Question 2.3: (a) Derive an expression for the displacement response, z, w......

Part  (a)  :
In this case the equation to be solved is

m \ddot{z} +c\dot{z} + kz = P        (A)

Equation (A) can be written in a form similar to Eq. (2.20),

\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n} z = 0         (2.20)

sometimes known as the ‘standard form’:

\ddot{z} + 2\gamma \omega _{n}\dot{z} + \omega ^{2}_{n} z = P / m         (B)

Since \gamma <1, the complementary function z = e^{-\gamma \omega _{n}t}(A \cos \omega _{d}t + B \sin \omega _{d}t ), Eq. (2.28), is appropriate. For the particular integral, from Table 2.1, we try z = C, which makes \dot{z} = 0 and  \ddot{z} = 0. Substituting these into Eq. (B) gives  C=P/m\omega ^{2}_{n}, so the complete solution is

z = e^{-\gamma \omega _{n}t}(A \cos \omega _{d}  t + B \sin \omega _{d}  t ) +P/(m   \omega ^{2}_{n} )         (C)

Differentiating

\dot{z} = e^{-\gamma \omega _{n}t}[(B\omega _{d} – A\gamma \omega _{n})\cos \omega _{d}t  –  (A\omega _{d} + B\gamma \omega _{n}) \sin \omega _{d}t]          (D)

Inserting the initial conditions z = 0 and \dot{z} = 0 at t = 0 into Eqs (C) and(D), and solving for the constants A and B gives, using Eqs (2.16) and (2.26),

\omega _{n} = \sqrt{\frac{k}{m} }         (2.16)

\omega _{d} = \omega _{n}\sqrt{(1-\gamma ^{2})}          (2.26)

A = -P/(m\omega ^{2}_{n} ) = – P/k           (E)

and

B = – P\gamma \omega _{n} / (m\omega ^{2}_{n} \omega _{d}) = – \frac{P\gamma }{k\sqrt{1-\gamma ^{2}} }           (F)

Inserting Eqs (E) and (F) into Eq. (C), gives the solution:

z = \frac{P}{k}\left[1- e^{-\gamma \omega _{n}t} \left(\cos \omega _{d}t + \frac{\gamma }{\sqrt{(1-\gamma ^{2})} } \sin \omega _{d}t \right) \right]         (G)

Part  (b)  :
Now the displacement of the system when a static load, equal in magnitude to P, is applied, is z_{s}, say, where z_{s} = P/k. Dividing both sides of Eq. (G) by z_{s}, we have a non-dimensional version:

\frac{z}{z_{s}} = 1- e^{-\gamma \omega _{n}t} \left(\cos \omega _{d}t + \frac{\gamma }{\sqrt{1-\gamma ^{2}} } \sin \omega _{d}t \right)         (H)

This is plotted in Fig. 2.5, with \gamma = 0.1 and \omega _{n} = 10. It is seen that the displacement approaches twice the static value, before settling at the static value.