Oscillations and SHM

Oscillations and SHM

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Simple Harmonic Motion (SHM) is a periodic motion where the restoring force obeys F = −kx, making acceleration a = −ω²x. The displacement follows x(t) = A sin(ωt + φ), where A is amplitude, ω = √(k/m) is angular frequency, and φ is phase. Time period T = 2π/ω = 2π√(m/k) for a spring-mass system and T = 2π√(L/g) for a simple pendulum. Total energy is constant: E = ½kA².

At the mean position, velocity is maximum and acceleration is zero; at the extremes, the body is momentarily at rest but acceleration peaks at −ω²A. Phase: displacement leads velocity by π/2, and displacement leads acceleration by π. In ECAT MCQs, watch for ω vs f confusion and the trap of adding the π-phase when comparing x and v graphs.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Defining equation and solution

The defining test for SHM is a = −ω²x. Starting from F = −kx and Newton’s second law, m(d²x/dt²) = −kx, which is the differential equation whose general solution is x(t) = A sin(ωt + φ). Here ω = √(k/m) is fixed by the system, while A and φ are set by initial conditions (x₀ and v₀).

Velocity, acceleration, and phase

Differentiating gives v(t) = Aω cos(ωt + φ) and a(t) = −Aω² sin(ωt + φ) = −ω²x. Hence:

• x leads v by π/2 (90°)
• x leads a by π (180°); v leads a by π/2
• |v|max = Aω occurs at the mean position; |a|max = Aω² occurs at the extremes.

Energy interchange

Total mechanical energy E = ½kA² is conserved in ideal SHM. Splitting it:

• KE(t) = ½kA² cos²(ωt + φ)
• PE(t) = ½kA² sin²(ωt + φ)

KE and PE oscillate at twice the frequency of displacement, and their time-average over a cycle is each equal to E/4.

Spring-mass vs pendulum

System	ω	Period T
Spring-mass	√(k/m)	2π√(m/k)
Simple pendulum (small θ)	√(g/L)	2π√(L/g)

Note that the pendulum’s T is independent of mass and amplitude (for θ ≲ 10°); only L and g matter.

Typical ECAT question patterns

1. Finding T or f from given k and m (or L and g) — usually a 1-mark direct plug.
2. Reading off A, T, φ from a displacement-time graph.
3. Comparing KE and PE at a stated position x = (A/2) or phase ωt.
4. Phase-shift problems: “by how much does velocity lead displacement?”

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Projection-of-circle viewpoint

SHM is the shadow of uniform circular motion projected onto a diameter. A particle moving with constant angular speed ω around a circle of radius A has x-coordinate A cos(ωt + φ); this is mathematically identical to the SHM solution. The connection explains why ω appears in radians per second and why phase φ maps onto the starting angle on the circle.

Damped and forced oscillations

In a real medium, a resistive force −b(dx/dt) introduces damping. Solutions decay as x(t) = A₀ e^(−bt/2m) cos(ω′t + φ), where ω′ = √(ω₀² − (b/2m)²). The amplitude envelope falls exponentially while the (slightly reduced) angular frequency persists. Three regimes:

• Underdamped (b < 2√(km)): oscillatory decay.
• Critically damped (b = 2√(km)): fastest non-oscillatory return.
• Overdamped (b > 2√(km)): slow non-oscillatory return.

Adding a periodic driving force F₀ sin(ωd t) gives a steady-state solution at the driving frequency; amplitude peaks when ωd = ω₀ — the resonance condition, central to ECAT conceptual questions on tuning circuits and Tacoma-Narrows-type examples.

Edge cases and common traps

• A body on a frictionless incline attached to a spring still obeys SHM with the same ω = √(k/m); gravity merely shifts the equilibrium, not ω.
• For a pendulum, the small-angle approximation sin θ ≈ θ (in radians) is what makes T independent of amplitude. For θ ≈ 30°, the true period is ~1.7% longer — a frequent trap.
• Don’t confuse angular frequency ω (rad/s) with ordinary frequency f (Hz). The factor 2π between them trips up many students.

Worked example

A 0.5 kg block attached to a spring (k = 200 N/m) is pulled 0.1 m and released.

• ω = √(200/0.5) = √400 = 20 rad/s
• T = 2π/20 ≈ 0.314 s
• f = 1/T ≈ 3.18 Hz
• Total energy E = ½(200)(0.1)² = 1.0 J
• Maximum speed vmax = Aω = 0.1 × 20 = 2 m/s

Practice prompts

1. A simple pendulum of length 1.0 m has T ≈ 2.01 s on Earth. What is its length on a planet where T = 4.0 s? (Answer: L ≈ 4.0 m.)
2. At what displacement (as a fraction of A) is KE equal to PE in SHM? (Answer: x = A/√2.)

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