← Mechanical Systems
6 Simple harmonic motion (SHM)
SHM is motion where the restoring force (and so acceleration) is proportional to displacement and directed back toward equilibrium: a = −ω²x.Related in other subjects
SHM is genuinely a calculus problem wearing a physics costume:
Calculus: Differentiation methods (91578) a = −ω²x is a statement about DERIVATIVES: acceleration is the second derivative of displacement, so SHM is really the differential equation d²x/dt² = −ω²x. When you differentiate x = A cos(ωt) twice using the chain rule, you get −ω²A cos(ωt) = −ω²x, which is exactly why cosine describes SHM. The chain rule you practise in Calculus is the tool that proves the Physics. Calculus: Integration methods (91579) Going the other way, integrating acceleration gives velocity and integrating again gives displacement. That is also how the energy result E = ½kA² is derived, by integrating the restoring force F = −kx over displacement (work done = ∫F dx). Defining relation a = −ω² x The negative sign = restoring (opposes displacement). Displacement / velocity x = A cos(ωt) · v = ±ω√(A² − x²) · v<sub>max</sub> = ωA A = amplitude. v is max at equilibrium, zero at the extremes. Spring period T=2\pi\sqrt{\frac{m}{k}} k = spring constant. Independent of amplitude. Simple pendulum period T=2\pi\sqrt{\frac{L}{g}} Small angles only. Independent of mass. Energy in SHM E<sub>total</sub> = ½ k A² = ½mv² + ½kx² Energy swaps between kinetic and potential; total is constant. Derivation Show that a mass on a spring gives T = 2π√(m/k) Derive the period of a mass m on a spring of constant k. Restoring force (Hooke): F = −kx. Newton’s 2nd law: ma = −kx ⟹ a = −(k/m) x. Compare with SHM a = −ω²x ⟹ ω² = k/m, so ω = √(k/m). T = 2π/ω = 2π√(m/k). ∎ ✓ T = 2π√(m/k): note it’s independent of amplitude. Derivation Show that a simple pendulum gives T = 2π√(L/g) Derive the period of a pendulum of length L for small swings. Restoring force along the arc: F = −mg sinθ. For small θ, sinθ ≈ θ = x/L. So F ≈ −mg(x/L). Newton: ma = −(mg/L)x ⟹ a = −(g/L)x. Compare a = −ω²x ⟹ ω² = g/L. T = 2π/ω = 2π√(L/g). ∎ ✓ T = 2π√(L/g): independent of the bob’s mass; only valid for small angles. Derivation Bifilar suspension (torsional SHM) A uniform bar (moment of inertia I, mass m) hangs from two vertical threads length L, separation d. Show the torsional period. Twist the bar by small angle θ about the vertical axis. Each thread tilts by φ, where the attachment moves (d/2)θ = Lφ ⟹ φ = dθ/2L. Each thread tension ≈ mg/2; its horizontal restoring component ≈ (mg/2)φ, acting at radius d/2. Restoring torque (two threads): τ = −2 · (mg/2)φ · (d/2) = −(mgd²/4L) θ. Iα = τ ⟹ ω² = mgd² / (4LI), so T = 2π√(4LI / (mgd²)) = (4π/d)√(LI/(mg)). ∎ ✓ T = (4π/d)√(LI/(mg)). Used to measure I (or g): T ∝ √L and T ∝ 1/d.Damping
- A resistive force (friction, air resistance) removes mechanical energy from a real oscillator as heat.
- The AMPLITUDE decreases over successive cycles — the envelope of the graph shrinks toward zero.
- For LIGHT damping, the period stays essentially unchanged even as the amplitude falls — only the envelope shrinks, not the timing.
- On a graph: still a cosine/sine shape, but each peak is lower than the last, tracing an exponentially decaying envelope.
Driven oscillations and resonance
- Every oscillator has a natural frequency f₀: the one it swings at when left alone. For a pendulum, f₀ = (1/2π)√(g/L).
- A push that repeats is a driving force. When its frequency equals f₀, each push arrives in step with the motion and adds to it. That is resonance.
- In energy terms, the driver does work on the oscillator every cycle. At resonance it puts energy in faster than damping takes it out, so the stored energy, and with it the amplitude, builds up until the losses catch up.
- Away from f₀ the pushes drift out of step, some of them work against the motion, and the amplitude stays small.