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The time period of a particle in simple harmonic motion is T. Assume potential energy at mean position to be zero. After a time of $T/6$ it passes its mean position, its

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KE = PE

acceleration will be nearly

86%

of its maximum acceleration

displacement will be half its amplitude

velocity will be half its maximum velocity

answer is A, C.

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Detailed Solution

The time period of the particle is ‘T’. After sec the displacement of body

$y = A \sin ω t \Rightarrow y = A \sin \frac{2 π}{T} \frac{T}{6} = \frac{\sqrt{3} A}{2}$

velocity $v = ω \sqrt{A^{2} - \frac{{3A}^{2}}{4}} = \frac{ω A}{2} = \frac{V_{\max}}{2}$

Acceleration $a = | - ω^{2} y |$

$a = ω^{2} \frac{\sqrt{3} A}{2}, a_{max} = ω^{2} A, \frac{a}{a_{max}} = \frac{\sqrt{3}}{2}$

$100 \times \frac{a}{a_{max}} = \frac{\sqrt{3}}{2} \times 100 = 86.6 %$

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