FORCED VIBRATION DUE TO SUPPORT EXCITATION.

By newton’s 2nd law of motion the equation can be written as

$m\ddot{x} \ + \ c(\dot{x} – \dot{y}) \ + \ k(x-y) \ = \ 0$

i.e. $m \ddot{x} \ + \ c\dot{x} \ + \ kx \ = \ c\dot{y} \ + \ ky$

but y = Y sin wt

$\dot{y}$ = Yw cos wt

Substituting in above equation we get

$m\ddot{x} + c\dot{x} + kx = Y [ cw \ cos \ wt + k \ sin \ wt]$ - - - - (1)

Let $tab \ \alpha \ = \ \frac{cw}{k}$

From the triangle, $sin \ \alpha \ = \ \frac{cw}{n}$ and $cos \ \alpha \ = \ \frac{k}{n}$

Multiplying and dividing by ‘n’ on RHS of equation no. (1) we get

$m\ddot{x} + c\dot{x} + kx= n \ Y [ \frac{cw}{n} \ cos \ wt \ + \ \frac{k}{n} \ sin \ wt]$

$m\ddot{x} + c\dot{x} + kx = \sqrt{c^2 w^2 + k^2 \ Y} [sin \ \alpha \ cos \ wt \ + \ cos \ \alpha \ sin \ wt]$

$T \sqrt{c^2 w^2 + k^2 } sin \ (wt \ + \ \alpha)$ - - - (2)

Above equation is same as that of

$m\ddot{x} + c\dot{x} + kx = F_0 \ sin \ wt$ whose solution was written as

$X \ = \ X \ sin \ (wt - \phi)$

Then from equation (19) we can write the solution.

$x \ = \ X \ sin [ (wt + \alpha) - \phi]$

$= X \ sin \ [wt - \phi + a]$

Then steady state amplitude is given by equation.

$X \ = \ \frac{(Y \sqrt{c^2 w^2 + k^2)} / k}{\sqrt{ (1- r^2)^2 + (\ \xi r)^2}}$

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

The variation of $(\frac{x}{y})$ versus r is shown below.

NOTE: The motion of mass lags that of the support through an angle $(\phi - \alpha)$ Where $tan \ \phi \ = \ \frac{2 \xi \ r}{1 – r^2}, \ tan \ \alpha \ = \ \frac{cw}{k} = 2 \xi \ r$