A particle executes simple harmonic motion with an amplitude of $4 \mathrm{cm}$. At the mean position, velocity of the particle is $10 \mathrm{cm} {\mathrm{s}}^{-1}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{cm} {\mathrm{s}}^{-1}$ is $\sqrt{\alpha } \mathrm{cm}$, where $\alpha =$______.

The formula to calculate the velocity of the particle at the mean position is given by

$V=A\omega ...\left(1\right)$

From equation (1), it follows that

$$\begin{gathered} 10=4\times \omega \\ \Rightarrow \omega =\frac{5}{2} \mathrm{rad} {\mathrm{s}}^{-1} \end{gathered}$$

The formula to calculate the velocity of the particle at any instantaneous position $x$ can be expressed as

$v=\omega \sqrt{A^2-x^2} ...\left(2\right)$

From equation (2), it follows that

$$\begin{gathered} 5=\frac{5}{2}\sqrt{4^2-x^2} \\ \Rightarrow x^2=16-4 \\ =12 \\ \Rightarrow x=\sqrt{12} \mathrm{cm} \end{gathered}$$

Hence, $\alpha =12$.