A body undergoing S.H.M. has a maximum velocity of $1 m s^{- 1}$ and a maximum acceleration of $4 m$ $s^{- 2}$ .Its amplitude in metres is :

1 m

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0.75 m

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0.5 m

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0.25 m

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Solution

The correct option is D $0.25 m$
$v_{m a x} = \frac{1 m}{s} = ω A$
$a_{m a x} = \frac{4 m}{s} = - ω^{2} A$
$a_{m a x} = - ω . v_{m a x}$
$ω = 4 r a d / s e c$
$A = \frac{1}{4} = 0.25$

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