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$^{-1}$.

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:

$$T={2\pi}{\sqrt{\frac{10}{250}}}={2\pi}{\sqrt{\frac{1}{25}}}={2\pi}{\frac{1}{5}}=\frac{2\pi}{5}$$

$$T \approx 1.257 \text{ s}$$