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 sinωnt + B cosωnt

where ωn 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:   ωn1 = 2.24 rad/s

     case 2:   ωn2 = 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 cosωnt


Case 1 Results

Case 1 Results


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

     dy/dt = -20 ωn sinωnt

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

     d2y/dt2 = -20 ωn2 cosωnt

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,

     gmax = 20/32.2 ωn2 = 0.621ωn2 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.

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