Answer
$f=0.900Hz$
Work Step by Step
We can determine the natural frequency of vibration as follows:
$\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$
$\implies \omega_n=\sqrt{\frac{3.6}{0.1125}}=4\sqrt{2}rad/s$
Now $f=\frac{\omega_n}{2\pi}$
We plug in the known values to obtain:
$f=\frac{4\sqrt{2}}{2\pi}$
$\implies f=0.900Hz$
