If the displacement, velocity and acceleration of particle in $S H M$ are $1 c m, 1 c m / s e c$ and $1 c m / s e c^{2}$ respectively its time period (in secs) will be:

π

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0.5 π

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2 π

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1.5 π

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Solution

The correct option is C $2 π$
$x = 1 c m$
$v = 1 c m / s$
$a = 1 c m / s$
We know that $a = - w^{2} x$
$\Rightarrow | a | = w^{2} | x |$
$\Rightarrow w^{2} = \frac{| a |}{| x |} = \frac{1 c m / s^{2}}{1 c m} = 1 \frac{1}{s^{2}}$
$\Rightarrow w = 1 r a d / s$
time period $T = \frac{2 π}{w} = \frac{2 π}{1} = 2 π s e c$

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