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 $\alpha=$ ________.
Answer (integer)
12
Solution
<p>$$\begin{aligned}
& \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\
& \quad \omega=\frac{5}{2} \\
& \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\
& 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\
& \mathrm{x}=\sqrt{12} \mathrm{~cm}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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