🧪 Spring-Mass Simple Harmonic Motion

📈 The SHM Equation

Newton's second law for a spring-mass system gives:

• F = −kx  →  mẍ = −kx
• Rearranged: ẍ = −ω₀²x, where ω₀ = √(k/m)
• Solution: x(t) = A cos(ω₀t + φ)
• Velocity: v = −Aω₀ sin(ω₀t + φ)
• Acceleration: a = −Aω₀² cos(ω₀t + φ) = −ω₀²x

The period T = 2π / ω₀ = 2π √(m/k) is completely independent of amplitude A. This isochronous property is what makes pendulum clocks so accurate.

⚡ Energy in SHM

Energy constantly transfers between kinetic and potential forms, but total mechanical energy E is conserved (undamped case):

• KE = ½mv²
• PE = ½kx²
• Total E = ½kA² = constant

At maximum displacement (x = ±A): KE = 0, PE = ½kA². At equilibrium (x = 0): KE = ½kA², PE = 0. The energy graph traces a perfect sinusoid shifted by 90° relative to the displacement graph.

🌊 Damped Oscillations

Real systems lose energy to friction and air resistance. Adding a damping term −bv gives:

• mẍ = −kx − bẋ
• γ = b/(2m) — damping rate
• ω_d = √(ω₀² − γ²) — damped frequency
• x(t) = A e^−γt cos(ω_d t + φ)

Critical damping (b = 2√(km)) gives the fastest return to rest without oscillation. Underdamping (b < 2√(km)) gives decaying oscillations; overdamping (b > 2√(km)) gives a slow creep back to equilibrium.

🎶 Resonance & Driving Force

When an external periodic force F₀ cos(ωt) drives the system:

• mẍ = −kx − bẋ + F₀ cos(ωt)
• Steady-state amplitude: A = F₀ / √[(k−mω²)² + (bω)²]
• Resonance at ω ≈ ω₀: amplitude peaks
• Q-factor: Q = ω₀m / b = (1/b)√(km)

High Q (low damping) → sharp, tall resonance peak. Low Q (high damping) → broad, short peak. Enable the driving force in the simulator and tune the frequency to ω₀ = √(k/m) to witness resonance!

📈 Underdamped (b < 2√km)

Most real oscillators — car suspension (ideally under-damped for ride comfort), clock pendulums, guitar strings, LC circuits, and atomic vibrations. The system oscillates with exponentially decaying amplitude. Amplitude envelope: A e^−γt. The energy decays as E ∝ e^−2γt.

⚙ Critically Damped (b = 2√km)

The system returns to equilibrium in the shortest possible time without oscillating. Used in door closers, galvanometers, and car dampers where overshoot must be avoided. Mathematically: x(t) = (C₁ + C₂t) e^−γt. The critically damped case is the boundary between oscillatory and non-oscillatory motion.

🐌 Overdamped (b > 2√km)

The system returns to equilibrium slowly, creeping back without oscillation. Heavy molasses, overdamped shock absorbers, and some biological systems. Two real decay rates: γ ± √(γ² − ω₀²). As b → ∞, return time → ∞. Not usually desirable in engineering control systems where fast response is needed.

⌛ Timekeeping

Pendulum clocks (T = 2π√(L/g)), quartz crystal oscillators (atomic SHM at ~32 768 Hz), and atomic clocks (microwave-frequency electron transitions) all exploit isochronous oscillation for precise timekeeping. The GPS system relies on atomic clocks accurate to nanoseconds per day.

🚗 Vehicle Suspension

Car suspension is a spring-mass-damper system. Engineers choose spring constant k and damping b for comfort (underdamped, ~1 Hz natural frequency) vs. performance (stiffer, less body roll). Underdamped suspension bounces; overdamped suspension isn't absorbing road shocks effectively. The optimal is slightly underdamped.

🧭 Molecular Vibrations

Chemical bonds act as springs between atoms. Diatomic molecules vibrate at infrared frequencies (10¹² – 10¹³ Hz) — IR spectroscopy identifies molecular bonds by their characteristic SHM frequencies. Stretching and bending modes are described by normal-mode analysis, an extension of the SHM framework.

🔌 Electrical Circuits (LC)

In an LC circuit, charge q oscillates with angular frequency ω = 1/√(LC), the direct electrical analogue of ω₀ = √(k/m). Inductance L maps to mass, capacitance C maps to 1/k, and resistance R to damping b. Radio tuners exploit LC resonance to select specific broadcasting frequencies.

📈 Step 1: Angular Frequency

ω₀ = √(k/m) = √(20/0.5) = √40 ≈ 6.32 rad/s

📈 Step 2: Period

T = 2π/ω₀ = 2π/6.32 ≈ 0.994 s ≈ 1 second

📈 Step 3: Maximum Speed

v_max = Aω₀ = 0.3 × 6.32 ≈ 1.90 m/s at x = 0

📈 Step 4: Total Energy

E = ½kA² = ½ × 20 × 0.09 = 0.90 J

Check: ½mv_max² = ½ × 0.5 × 1.90² = 0.90 J ✓