Let z represent relative of mass with respect to support then Z = X – Y i.e. X = Z + Y

We know the equation of motion, mx + cx + kx = cy + ky

Substituting values of x, x, x, in above equation we get

M$(\ddot{z} + \ddot{y}) + c(\dot{z} + \dot{y}) + k(z + y) = c\dot{y} + ky$

$\therefore m\ddot{z} + m\ddot{y} + c\dot{z} + c\dot{y} + kz + ky = c\dot{y} + ky$

$\therefore m\ddot{z} + c\dot{z} + kz = -m\ddot{y}$

Y = Y sin wt

$\therefore \ddot{y} = - Y sin wt$

Substituting in above equation

$M\ddot{z} + c\dot{z} + kz = mw^2 \ Y \ sin \ wt$

Above equation is same as:

$M\ddot{x} + c\dot{x} + kx = F_0 \ sin \ wt$

Hence the relative amplitude

$Z = \frac{(mw^2Y/K)}{\sqrt{(1- r^2) + (2 \xi r)^2}}$

$\therefore$ $\frac{z}{y} = \frac{r^2}{\sqrt{(1-r^2)^2 + (2 \xi r)^2}}$

The variation of $(\frac{z}{y})$ versus r is as shown in figure 13.

The angle of phase lag between the excitation and the relative displacement is:

$tan \ \phi \ = \ \frac{2 \xi r}{1 – r^2}$