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CGP EDU Academic Team
Published on: September 12, 2026
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is
$(a) \frac{A}{2} \quad (b) \frac{A}{\sqrt{2}} \quad (c) \frac{A \sqrt{3}}{2} \quad (d) \frac{2A}{\sqrt{3}}$
Text Solution
Verified by ExpertsThe correct answer is:
C
$V_{\max} \Rightarrow V = \frac{\omega A}{2} = \omega \sqrt{A^2 - y^2}$ $\Rightarrow A^2 - y^2 = \frac{A^2}{4} \Rightarrow y^2 = \frac{3A^2}{4} \Rightarrow y = \frac{\sqrt{3}A}{2}$
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