Simple Harmonic Motion

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Simple Harmonic Motion (SHM) is a back-and-forth oscillatory motion in which the restoring force is directly proportional to displacement from equilibrium and opposite in direction. Mathematically: F = -kx, where k is the force constant (N m⁻¹) and x is displacement from equilibrium.

Key formulas to memorise:

Period of mass–spring system: T = 2π√(m/k), where m is mass and k is spring constant.
Period of simple pendulum (small angle): T = 2π√(L/g), where L is string length and g ≈ 9.8 m s⁻².
Displacement equation: x = A sin(ωt + φ), where A is amplitude, ω = 2π/T, and φ is initial phase.

Exam pointers: (1) Period is independent of amplitude in true SHM. (2) Velocity leads displacement by 90°; acceleration leads velocity by another 90° (180° out of phase with x). (3) Total energy E = ½kA² = ½mv² + ½kx².

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Conditions for SHM

For motion to qualify as SHM, two conditions must hold simultaneously: (i) the restoring force must be linear in displacement, and (ii) the force must point opposite to displacement. Hooke’s law for a spring (F = -kx) satisfies both. Any system whose effective restoring “force” follows F = -kx will oscillate sinusoidally with angular frequency ω = √(k/m).

Kinematics of SHM

Taking x = A sin(ωt + φ), differentiating twice gives:

Velocity: v = Aω cos(ωt + φ) → maximum v_max = Aω at x = 0.
Acceleration: a = -Aω² sin(ωt + φ) = -ω²x → maximum a_max = Aω² at x = ±A.

This confirms a = -ω²x, the defining kinematic identity of SHM. In NABTEB multiple-choice items, students are often asked to identify which graph shows 90° phase lead/lag between x, v, and a.

Period Formulas

System	Period T	Independent of
Mass on spring	2π√(m/k)	Amplitude, gravity
Simple pendulum (small angle)	2π√(L/g)	Mass, amplitude (small)

Energy in SHM

Total mechanical energy is conserved: E = ½kA² = ½mω²A². At any instant, E = ½mv² + ½kx². Kinetic energy peaks at equilibrium; potential energy peaks at the extremes.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation: Pendulum Period

For a pendulum bob of mass m on string of length L displaced by small angle θ, the restoring force is F = -mg sin θ ≈ -mg θ (small angle). Since arc length x = Lθ, restoring force becomes F = -(mg/L)x, giving effective k = mg/L. Substituting into T = 2π√(m/k) yields T = 2π√(L/g). This approximation holds only when θ < ~15°; beyond that, the period lengthens because sin θ ≠ θ.

Common Mistakes

Dropping the negative sign in F = -kx — examiners use this to test whether students recognise directionality.
Applying T = 2π√(L/g) to a spring or vice versa — each formula belongs to one system.
Conflating angular frequency ω (rad s⁻¹) with frequency f (Hz): they differ by the factor 2π.

Connection to Circular Motion

SHM is the projection of uniform circular motion onto a diameter. A point moving on a circle of radius A with angular speed ω projects onto SHM with x = A cos(ωt). This analogy explains the 90° phase relationships.

Worked Example

A 0.50 kg mass on a spring oscillates with T = 0.80 s. Find k. Solution: T = 2π√(m/k) → k = 4π²m/T² = 4π²(0.50)/(0.64) ≈ 30.8 N m⁻¹.

Practice Prompts

A pendulum of length 1.00 m on Earth (g = 9.8 m s⁻²) — calculate T and frequency f.
Show that for a mass-spring system, doubling the amplitude does not change the period.

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Simple Harmonic Motion