Question:
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement \(d\). Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement \(d\). Which one of the following represents these correctly?
(graphs are schematic and not drawn to scale)
(graphs are schematic and not drawn to scale)
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For a pendulum (PE measured from the mean position) \(PE\propto d^2\) is a minimum at \(d=0\), while \(KE = E-PE\) is a maximum at \(d=0\) and zero at the extremes.
Updated On: Nov 11, 2026
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The Correct Option is D
Solution and Explanation
Total energy \(E\) is constant. Taking PE zero at the mean position, \(PE = \frac12 kd^2\) (upward parabola, minimum at \(d=0\), maximum \(E\) at the extremes \(d=\pm A\)).
\(KE = E - PE = \frac12 k(A^2-d^2)\): downward parabola, maximum \(E\) at \(d=0\) and zero at \(d=\pm A\).
Option (1): KE upward and PE downward below the axis - wrong. Option (2): cusped curves - wrong shapes. Option (3): KE is minimum and PE maximum at \(d=0\), i.e. interchanged - wrong. Option (4): KE is an inverted parabola with maximum at \(d=0\) and zero at the ends, PE is an upright parabola with minimum at \(d=0\) - correct.
Answer: (4) graph with KE as an inverted parabola and PE as an upright parabola
\(KE = E - PE = \frac12 k(A^2-d^2)\): downward parabola, maximum \(E\) at \(d=0\) and zero at \(d=\pm A\).
Option (1): KE upward and PE downward below the axis - wrong. Option (2): cusped curves - wrong shapes. Option (3): KE is minimum and PE maximum at \(d=0\), i.e. interchanged - wrong. Option (4): KE is an inverted parabola with maximum at \(d=0\) and zero at the ends, PE is an upright parabola with minimum at \(d=0\) - correct.
Answer: (4) graph with KE as an inverted parabola and PE as an upright parabola
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