The equation of motion of a particle executing S.H.M. where letters have usual meaning is:

\frac{d^{2} x}{d t^{2}} = \frac{k}{m} x

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\frac{d^{2} x}{d t^{2}} = + ω^{2} x

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\frac{d^{2} x}{d t^{2}} = - ω^{2} x^{2}

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\frac{d^{2} x}{d t^{2}} = - k m x

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Solution

The correct option is A $\frac{d^{2} x}{d t^{2}} = \frac{k}{m} x$
We know that accele ratio is its velocity per unit time. The differential equation of a particular SHM is $\frac{d^{x}}{d t^{2}} + (\frac{k}{m}) x = 0$
where, $\frac{d^{2}}{d t^{2}}$ is the acceleration of the particle $x$ is the displacement of the particle $m$ is the mass of the particle and $k$ is force constant. We knot that $\frac{k}{m} = w^{2}$ where $w$ is the angular frequency.
Therefore $\frac{d^{2} x}{d t^{2}} + w^{2} x = 0$
Hence, acceleration of SHM $= \frac{d^{2} x}{d t^{2}} = w^{2} x$
The negative sign indicates that acceleration and displacement are in opposite direction of each other.
Hence, the answer is $\frac{d^{2} x}{d t^{2}} = \frac{k}{m} x .$

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