An object executing SHM has an acceleration of 0.2m/s² at a distance of 0.05m from the mean position. Calculate the period of SHM.

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5.764sec

4.654sec

3.142sec

2.764sec

answer is C.

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Detailed Solution

$\begin{array}{l} a = 0 .2 m / \sec^{2}, T = ? \\ y = 0 .05 m \\ A = ω^{2} y \Rightarrow ω^{2} = \frac{0 .2}{0 .05} = \frac{20}{5} = 4 \Rightarrow ω = 2 rad / s \\ \frac{2 π}{T} = 2 \Rightarrow π = T T = 3.142 \sec \end{array}$

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