Oscillations Question 2
Question 2 - 2024 (27 Jan Shift 1)
A particle executes simple harmonic motion with an amplitude of $4 cm$. At the mean position, velocity of the particle is $10 cm / s$. The distance of the particle from the mean position when its speed becomes $5 cm / s$ is $\sqrt{\alpha} cm$, where $\alpha=$
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Answer: (12)
Solution:
$V _{ut \text { mean position }}=A \omega \Rightarrow 10=4 \omega$
$\omega=\frac{5}{2}$
$v=\omega \sqrt{A^{2}-x^{2}}$
$5=\frac{5}{2} \sqrt{4^{2}-x^{2}} \Rightarrow x^{2}=16-4$
$x=\sqrt{12} cm$
