One equation — x = A·sin(ωt+φ) — drives every oscillator. Spin the phasor, load the spring, swing the pendulum, and watch energy slosh between kinetic and potential.
01
The oscillation & its phasor
reference circle → sine
Period T
{{ Ttxt }}
Angular ω
{{ Wtxt }}
Max speed Aω
{{ Vmaxtxt }}
Max accel Aω²
{{ Amaxtxt }}
The projection of a point moving in a circle at constant ω is SHM. So x = A·sin(ωt+φ), and differentiating gives v = Aω·cos(ωt+φ) and a = −ω²x — the defining signature: acceleration is proportional to displacement and points back to the mean position. At the mean position speed is maximum; at the extremes it is zero and acceleration is maximum.
Superposing two SHMs. Two SHMs of the same ω along the same line add like phasors — the resultant amplitude for a phase difference δ is A_R = √A₁² + A₂² + 2A₁A₂cosδ. Two perpendicular equal-ω SHMs instead trace Lissajous figures: a straight line (δ = 0), a circle (δ = π/2, equal amplitudes), or an ellipse in between.
02
Spring–mass oscillator
T = 2π√mkeff = {{ TspTxt }}
keff = {{ keffTxt }}. Springs in series are softer (k/2 for two equal springs) → slower; in parallel stiffer (2k) → faster. A vertical spring has the same T — gravity only shifts the equilibrium, it does not change k or m.
why k_eff?
Series — the same force F runs through both springs, so each stretches x=F/k. Total x = F/k + F/k = 2F/k, giving k_eff = F/x = k/2 (softer, longer T).
Parallel — both share the same extension x, so the total force F = kx + kx = 2kx, giving k_eff = 2k (stiffer, shorter T).
03
Simple pendulum
T = 2π√Lg = {{ TpdTxt }}
For small θ, the restoring torque gives α = −(g/L)θ → SHM with ω = √g/L. Independent of mass and (small) amplitude. Beyond ~15° the sinθ ≈ θ approximation starts to tell: the true period runs about +0.4 % at 15°, +1.7 % at 30° and +7 % at 60° (the θ₀²/16 correction).
Try g = 1.6 (Moon) or 24 (deep dive): the swing visibly slows or quickens while L is fixed.
04
Physical pendulum & angular SHM
Any rigid body pivoted off its centre of mass swings as a physical (compound) pendulum. Its period needs the moment of inertia about the pivot, obtained from the centre of mass by the parallel-axis theoremI = Icm + md². The bob of the matching simple pendulum sits at the centre of oscillation.
rigid body · pivot on it
Icm
{{ IcmTxt }}
I = Icm+md²
{{ IpivTxt }}
Period T
{{ TphysTxt }}
Equiv. length L=I/md
{{ LeqTxt }}
T = 2π√Imgd
The period is minimum when d equals the radius of gyration k=√Icm/m = {{ kGyrTxt }}; it blows up as d→0 (pivot at CoM, no restoring torque). Pivot and centre of oscillation O are interchangeable.
Torsional oscillator · T = 2π√I/κ
Period
{{ TtorsTxt }}
05
Energy in SHM
Total E
{{ EtotTxt }}
KE
—
PE
—
Total energy E = ½kA² = ½mω²A² is constant. PE = ½kx² (the parabola) and KE = ½k(A²−x²) swap twice per cycle, so each oscillates at 2ω. Averaged over a period, ⟨KE⟩ = ⟨PE⟩ = E/2. KE = PE when x = ±A/√2.
06
Going deeper — damping & resonance
Real oscillators lose energy. Add a damping force −bv, and drive one at frequency Ω to see resonance — the crux of countless JEE-Advanced problems.
Damped free oscillation · underdamped
Amplitude decays as A₀e^(−bt/2m); oscillation frequency drops to ω′=√ω₀²−(b/2m)². Past critical, it just crawls back.