The amplitude and time period of a particle of mass 0.1 kg executing simple harmonic motion are 1m and 6.28s, respectively. Find its (i) angular frequency, (ii) acceleration and (iii) velocity at a displacement of 0.5m.

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$1 rad / \sec, - 0 .5 {ms}^{- 2}, \frac{\sqrt{3}}{2} {ms}^{- 1}$

$1 rad / \sec, - 0 .5 {ms}^{- 2}, \sqrt{3} {ms}^{- 1}$

$1 rad / \sec, - 1 {ms}^{- 2}, \sqrt{3} {ms}^{- 1}$

$0 .5 rad / \sec, - 0 .5 {ms}^{- 2}, \frac{\sqrt{3}}{2} {ms}^{- 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} m = 0 .1 kg; A = 1 m; T = 6 .28 \sec; y = 0 .5 m \\ \Rightarrow ω = \frac{2 π}{I} = \frac{2 \times 3 .14}{6 .28} = 1 rad / \sec \\ \Rightarrow Acceleration a = - ω^{2} y \\ a = - {[\frac{2 π}{T}]}^{2}; y = {[\frac{2 \times 3 .14}{6 .28}]}^{2} \times 0 .5 = 1^{2} \times 0 .5 \end{array}$
$\begin{array}{l} = - 1^{2} \times 0 .5 = - 0 .5 m / s^{2} \\ \Rightarrow velocity = ω \sqrt{A^{2} - y^{2}} = (1) \sqrt{1^{2} - (0 .5)^{2}} \\ = \sqrt{1 - {(\frac{1}{2})}^{2}} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2} m / \sec \end{array}$

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