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If the displacement (y in m) and velocity $({v in ms}^{- 1})$ of a particle executing SHM are related by the equation ${4v}^{2} = 16 - y^{2}$ . Find maximum acceleration of body.

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2 ms^-2

3 ms^-2

1 ms^-2

4 ms^-2

answer is A.

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Detailed Solution

$\begin{array}{l} 4 V^{2} = 16 - y^{2}; V^{2} = \frac{1}{4} (16 - y^{2}) \\ V = \frac{1}{2} \sqrt{16 - y^{2}}; ω = \frac{1}{2} rad / s \\ A^{2} = 16; A = \pm 4 m \end{array}$
a maximum = ${Aw}^{2}$
= $4 \frac{1}{4} {=1 ms}^{- 2}$

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