The Electrical-Mechanical Analogy — Question 4

Question 4

Energy scales consistently under a position-to-charge conversion

An undamped analogue uses $q=ax$ with $a>0$, electrical time t equal to mechanical time, and $C=m/(Lk)$. The mechanical oscillator has energy $E=\tfrac12m\dot x^2+\tfrac12kx^2$.

1. Find the ratio of electrical to mechanical energy.

2. Find the ratio of capacitor voltage to restoring spring-force magnitude at nonzero displacement.

Question 4 - Solution

Energy scales consistently under a position-to-charge conversion

(a) Method and calculation

$$U=\tfrac12La^2\dot x^2+\frac{a^2x^2}{2C}=\frac{La^2}{m}E.$$$$\boxed{U/E=La^2/m}.$$

(b) Interpretation and further calculation

$$|v_C|=a|x|/C,\quad |F_s|=k|x|.$$$$\boxed{|v_C|/|F_s|=a/(Ck)=La/m}.$$

Physical check

The voltage-to-force conversion differs from the energy conversion because charge and displacement have their own scale factor.