SHM

SHM

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

• Simple Harmonic Motion (SHM) is periodic motion where the restoring force is directly proportional to displacement and always directed toward the mean position: F = −kx.
• The standard displacement equation is x(t) = A sin(ωt + φ), where A is amplitude, ω is angular frequency (rad/s), and φ is the phase constant set by initial conditions.
• Must-know formulas: T = 2π√(m/k) for a spring-mass system and T = 2π√(l/g) for a simple pendulum (valid for small oscillations, θ < ~15°).
• Total energy is conserved: E = ½kA² — fully kinetic at the mean position, fully potential at the extremes.
• Velocity leads displacement by π/2; acceleration leads displacement by π (they are in antiphase).
• CUET pointer: expect 1–2 MCQs mixing spring and pendulum formulae with conceptual phase or energy questions.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Defining Condition

A body executes SHM when the net force on it obeys F = −kx (Hooke’s-law form), giving acceleration a = −ω²x where ω² = k/m. The negative sign encodes the restoring nature — the force and acceleration always point toward the equilibrium (mean) position, opposing the displacement. Uniform circular motion is periodic but is not SHM; only one of its projections (e.g., the x-component of a point on the rim) traces SHM.

Kinematic Description

Displacement, velocity, and acceleration are all sinusoidal with the same ω:

• x(t) = A sin(ωt + φ)
• v(t) = Aω cos(ωt + φ) → v = ω√(A² − x²) when φ = 0
• a(t) = −Aω² sin(ωt + φ)

This gives the phase ladder: x → v leads by π/2 → a leads x by π (a and x are in antiphase). At the extreme position (x = ±A): v = 0, a = ±Aω² (maximum). At the mean position (x = 0): v = ±Aω (maximum), a = 0.

Energy in SHM

Mechanical energy is conserved:

• KE = ½mω²(A² − x²)
• PE = ½kx²
• E_total = ½kA² — independent of time; depends on A and k only.

Standard Systems

System	Time Period	Key Parameters
Spring-mass (horizontal/vertical)	T = 2π√(m/k)	T ∝ √m, T ∝ 1/√k, independent of A and g
Simple pendulum (small θ)	T = 2π√(l/g)	T ∝ √l, T ∝ 1/√g, independent of mass and amplitude
Liquid in a U-tube	T = 2π√(L/2g)	L = total length of liquid column

For a pendulum, the small-angle approximation sin θ ≈ θ (rad) is what linearises the restoring force; the formula fails for large amplitudes.

Typical CUET Question Patterns

• Numerical T from given m, k (or l, g); sometimes combined with frequency f = 1/T.
• Identifying which graphs of x, v, a vs t correctly show the π/2 and π phase relations.
• Energy ratio at a given x: KE/E = 1 − (x/A)², PE/E = (x/A)².

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Subtleties

• Amplitude independence of T (isochronism) holds exactly for an ideal spring-mass system. For a real pendulum, T increases slowly with amplitude; the first correction is T ≈ 2π√(l/g) · (1 + θ₀²/16), but CUET tests only the small-angle form.
• Vertical spring: T is still 2π√(m/k); gravity merely shifts the equilibrium by mg/k but does not change ω.
• Two springs in series: effective k = k₁k₂/(k₁ + k₂) → T increases. In parallel: k_eff = k₁ + k₂ → T decreases.
• Phase constant φ is fixed by x(0) and v(0). If the body starts from the positive extreme, x = A cos(ωt); if released from the mean position, x = A sin(ωt). Mixing these up is a frequent CUET trap.
• LC circuit analogy: charge on a capacitor obeys q(t) = Q₀ cos(ωt) with ω = 1/√(LC) — mathematically identical to SHM, useful for cross-topic MCQs in CUET’s mixed Physics sections.
• Bar magnet in uniform B: the magnet oscillates torsionally with T = 2π√(I/MB), connecting SHM to magnetism.

Common Mistakes

• Writing F = kx (missing the minus sign), which describes acceleration away from equilibrium — not SHM.
• Taking v as max at x = ±A; v is actually zero there.
• Using T = 2π√(l/g) when θ > ~15°; the restoring torque is −mgL sin θ, not −mgLθ.
• Confusing ω (rad/s) with f (Hz) — they differ by 2π.

Worked Micro-Example

A 0.5 kg mass on a spring (k = 200 N/m) is pulled 0.1 m from equilibrium and released. Find T, v_max, and a_max.

• ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s
• T = 2π/ω = 0.314 s
• v_max = Aω = 0.1 × 20 = 2 m/s
• a_max = Aω² = 0.1 × 400 = 40 m/s²

Practice Prompts

1. A pendulum of length 1 m is taken to a planet where g = 4 m/s². Find its new time period and compare with Earth (g = 9.8 m/s²).
2. For a spring-mass system with m = 2 kg, k = 8 N/m, at what displacement is KE equal to PE?

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