Question 169: Oscillations

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{V}_{	ext {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \
& \quad \omega=\frac{5}{2} \
& \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \
& 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \
& \mathrm{x}=\sqrt{12} \mathrm{~cm}
\end{aligned}$$

Oscillations

Integer / Numeric Answer Type