For a particle executing $S.H.M.,\, x =$ displacement from equilibrium position, $v =$ velocity at any instant and $a =$ acceleration at any instant, then
A$a-v$ graph is an ellipse
B$v-x$ graph is an ellipse
C$a-x$ graph is a straight line
D
all of the above
Diffcult
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D
all of the above
d $x=A \sin (\omega t+b)$
$\boldsymbol{v}=A \omega \cos (\omega t+b)$
$a=-A \omega^{2} \sin (\omega t+b)$
so $x^{2}+v^{2} / \omega^{2}=A^{2} \ldots$ ellipse
$\boldsymbol{a}=-\boldsymbol{\omega}^{2} \boldsymbol{x} \ldots \ldots$ straight line
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