A motion transmission system is modeled as the following suspension system where the excitation z(t) is a vertical displacement excitation at the end-point C of the vertical spring k. The objective is to examine the vertical motion x(t) of the mass m. The bar OAB is assumed to be massless and it rotates about the fixed point O. The system is initially at a horizontal equilibrium. A) Write the linearized equations of motion of the system and determine the transfer function. G(s) = X(s)/Z(s).
B) Assume that the input z(t) is a unit step displacement input. Determine the motion x(t) of the mass m as a function of the system parameters.

(A) 2k/s² M+k (B) C = -M/k

Explanation:

Solution

Given that:

(A) Write the linearized equations of motion of the system determine the transfer function. G(s) = X(s)/Z(s).

Now

G(s) = X(s)/Z(s)

F(t) =  M. d² X(t)/dt² + K X(t)(1)

We convert the equation in the s domain

F(s) = s² M X(s) +kX (s)

F(s) =X(s)  [M s² + k]

F = mgl

F(s) =X(s)  [M s² + k]

mgl = X(s) [s² M +k]

F(t) = KZ(t) (2)

Thus

mgl/2 = kZ(s)

Z(s) = mgl/2k

So

X(s)/Z(s) = mgl/[s² m+k]/mgl/2k

Mgl/[s² M+k] *2k/mgl

Now

X (s)/Z(s) = 2k/s² M+k

(B) The next step is to determine  the motion x(t) of the mass m as a function of the system parameters.

Now,

Let assume that input z(t) is a unit step

Thus

X(s) =1/s  2k/s² M+k

=1/s 2k/M[s² + k}

So,

I/s = 2k/M [ s +k/MJ] [s- k/M J]

By applying practical fractions we have the following :

A/s + B/ [s +k/MJ] +C/  [s- k/M J]

Thus,

S = 0

A (k/MJ + s) (J -k/M + J) = 2k/M

A ( 0 +k/MJ) ( 0 -k/MJ) = 2k/M

so,

A (k/MJ) (- k/MJ) = 2k/M

A = (k²/M²) = 2k/M

A = 2M/k

Now,

we put s  = k/M J

C * k/M J ( 2k/MJ ) = 2k/M

C (2k²/m²) = 2k/M

C (k/M)

C = -M/k

answer:

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