For a particle executing S.H.M.,, x = displacement from equilibrium position, v = velocity at any instant and a = acceleration at any instant, then

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

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 $v^{2}+a^{2} / \omega^{2}=A^{2} \omega^{2} \ldots . .$ ellipse

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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