Question

A single degree of freedom spring-mass system is subjected to a harmonic force of constant ampitude. For an excitation frequency of $\sqrt{\frac{3k}{m}}$, the ratio of the amplitude of steady state response to the static deflection of the spring is

Answer 1

$\frac{A}{x_0}=$ Magnification factor
$\frac{A}{F_0/k}=\frac{1}{\sqrt{1-{\left(\frac{\omega }{W_n}\right)}^2+{\left(2\zeta \frac{\omega }{{\omega }_n}\right)}^2}}$
$\because \zeta =0$
$\frac{A}{F_0/k}=\frac{1}{{\left(\frac{\omega }{{\omega }_n}\right)}^2-1}$
$\omega =\sqrt{\frac{3k}{m}},{\omega }_n=\sqrt{\frac{k}{m}}$
$\frac{\omega }{{\omega }_n}=\sqrt{\frac{3k}{m}}\times \sqrt{\frac{m}{k}}=\sqrt{3}$
$\frac{A}{F_0/k}=\frac{1}{(\sqrt{3}{)}^2-1}=\frac{1}{2}$