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Q.

The equation for the displacement of a particle executing SHM is y = {5 sin 2 $π$ t} cm. Find (i) the velocity at 3cm from the mean position, (ii) acceleration after 0.5s after leaving the mean position.

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a

$8 π {cms}^{- 1}, zero$

b

$6 π {cms}^{- 1}, zero$

c

$4 π {cms}^{- 1}, zero$

d

$2 π {cms}^{- 1}, zero$

answer is A.

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Detailed Solution

$\begin{array}{l} y = 5 \sin 2 πt \\ A = 5 cm ω = 2 π \Rightarrow T = 1 \sec y = 3 cm \\ v = ω \sqrt{A^{2} - y^{2}} = \frac{2 π}{T} \sqrt{25 - 9} \\ v = 2 π \times 4 = 8 πcm / \sec \\ a = ω^{2} y \Rightarrow after 0 .5 \sec it is at mean position \\ \Rightarrow y = 0 a = 0 \end{array}$

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