Calculating time period of oscillation of a mass on a spring I have the question:

"A mass of $10$ kg bounces up and down on a spring. The spring constant is $250 $  N m$^{-1}$.
  Calculate the time period of the oscillation."

I know that time period $T = 1/f$. However I am not sure how I would work out the time period using the spring constant $250$ N m$^{-1}$.
 A: For a spring, we know that $F=-kx$, where $k$ is the spring constant.
Therefore, from $F=ma$, we deduce that:
$$a=-\frac{k}{m} x$$
We let $\omega^2=\frac{k}{m}$.
Thus, $a=-\omega^2 x$.
Therefore:
$$-\omega^2 x=-\frac{k}{m} x$$
$$\omega=\sqrt{\frac{k}{m}}$$
From the laws of Simple Harmonic Motion, we deduce that the period $T$ is equal to:
$$T=\frac{2\pi}{\omega}$$
Hence, we derive the following relation:
$$T={2\pi}{\sqrt{\frac{m}{k}}}$$
Therefore, we substitute $m=10$ and $k=250$ to obtain the solution:
$$T={2\pi}{\sqrt{\frac{10}{250}}}={2\pi}{\sqrt{\frac{1}{25}}}={2\pi}{\frac{1}{5}}=\frac{2\pi}{5}$$
$$T \approx 1.257 \text{ s}$$
A: Note $f = \dfrac{\omega}{2\pi}$, where $\omega$ is the angular frequency. Let $m$ denote the mass and $k$ the spring constant. Since $\omega^2 = \dfrac{k}{m}$ and $T = \dfrac{1}{f}$, then $$T = 2\pi \sqrt{\frac{m}{k}}$$
A: The equation of motion for  a simple harmonic oscillator,is
$$
\frac{d^2 x}{dt^2}=-\frac{k}{m}x
$$
The period is given by:
$$
T=2\pi\sqrt{\frac{m}{k}}
$$
where $m$ is the mass  and $k$ is the elastic cosnstant of the spring.
A: $\omega^2=\frac{k}{m}$ by elementary physics.  To see this write the equation of motion for your system:
$$
m\ddot x=-k x
$$
and solve the differential equation assuming $x(t)=A\cos(\omega t+\phi)$ with $\omega,\phi$ constants. 
