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Consider the microgravity isolation suspension system shown below to be used for

ID: 2326611 • Letter: C

Question

Consider the microgravity isolation suspension system shown below to be used for isolating sensitive experiments in space. The system consists of an outer frame of mass M = 10 kg and an inner experiment box of mass m = 4 kg. The inner box can slide vertically inside the frame without friction and its motion is isolated by 4 springs of stiffness k = 500 N/m. The outer frame is excited by a vertical disturbance force f(t). a) Write the equations of vertical motion of the outer frame of mass M and the inner experiment box of mass m. b) Find the transfer function from the disturbance force f(t) to the vertical displacement x(t) of the inner box. c) Consider a sinusoid disturbance input force f(t) = 4 sin(5t) (in N). Determine the steady-state amplitude of the inner box displacement |x|.

Explanation / Answer

GIVEN:- M = 10kg, m = 4kg, Stiffness k = 500N/m

PART (a):-

When a disturbing force f(t) is applied vertically upwards on the outer frame, the inner box slides downwards by a displacement x(t). To counter this force f(t) and equal but opposite reaction is produced in the inner box = R(t) causing an acceleration a. This reaction is balanced by the spring force f(s) of 4 springs acting on the inner box.

Summing the forces in the y-direction for inner box, we get-->

R(t) = -4 . f(s)

So, m.x(t)" = -4.k.x(t) which gives m.x(t)"+ 4.k.x(t) = 0 (Equation of motion for inner box)

Also since f(t) = R(t), So, M.x(t)" = -4.k.x(t) which gives

M.x(t)" + 4.k.x(t) = 0 (Equation of motion for outer frame)

PART (b):-

Referring standard books we find the transfer function for the inner box, given by x(t) = A. sin(n.t + ) where A & are constants and n is the natural frequency.

PART (c):-

Input force f(t) = 4 sin(5t). TO FIND Steady state amplitude of inner box displacement IxI = ?

So, m.x(t)" = -m.A.n^2.sin(nt) is comparable with m.x(t)" = 4 sin(5t)

That is -m.A.n^2 = 4, which gives A = -4/ (4x5^2)   

So Amplitude A = - 0.04 (ANSWER)