Damped Simple Harmonic Motion — Question 3

Question 3

Instantaneous mechanical energy does not follow a pure exponential

A viscously damped oscillator has mechanical energy $E=\tfrac12m\dot x^2+\tfrac12kx^2$. It is underdamped with decay constant $\beta=b/(2m)>0$ and damped period $T_d$.

1. Derive the exact instantaneous energy-loss law. Use a turning point to test the proposed exact law $E(t)=E(0)e^{-2\beta t}$.

2. Show why energies at corresponding phases one full damped period apart nevertheless have the exact ratio $e^{-2\beta T_d}$.

Question 3 - Solution

Instantaneous mechanical energy does not follow a pure exponential

(a) Derivation and calculation

$$E=\tfrac12mv^2+\tfrac12kx^2,\quad\dot E=v(m\dot v+kx).$$$$m\dot v+kx=-bv\Rightarrow\boxed{\dot E=-bv^2\le0}.$$$$v=0,\ x\ne0:\quad\dot E=0,\qquad\frac{d}{dt}(E_0e^{-2\beta t})=-2\beta E<0.$$Therefore the pure exponential is not the exact instantaneous energy law.

(b) Result and interpretation

Periodicity of the trigonometric factors gives$$x(t+T_d)=e^{-\beta T_d}x(t),\quad v(t+T_d)=e^{-\beta T_d}v(t).$$$$\boxed{E(t+T_d)=e^{-2\beta T_d}E(t)}.$$This exact relation compares corresponding phases; a weak-damping cycle average may approximate a continuously exponential energy.

Physical check

Energy decreases monotonically, with horizontal tangents at nonzero turning points.