A 5-kg block is suspended vertically between a spring pulley and dashpot damper. The spring remains in tension at all times and the pulley is small and frictionless. If the block is displaced downward 10 cm below its equilibrium position and then released with an upward velocity, v₀, of 5 m/s (t = 0), what is the earliest time when the velocity of the block is zero.

Force Diagram

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
      ω_n = (800)^0.5 = 28.28 rad/s
      ζ = 30/[2(1)(28.284)] = 0.5303
      ω_d =ω_n [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