## Engineering Mechanics: Statics & Dynamics (14th Edition)

$f=0.900Hz$
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: $f=\frac{4\sqrt{2}}{2\pi}$ $\implies f=0.900Hz$