Question

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. $v-x$ graph is a circle
• B. $v-x$ graph is an ellipse
• C. $a-x$ graph is a straight line
• D. $a-v$ graph is an ellipse

Answer 1

Answer: (B), (C), (D)

The correct options are
B $v-x$ graph is an ellipse
C $a-x$ graph is a straight line
D $a-v$ graph is an ellipse

$x=A\sin (\omega t+b)$
$v=A\omega \cos (\omega t+b)$
$a=-A{\omega }^2\sin (\omega t+b)$
so $x^2+v^2/{\omega }^2=A^2$ .....ellipse
$a=-{\omega }^{2}x$ ......straight line
$v^2+a^2/{\omega }^2=A^2{\omega }^2$ ..... ellipse