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(D) The acceleration of a particle in $S.H.M.$ is given by the relation $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displ

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$1$

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(D) The acceleration of a particle in $S.H.M.$ is given by the relation $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement.For the particle to be at $-x_{max}$,we substitute $x = -x_{max}$ into the equation:$a = -\omega^2 (-x_{max}) = +\omega^2 x_{max}$.This shows that at $x = -x_{max}$,the acceleration $a$ must be at its maximum positive value.Looking at the provided graph,point $1$ corresponds to the maximum positive value of acceleration.Therefore,point $1$ corresponds to the particle being at $-x_{max}$.

The acceleration $a$ of a particle undergoing $S.H.M.$ is shown in the figure. Which of the labelled points corresponds to the particle being at $-x_{max}$?

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