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Lesson 4: Mechanical vs Electrical Analogies

Estimated time: 30-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

The Core Equations Side by Side

The mathematical structure is identical:

Mechanical: mx'' + cx' + kx = F(t)

Electrical: Lq'' + Rq' + (1/C)q = E(t)

Every concept, solution method, and qualitative behavior in one system has a direct counterpart in the other.

The Force-Voltage Analogy Table

Force-Voltage Analogy: Each mechanical quantity maps to an electrical quantity so that the governing equations are structurally identical.

MechanicalSymbolElectricalSymbol
DisplacementxChargeq
Velocityx' = vCurrentq' = i
MassmInductanceL
Damping coefficientcResistanceR
Spring constantk1/Capacitance1/C
Applied forceF(t)Applied voltage (EMF)E(t)
Kinetic energy ½mv²Magnetic energy ½Li²
Potential energy ½kx²Electric energy ½q²/C

Damping Classification in Both Domains

TypeMechanical ConditionElectrical Condition
Overdampedc² > 4mkR² > 4L/C
Critically dampedc² = 4mkR² = 4L/C
Underdampedc² < 4mkR² < 4L/C

Natural Frequency: Mechanical: ω0 = √(k/m). Electrical: ω0 = 1/√(LC). Both describe oscillation in the absence of damping.

Worked Examples: Same Equation, Two Interpretations

Example 1: One Equation, Two Systems

Consider u'' + 6u' + 8u = 0, u(0) = 3, u'(0) = 0.

Mechanical reading: m=1 kg, c=6 Ns/m, k=8 N/m. Mass displaced 3 m from equilibrium, released from rest.

Electrical reading: L=1 H, R=6 Ω, 1/C=8 (so C=0.125 F). Initial charge q(0)=3 C, initial current i(0)=0.

Solution: r²+6r+8 = (r+2)(r+4)=0. D = 36-32 = 4 > 0: overdamped.

u = c1e-2t + c2e-4t. u(0) = c1+c2=3. u'(0)=-2c1-4c2=0, so c1=2c2. Then 3c2=3, c2=1, c1=2. u(t)=2e-2t+e-4t.

Physical interpretation: In both systems, no oscillation; the response decays monotonically to zero.

Example 2: Translating a Problem

A spring-mass system has m = 0.5 kg, c = 3 Ns/m, k = 4 N/m. Find the analogous RLC circuit.

Step 1: L = m = 0.5 H.

Step 2: R = c = 3 Ω.

Step 3: 1/C = k = 4, so C = 0.25 F.

Answer: L = 0.5 H, R = 3 Ω, C = 0.25 F. Both systems satisfy 0.5u'' + 3u' + 4u = 0.

Example 3: Resonance in Both Domains

A mechanical system with m=1, k=25 is driven at ω=5 with no damping. What is the analogous circuit?

Mechanical: x'' + 25x = F0cos(5t). ω0 = 5 = ω. Pure resonance.

Electrical analog: L=1 H, 1/C=25 so C=0.04 F, R=0. ω0=1/√(1·0.04) = 1/0.2 = 5. Driving E(t)=E0cos(5t) at the natural frequency. This LC circuit also exhibits resonance: charge grows without bound as qp = (E0/10) t sin(5t).

Why the Analogy Matters

The analogy is more than a mathematical curiosity. It allows engineers to:

Extending the Analogy: Energy

Energy Correspondence: Kinetic energy ½mv² corresponds to magnetic energy ½Li². Potential energy ½kx² corresponds to electric energy q²/(2C). Energy dissipated by damping (cv²) corresponds to energy dissipated by resistance (Ri²).

Check Your Understanding

1. In the force-voltage analogy, what electrical component corresponds to a heavier mass?

Answer: A larger inductance L. Mass and inductance both resist changes in motion/current (inertia).

2. A circuit has L=2, R=0, C=1/18. What is the natural frequency? What spring-mass system is analogous?

Answer: ω0 = 1/√(LC) = 1/√(2/18) = 1/√(1/9) = 3 rad/s. Analogous: m=2, k=18 (so ω0=√(18/2)=3).

3. If adding oil to a mechanical system increases damping, what is the electrical equivalent?

Answer: Increasing the resistance R. More resistance dissipates more energy, just like more viscous damping in the mechanical system.

Key Takeaways

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