The Electrical-Mechanical Analogy — Question 2

Section 31-2 · Physics pages 4994–4995

Question 2 Matching a mechanical time scale does not require matching its units A mechanical oscillator has m=.600 kg and k=150 N/m. An electrical analogue is to use L=.0800 H and have the same natural angular frequency. m k x Schematic; labels, not drawing lengths, specify the data. (a) Find the required capacitance. (b) Compare the effect of doubling mechanical mass with doubling electrical inductance.
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Question 2 - Solution Matching a mechanical time scale does not require matching its units m k x (a) Method and calculation ω2 = k/m = 250 s−2 = 1/(LC). C = m/(kL) = .0500 F . (b) Interpretation and further calculation ωm ∝m−1/2, ωe ∝L−1/2. ωnew/ωold = 1/ √ 2 in both cases . Physical check Corresponding coefficients have different physical units but the same role in the differential equation.

Question 2

Matching a mechanical time scale does not require matching its units

A mechanical oscillator has m=.600 kg and k=150 N/m. An electrical analogue is to use L=.0800 H and have the same natural angular frequency.

Diagram shown in the original SVG worksheet.

  1. Find the required capacitance.

  2. Compare the effect of doubling mechanical mass with doubling electrical inductance.

Question 2 - Solution

Matching a mechanical time scale does not require matching its units

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ω2=k/m=250s−2=1/(LC).\omega^2=k/m=250\,s^{-2}=1/(LC).C=m/(kL)=.0500F.\boxed{C=m/(kL)=.0500\,F}.

(b) Interpretation and further calculation

ωm∝m−1/2,ωe∝L−1/2.\omega_m\propto m^{-1/2},\quad\omega_e\propto L^{-1/2}.ωnew/ωold=1/2in both cases.\boxed{\omega_{\mathrm{new}}/\omega_{\mathrm{old}}=1/\sqrt2\quad\text{in both cases}}.

Physical check

Corresponding coefficients have different physical units but the same role in the differential equation.