A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at x = 0.280 m, the acceleration of the block is $\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{30}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What is the frequency of the motion?

Frequency (f) is the number of waves that pass through a point in a given period of time (T). It is inversely proportional to the time period.

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{T}}$

The angular frequency, is 2 times the frequency;

$\omega =2\pi f=\frac{2\pi }{T}\left(1\right)$

Now, in terms of force constant or the restoring force (k) and mass (m) of the block;

role="math" localid="1664267743976" $\omega =\sqrt{\frac{k}{m}}\left(2\right)$

Therefore, from the equation (1) & (2), you get the frequency which is,

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\left(3\right)$

If ${\mathit{F}}_{\mathbf{x}}$, the restoring force in periodic motion is directly proportional to the displacement x and k be the force constant, the motion is called simple harmonic motion (SHM), also known as Hooke’s law.

So, you can apply Newton’s second law of motion in the x-direction, as the restoring force is directly proportional to the displacement of the block,

$F_x=m{a}_x\mathbf{}\mathbf{}\mathbf{\left(}\mathbf{5}\mathbf{\right)}$

here, $F_x$is force acting on the block, m is mass of the block and $a_x$is acceleration.

So, from the equation (4) & (5), you get,

$-kx=m{a}_x$

Divide both sides by -x,

$\frac{k}{m}=\frac{-a_x}{x}\left(6\right)$

$f=\frac{1}{2\pi }\sqrt{\frac{-a_x}{x}}$

Given,localid="1664268409480" $a_x$the acceleration is $5.30m/s^2$and displacement, x is 0.280 m

$$\begin{gathered} f=\frac{1}{2\pi }\sqrt{\frac{-\left(-5.30m/s^2\right)}{0.280m}} \\ f=\frac{1}{2\pi }\sqrt{18.93s^{-2}} \\ f=0.692Hz \end{gathered}$$

Hence, the frequency of the motion is $0.692Hz$