A particle moves along the x-axis under a conservative force. Its potential energy is U = 5x² − 20x + 2 J, with x in metres. It is released from rest at x = −3 m.
Physics · Simple Harmonic Motion
JEE Advanced

Question

A particle moves along the x-axis under a conservative force. Its potential energy is U = 5x² − 20x + 2 J, with x in metres. It is released from rest at x = −3 m.

G1 Physics · Accepted Answer

For U=5x²−20x+2=5(x−2)²−18, the stable equilibrium is x=2 m. Released from rest at x=−3 m, the displacement from equilibrium is −5 m, so amplitude=5 m. Correct answer to Sub 1: C. The opposite extreme is x=2+5=7 m, so maximum x-coordinate is 7 m. Correct answer to Sub 2: B.