Question

A particle executes simple harmonic motion with an amplitude of $4cm.$At the mean position the velocity of the particle is $10cm/s$ . The distance of the particle from the mean position when its speed becomes $5cm/s$ is

• A. $\sqrt{3}cm$
• B. $\sqrt{5}cm$
• C. $\pm$2$\left(\sqrt{3}\right)cm$
• D. 2$\left(\sqrt{5}\right)cm$

Answer 1

The correct option is C $\pm$2$\left(\sqrt{3}\right)cm$

Given$,$

Amplitude $,$ $A=4cm$

velocity of the particle at mean position $,$ $v₀=10cm/s$

We know the relation between speed of particle $,$ amplitude and distance of particle from mean position is given by$,$

$v=\omega \surd A²-x²$

Here ω is angular frequency$,$

speed at mean position $,$ $v=\omega \sqrt{\left(A^2-X^2\right)}$

speed at x distance from mean position $,$ $v=\omega \left(A_2-X^2\right)=5cm/s$

So$,$ $\frac{\omega A}{\omega \sqrt{\left(A^2-X^2\right)}}=\frac{10}{5}$

$\Rightarrow \frac{A}{\sqrt{\left(A^2-X^2\right)}}=2$

$\Rightarrow \frac{A}{\sqrt{\left(4^2-X^2\right)}}=2$

$\Rightarrow 4=4^2-X^2$

$\Rightarrow X^2=12$

$\Rightarrow X=\pm 2\sqrt{3}cm$

Hence,

option $\left(C\right)$ is correct answer.