Physics · Oscillations · NEET
We measure PE from the mean (equilibrium) position, where x = 0. Since PE = (1/2) m w^2 x^2, putting x = 0 gives PE = 0. The mean position is the reference where the restoring force is zero, so no energy is stored there. All the energy at the mean position is kinetic (the particle moves fastest there).
At the extreme position the displacement is largest, x = A (the amplitude). PE = (1/2) m w^2 x^2 becomes PE_max = (1/2) m w^2 A^2. Here the particle stops for an instant (velocity = 0), so kinetic energy is zero and all the energy is potential. This maximum PE equals the total energy E of the SHM.
Displacement is x = A sin(wt + phi). Substituting into PE = (1/2) m w^2 x^2 gives PE = (1/2) m w^2 A^2 sin^2(wt + phi). Because sin^2 is always positive and repeats twice per cycle, PE varies at double the frequency of the SHM. That double-frequency idea is explained in the next concept.
When you plot PE against displacement x, the equation PE = (1/2) m w^2 x^2 is of the form y = c x^2, which is a parabola opening upward (a U or smile shape). It is symmetric about x = 0. Only when you plot PE against time does it look like a sin^2 curve. Do not confuse the PE-vs-x graph (parabola) with the PE-vs-t graph (sin^2 hump).
No. The formula PE = (1/2) k x^2 comes from the restoring force F = -kx, but SHM can be mechanical (spring), gravitational (pendulum), or other types. The idea is the same: whenever the restoring force is proportional to displacement, the stored PE is proportional to x^2. For a pendulum the PE is gravitational, but it still follows the (1/2) m w^2 x^2 pattern for small angles.
The sum of the kinetic energy and potential energy of a simple pendulum bob is 0.02 J. The speed of the bob at its equilibrium position is approximately (mass of the bob = 20 g).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
PE = (1/2) k x^2 = (1/2) m w^2 x^2, where x is displacement from the mean position, k is the force constant, m is mass, and w is angular frequency.
At the extreme positions (x = A), where PE_max = (1/2) m w^2 A^2. This maximum equals the total energy of the SHM.
At the mean position (x = 0), where PE = 0. At this point the kinetic energy is maximum.
It is an upward-opening parabola (a smile shape) symmetric about x = 0, because PE is proportional to x squared.
Total energy E = (1/2) m w^2 A^2 is constant. At any point E = KE + PE, so PE = E - KE. PE equals E at the extremes and is zero at the mean position.
PE = (1/2) m w^2 A^2 sin^2(wt + phi), so it varies at twice the frequency of the SHM itself. This is covered in the next concept.
