Oscillations Question 2
Question 2 - 2024 (27 Jan Shift 1)
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 $\alpha=$
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Answer: (12)
Solution:
$\mathrm{V}_{\mathrm{ut} \text { mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega$
$\omega=\frac{5}{2}$
$\mathrm{v}=\omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}}$
$5=\frac{5}{2} \sqrt{4^{2}-\mathrm{x}^{2}} \Rightarrow \mathrm{x}^{2}=16-4$
$\mathrm{x}=\sqrt{12} \mathrm{~cm}$
