Problem Parameters

A free-body diagram can be used to identify all the forces on a the platform. The weight, mg, will act on the spring but the spring force will counter-act the weight, so the weight actually cancels. The spring deflection will cause a force, ky, that will accelerate the platform. The final equation of motion becomes,

The general solution of this differential equation is

      y(t) = A sinω_nt + B cosω_nt

where ω_n represents the natural frequency and is defined as

For the two cases, k₁ = 5,000 and k₂ = 10,000, the natural frequencies are found:

     case 1:   ω_n1 = 2.24 rad/s

     case 2:   ω_n2 = 3.16 rad/s

The initial conditions for this particular situation are

     y(0) = -20 ft
     dy(0)/dt = 0

These two conditions can be used to determined the constants A and B. The final equation becomes

     y(t) = -20 cosω_nt

Case 1 Results

Case 1 Results

The velocity equation is determined by taking the time derivative of the displacement equation,

     dy/dt = -20 ω_n sinω_nt

Likewise, the acceleration equation is found taking the second time derivative of the displacement,

     d²y/dt² = -20 ω_n² cosω_nt

Therefore, the maximum acceleration is directly related to the square of the natural frequency, or

This can be converted to g forces by dividing by the acceleration of gravity,

     g_max = 20/32.2 ω_n² = 0.621ω_n² g's

The motion of the platform and its acceleration are plotted at the left for the two spring constants of 5,000 and 10,000 lb/ft.

The 10,000 lb/ft case is unacceptable since the acceleration exceeds the safety limit of 4 g's.