Question

The amplitude of a particle executing S.H.M. is $4\mathrm{cm}$. At the mean position, the speed of the particle is $16{\mathrm{cms}}^{-1}$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}{\mathrm{cms}}^{-1}$, will be

Answer 1

Step 1: Given data

The amplitude of the particle, $A=4\mathrm{cm}$

At the mean position, the velocity will be maximum. Thus, maximum velocity, ${\mathrm{V}}_{\max }=16{\mathrm{cms}}^{-1}$.

Let $y$ be the position of the particle from its mean and $\mathrm{V}$ be its velocity at $y$.

From the given, $\mathrm{V}=8\sqrt{3}{\mathrm{cms}}^{-1}$.

Step 2: Calculate angular frequency

We know that, in SHM, the velocity of the particle is given as,

$\mathbf{V}\mathbf{=}\mathit{\omega }\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}$

Where $\omega$ is the angular frequency,

At the mean position, $y=0$

Thus, ${\mathrm{V}}_{\max }=\omega A$

Substituting the values for ${\mathrm{V}}_{\max }$ and $A$,

$\begin{array}{cc}\Rightarrow & 16=\omega \times 4\\ \Rightarrow & \omega =4\end{array}$

Step 3: Calculate the required position of the particle

We know that the velocity of the particle at the required position is $\mathrm{V}=8\sqrt{3}{\mathrm{cms}}^{-1}$

Substituting this in the main formula,

$\begin{array}{cc}\Rightarrow & 8\sqrt{3}=4\sqrt{4^2-y^2}\\ \Rightarrow & 192=16\left(16-y^2\right)\\ \Rightarrow & 12=16-y^2\\ \Rightarrow & y^2=4\\ \Rightarrow & y=2\mathrm{cm}\end{array}$

Therefore, the position of the particle from the mean when the velocity is $8\sqrt{3}{\mathrm{cms}}^{-1}$ is $2\mathrm{cm}$.