JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 17)
A particle executes S.H.M., the graph of velocity as a function of displacement is :
a parabola
a helix
an ellipse
a circle
Explanation
For a body performing SHM, relation between velocity and displacement
$$v = \omega \sqrt {{A^2} - {x^2}} $$
now, square both side
$${v^2} = {w^2}({A^2} - {x^2})$$
$$ \Rightarrow {v^2} = {w^2}{A^2} - {\omega ^2}{x^2}$$
$${v^2} + {\omega ^2}{x^2} = {\omega ^2}{A^2}$$
divide whole equation by $${{\omega ^2}{A^2}}$$
$${{{v^2}} \over {{\omega ^2}{A^2}}} + {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}} = {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}}$$
$${{{v^2}} \over {{{(\omega A)}^2}}} + {{{x^2}} \over {{{(A)}^2}}} = 1$$
above equation is similar as standard equation of ellipses, so graph between velocity and displacement will be ellipses.
$$v = \omega \sqrt {{A^2} - {x^2}} $$
now, square both side
$${v^2} = {w^2}({A^2} - {x^2})$$
$$ \Rightarrow {v^2} = {w^2}{A^2} - {\omega ^2}{x^2}$$
$${v^2} + {\omega ^2}{x^2} = {\omega ^2}{A^2}$$
divide whole equation by $${{\omega ^2}{A^2}}$$
$${{{v^2}} \over {{\omega ^2}{A^2}}} + {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}} = {{{\omega ^2}{x^2}} \over {{\omega ^2}{A^2}}}$$
$${{{v^2}} \over {{{(\omega A)}^2}}} + {{{x^2}} \over {{{(A)}^2}}} = 1$$
above equation is similar as standard equation of ellipses, so graph between velocity and displacement will be ellipses.
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