§ 13.2NCERT Class 11 · Physics · Chapter 13
Energy in Simple Harmonic Motion
Conservation of Mechanical Energy: During SHM, the energy of the oscillating particle continuously transforms between Kinetic Energy (K) and Potential Energy (U). In the absence of dissipative friction/damping forces, the Total Mechanical Energy (E = K + U) remains strictly constant at all points and at all times:
E = K + U = ½ k A2 = ½ m ω2 A2 = Constant
1. Kinetic Energy (K) of an Oscillator
A particle of mass m moving with velocity v has kinetic energy K = ½ m v2:
K as a function of Displacement x:
Substituting v = ω√(A2 − x2):
K(x) = ½ m ω2 (A2 − x2) = ½ k (A2 − x2)
- At Mean Position (x = 0): Kmax = ½ k A2 = E.
- At Extremes (x = ±A): K = 0.
K as a function of Time t:
Substituting v = Aω cos(ωt + φ):
K(t) = ½ m ω2 A2 cos2(ωt + φ)
Since cos2θ oscillates between 0 and 1 with angular frequency 2ω, kinetic energy completes two cycles in one complete time period of SHM!
2. Potential Energy (U) of an Oscillator
Work done against the restoring force F = −kx in displacing the particle by dx is stored as elastic potential energy (assuming U = 0 at x = 0):
U as a function of Displacement x:
Integrating dU = −F dx = kx dx:
U(x) = ½ k x2 = ½ m ω2 x2
- At Mean Position (x = 0): U = 0 (minimum).
- At Extremes (x = ±A): Umax = ½ k A2 = E.
U as a function of Time t:
Substituting x = A sin(ωt + φ):
U(t) = ½ k A2 sin2(ωt + φ)
Potential energy also oscillates with angular frequency 2ω (twice the displacement frequency).
3. Special Cross-Over Points (Where K = U)
Equal Energy Point: At what displacement is the kinetic energy equal to the potential energy?
K = U ⇒ ½ k (A2 − x2) = ½ k x2 ⇒ 2x2 = A2 ⇒ x = ± A / √2 ≈ ± 0.707 A
At this point, K = U = E / 2 = ¼ k A2.
4. Time Average vs Position Average of Energy
| Averaging Method | Average Kinetic Energy ⟨K⟩ | Average Potential Energy ⟨U⟩ | Total Energy ⟨E⟩ |
|---|---|---|---|
| Time Average (over one full period T) | ¼ k A2 = ½ E | ¼ k A2 = ½ E | E |
| Position Average (over −A to +A) | (2/3) E = (1/3) k A2 | (1/3) E = (1/6) k A2 | E |
High-Yield JEE Trap:
- If frequency of SHM is f, then the frequency of oscillation of Kinetic Energy and Potential Energy is 2f (frequency doubles).
- However, the frequency of Total Energy E is ZERO (because E is a time-independent constant!).
- If the potential energy at the mean position is non-zero (U0), then U(x) = U0 + ½ k x2 and Total Energy E = U0 + ½ k A2. Kinetic energy remains completely unaffected!
Your progress
Saved on this device only — no account, no sign-in.