## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

$f = 1.6~Hz$
The mass oscillates about the equilibrium point which is $x = 10~cm$ below the point where the mass was released. At this point, the force from the spring is equal in magnitude to the weight. We can write an expression for $\frac{k}{m}$: $kx = mg$ $\frac{k}{m} = \frac{g}{x}$ We can find the frequency of the oscillations; $f = \frac{1}{2\pi}~\sqrt{\frac{k}{m}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{g}{x}}$ $f = \frac{1}{2\pi}~\sqrt{\frac{9.80~m/s^2}{0.10~m}}$ $f = 1.6~Hz$