For a particle executing S.H.M.,where x is th | vedclass

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(D) For a particle in $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t

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all of the above

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(D) For a particle in $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$.The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$.The acceleration is $a = \frac{dv}{dt} = -A \omega^2 \sin(\omega t + \phi) = -\omega^2 x$.$1$. For the $v-x$ graph: Since $x = A \sin(\omega t + \phi)$ and $v = A \omega \cos(\omega t + \phi)$,we have $(x/A)^2 + (v/A\omega)^2 = 1$,which represents an ellipse.$2$. For the $a-x$ graph: Since $a = -\omega^2 x$,this is a linear equation of the form $y = mx$,which represents a straight line passing through the origin.$3$. For the $a-v$ graph: Since $v = A \omega \cos(\omega t + \phi)$ and $a = -A \omega^2 \sin(\omega t + \phi)$,we have $(v/A\omega)^2 + (a/A\omega^2)^2 = 1$,which represents an ellipse.Therefore,all the given statements are correct.

For a particle executing $S.H.M.$,where $x$ is the displacement from the equilibrium position,$v$ is the velocity at any instant,and $a$ is the acceleration at any instant,then:

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