Question 6.2: (a) Use assumed modes to transform the equations already der......

Part (a):
The equations for this system were derived in Section 6.1 as Eqs (6.2), (6.5) and (6.10), which are all the same. They are

\begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix} \begin{Bmatrix} \ddot{z}_{1} \\ \ddot{z}_{2} \end{Bmatrix} + \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} \\ -k_{2} & k_{2} \end{bmatrix} \begin{Bmatrix} z_{1} \\ z_{2} \end{Bmatrix} = \begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}              (A)

These equations were formed by taking the displacements, z_{1}  \mathrm{and}  z_{2}, of the individual masses, m_{1}  \mathrm{and}  m_{2}, as the generalized coordinates. This resulted in a coupled stiffness matrix, but an uncoupled mass matrix.
We can now transform the coordinates, using the transformation defined by Eq. (6.26):

\left\{z\right\} = \left[X\right] \left\{q\right\}                         (B)

Let us choose to represent the motion of the system by two coordinates, q_{1}  \mathrm{and}  q_{2}, where the actual displacements, z_{1}  \mathrm{and}  z_{2}, of the masses m_{1}  \mathrm{and}  m_{2} are given by the following transformation, corresponding to Eq. (B):

\begin{Bmatrix} z_{1} \\ z_{2} \end{Bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix}             (C)

The columns of the 2 × 2 matrix are simple examples of assumed modes; we have arbitrarily decided that the first mode is \begin{Bmatrix} 1 \\ 1 \end{Bmatrix}, i.e. the two masses move one unit upwards together, and the second mode is \begin{Bmatrix} 0 \\ 1 \end{Bmatrix}, i.e. m_{1} does not move at all, and m_{2} moves one unit upwards. From Fig. 6.3, we see that the first mode compresses k_{1}, but not k_{2}, and that the second mode compresses only k_{2}\mathrm{,  not}  k_{1}. We would therefore expect that using these modes would remove the stiffness coupling.
We can now use the expressions developed in Section 6.2.3 to transform the equations from the form of Eq. (6.24):

[M]\left\{\ddot{z} \right\} + [K]\left\{z\right\} = \left\{F\right\}                   (D)

to the form of Eq. (6.25):

[\underline{M} ]\left\{\ddot{q} \right\} + [\underline{K} ]\left\{q\right\} = \left\{Q\right\}                     (E)

The new mass and stiffness matrices are given by Eqs (6.31) and (6.33), respectively, i.e.,

[\underline{M}] = [X]^{T}[M][X]                          (F)

and

[\underline{K}] = [X]^{T}[K][X]                             (G)

where

[X]=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}              (H)

Substituting Eqs (F) and (G) into Eq. (E) gives the new equations:

[X]^{T}[M][X]\left\{\ddot{q} \right\}+[X]^{T}[K][X]\left\{q\right\} = \left\{Q\right\}              (I)

or written out in full, using Eq. (H):

\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{T}\begin{bmatrix} m_{1} & 0 \\ 0 & m_{2} \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{Bmatrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{Bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{T}\begin{bmatrix} (k_{1}+k_{2}) & -k_{2} \\ -k_{2} & k_{2} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix}             (J)

Multiplied out, these become

\begin{bmatrix} (m_{1}+m_{2}) & m_{2} \\ m_{2} & m_{2} \end{bmatrix}\begin{Bmatrix} \ddot{q}_{1} \\ \ddot{q}_{2} \end{Bmatrix} + \begin{bmatrix} k_{1} & 0 \\ 0 & k_{2} \end{bmatrix} \begin{Bmatrix} q_{1} \\ q_{2} \end{Bmatrix} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix}               (K)

This has had the desired effect of uncoupling the stiffness matrix, and the mass matrix is now coupled instead.
Part (b):
It is now easy to input actual external forces F_{1}  \mathrm{and}  F_{2} by using Eq. (6.39),

\left\{Q\right\} = [X]^{T}\left\{F\right\}                           (L)

to convert them to modal external forces {Q}. In this example:

\left\{Q\right\} = \begin{Bmatrix} Q_{1} \\ Q_{2} \end{Bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^{T}\begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\begin{Bmatrix} F_{1} \\ F_{2} \end{Bmatrix}                  (M)

If the equations were then solved to find the time histories of q_{1}  \mathrm{and}  q_{2}, Eq. (C) could then be used to find the local displacement time histories, z_{1}  \mathrm{and}  z_{2}