Answer
$x=\frac{mr\omega_{\circ}^2L^3}{48EI-M\omega_{\circ}^2L^3}$
Work Step by Step
The required amplitude can be determined as follows:
$K=\frac{P}{\delta}$
$\implies K=\frac{P}{\frac{PL^3}{48EI}}$
$\implies K=\frac{48EI}{L^3}$
We know that
$\omega_n=\frac{K}{m}$
$\omega_n=\frac{48EI/L^3}{M}$
$\implies \omega_n=\frac{48EI}{ML^3}$
As $F_{\circ}=mr\omega_{\circ}^2$
Now $x=|\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}|$
$\implies x=\frac{\frac{m\omega_{\circ}^2r}{48EI/L^3}}{1-(\frac{\omega_{\circ}}{\sqrt{\frac{48EI}{ML^3}}})^2}$
This simplifies to:
$x=\frac{mr\omega_{\circ}^2L^3}{48EI-M\omega_{\circ}^2L^3}$
