Thus, the tension force will be,
T = 2kx = 2(1,000)x = 2,000x
The dashpot force will be direction related to the displacement x and its velocity,
R = 150 v
This gives,
2,000x + 2,000x + 150 v = m a
4,000 x + 150 dx/dt = 5 d2x/dt2
d2x/dt2 + 30 dx/dt + 800 x = 0
The solution to this differential equation is,
where
ωn = (800)0.5 = 28.28 rad/s
ζ = 30/[2(1)(28.284)]
= 0.5303
ωd =ωn [1 - (0.5303)2]0.5 = 23.98 rad/s
The position and velocity of the block is,
x(t) =
A e-15.0 t cos(23.98 t - φ)
v(t) =
A [-15e-15.0 t cos(23.98 t - φ)
- 23.98 e-15.0 t sin(23.98 t - φ)]
At t = 0, the initial conditions are x = -0.1 m and v = 5 m/s, giving,
-0.1 =
A e0 cos(0 -φ)
5 = A [-15e0 cos(0 - φ) - 23.98 e0 sin(0 - φ)]
Solving for A and φ,
φ = 2.1715
A = 0.1769
Therefore, when v(t) = 0,
23.98 t - 2.171 = -0.5590
t = 0.06722 s
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