JEE Main 2021 (4 Aug, Shift 2) Physics Q19: Energy in SHM

Question

The potential energy function for a particle executing SHM is given by $V(x)=\frac{1}{2}kx^2$, where k is the force constant of the oscillator. Which of the following diagram correctly shows the position - momentum curve for the motion.

Accepted Answer

Option 1: [figure: P versus X graph; a closed ellipse centred at the origin with an arrow on its upper right part pointing anticlockwise; labels $\sqrt{2mE}$ at the top of the P axis with a small arrow beside it, $-\sqrt{2mE}$ at the bottom of the P axis with a small arrow beside it, $-\sqrt{\frac{2E}{K}}$ at the left end and $\sqrt{\frac{2E}{K}}$ at the right end of the X axis]

Answer

[figure: P versus X graph; a closed ellipse centred at the origin with an arrow on its upper right part pointing anticlockwise; labels $\sqrt{2mE}$ at the top of the P axis with a small arrow beside it, $-\sqrt{2mE}$ at the bottom of the P axis with a small arrow beside it, $-\sqrt{\frac{2E}{K}}$ at the left end and $\sqrt{\frac{2E}{K}}$ at the right end of the X axis]

Answer

[figure: P versus X graph; a closed ellipse centred at the origin with an arrow on its upper right part pointing anticlockwise; labels $+\sqrt{2mE}$ at the top of the P axis and $-\sqrt{2mE}$ at the bottom of the P axis; no labels on the X axis]

Answer

[figure: P versus X graph; an open arc to the right of the P axis, running from a point at height $\sqrt{2mE}$ through the origin to a point at height $-\sqrt{2mE}$, with dashed lines from those two heights on the P axis to the arc ends and a dashed vertical line joining the two ends; arrows on the arc point towards the origin]

Answer

[figure: P versus X graph; an open arc (right half of an ellipse) from $+\sqrt{2mE}$ on the P axis, through $\sqrt{\frac{2E}{K}}$ on the X axis, to $-\sqrt{2mE}$ on the P axis, with arrows pointing downwards (clockwise)]

JEE Main 4 August 2021, Shift 2, Physics Q19

Same topic in other shifts