Engineering
Physics
Conservation of Energy
Equilibrium
Basics of Simple Harmonic Motion
Question

Column-I represents potential energy graph for certain system. Column-II gives statements related to graphs.

Column I Column II
(P) If total energy is E3, it is not possible for the body to have any turning point in its motion.
(Q) For  a small displacement about point O potential energy function is quadratic in variable plotted on x-axis.
(R) For a small displacement about position O motion is simple harmonic.
(S) If total energy is Etotal < E2 particle executes periodic and oscillatory motion for all energy values greater than energy at O.
  (T) Point O is position of stable equilibrium
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Solution
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This question involves matching potential energy graphs (Column I) with correct physical interpretations (Column II). Key concepts:

Turning points occur where total energy E equals potential energy U. For E3 in graph 1, U ≤ E everywhere, so no turning points exist (matches P).

Near stable equilibrium (point O in graphs), U is approximately quadratic: U12kx2, leading to simple harmonic motion (matches Q, R, T).

For periodic motion, E must be between min(U) and max(U). In graph 4, if Etotal < E2 but > U(O), motion is bounded and oscillatory (matches S).

Final answer: 1-P, 2-QRT, 3-RT, 4-S