Energy in Simple Harmonic Motion (SHM)
Goal – Master the energy aspects of SHM to crack IIT‑JEE (Main & Advanced) and other competitive exams.
1. Quick Reference
| Quantity | Symbol | Expression (for a mass‑spring) | Units |
|---|---|---|---|
| Displacement | (x) | (A\cos(\omega t+\phi)) | m |
| Velocity | (v) | (-A\omega\sin(\omega t+\phi)) | m s⁻¹ |
| Acceleration | (a) | (-\omega^2x) | m s⁻² |
| Kinetic Energy | (K) | (\tfrac12 m v^2 = \tfrac12 m A^2 \omega^2 \sin^2(\omega t+\phi)) | J |
| Potential Energy | (U) | (\tfrac12 k x^2 = \tfrac12 m \omega^2 A^2 \cos^2(\omega t+\phi)) | J |
| Total Energy | (E) | (K+U = \tfrac12 m A^2 \omega^2) | J |
| Angular Frequency | (\omega) | (\sqrt{k/m}) (mass‑spring) or (\sqrt{g/L}) (simple pendulum) | rad s⁻¹ |
| Period | (T) | (2\pi/\omega) | s |
| Amplitude | (A) | Max displacement | m |
Observation: In an ideal SHM (no damping, no driving), total mechanical energy is constant.
2. Derivation of Energy Expressions
2.1. Mass–spring oscillator
- Equation of motion
[ m\ddot{x} + kx = 0 \quad \Rightarrow \quad \ddot{x} + \omega^2 x = 0, \quad \omega=\sqrt{\frac{k}{m}} ] - General solution
[ x(t) = A\cos(\omega t+\phi) ] - Velocity
[ v(t) = \dot{x} = -A\omega\sin(\omega t+\phi) ] - Kinetic Energy
[ K = \frac12 m v^2 = \frac12 m A^2 \omega^2 \sin^2(\omega t+\phi) ] - Potential Energy (elastic)
[ U = \frac12 k x^2 = \frac12 m\omega^2 A^2 \cos^2(\omega t+\phi) ] - Total Energy
[ E = K+U = \frac12 m A^2 \omega^2(\sin^2+\cos^2)=\frac12 m A^2 \omega^2 ] Constant – independent of (t).
Key Insight: The sum of (\sin^2) and (\cos^2) is always unity. Thus the mechanical energy is conserved.
2.2. Simple Pendulum (small angle)
- Equation
[ \ddot{\theta} + \frac{g}{L}\theta = 0 \quad \Rightarrow \quad \omega=\sqrt{\frac{g}{L}} ] - Energy
- Kinetic: (K = \tfrac12 m L^2 \dot{\theta}^2)
- Potential (relative to lowest point): (U = m g L (1-\cos\theta))
- Total: (E = \tfrac12 m L^2 \dot{\theta}^2 + m g L (1-\cos\theta))
For small (\theta), (\cos\theta \approx 1-\tfrac12\theta^2) →
[ U \approx \tfrac12 m g L \theta^2 = \tfrac12 k_{\text{eff}}x^2 ] with (x=L\theta). Thus reduces to mass‑spring form.
3. Energy in Non‑Ideal SHM
| Situation | Energy Behavior | Remark |
|---|---|---|
| Damped SHM (drag (c\dot{x})) | (E(t)=E_0 e^{- \gamma t}), (\gamma = c/(2m)) | Energy lost to friction/air resistance. |
| Forced SHM (drive (F_0\cos \omega_d t)) | Power input (P = F_0 v \cos(\omega_d t)) | Steady state: (E(t)) oscillates about a mean; resonance when (\omega_d\approx \omega_0). |
| Non‑linear restoring force (e.g., (kx + \alpha x^3)) | Energy still conserved if no damping, but (E = \tfrac12 m v^2 + \tfrac12 kx^2 + \tfrac14 \alpha x^4) | Non‑linear terms modify amplitude–frequency relation. |
Tricky Point – Many JEE questions disguise a damped/forced oscillator as a simple SHM. Look for explicit damping terms or external forces.
4. Important Formulae & Relations
| Relation | Use |
|---|---|
| (E = \frac12 m A^2 \omega^2) | Total energy in an ideal SHM |
| (K_{\max} = E = U_{\max}) | At turning points, all energy is potential; at equilibrium, all is kinetic |
| (K(t) = E \sin^2(\omega t+\phi)) | Kinetic fraction |
| (U(t) = E \cos^2(\omega t+\phi)) | Potential fraction |
| (K = \tfrac12 m v^2) | Kinetic energy |
| (U = \tfrac12 k x^2) | Elastic potential |
| (E_{\text{damped}}(t) = E |