Damped Simple Harmonic Motion — Question 8

Question 8

The quality factor and logarithmic decrement agree only in a limit

Define $Q=\omega_0/(2\beta)$ for an underdamped viscous oscillator. Its same-side logarithmic decrement is $\delta=\ln(x_n/x_{n+1})=\beta T_d$.

1. Derive the exact relation between $Q$ and $\delta$.

2. For $\delta=1.0$, compute $Q$ and compare with the weak-damping estimate $\pi/\delta$. Why is $Q>1/2$ necessary here?

Question 8 - Solution

The quality factor and logarithmic decrement agree only in a limit

(a) Derivation and calculation

$$Q=\frac{\omega_0}{2\beta},\quad\omega_d^2=\omega_0^2-\beta^2=4\beta^2(Q^2-1/4).$$$$\delta=\frac{2\pi\beta}{\omega_d}=\frac\pi{\sqrt{Q^2-1/4}}.$$$$\boxed{Q=\sqrt{\left(\frac\pi\delta\right)^2+\frac14}}.$$

(b) Result and interpretation

$$\delta=1:\quad\boxed{Q=3.18113},\qquad Q_{\mathrm{weak}}=\pi=3.14159.$$$$\frac{Q-Q_{\mathrm{weak}}}Q\times100\%=1.2430\%.$$$$\omega_d^2>0\iff\beta<\omega_0\iff\boxed{Q>1/2}.$$The approximation $Q\simeq\pi/\delta$ requires the $1/4$ term to be negligible.

Physical check

At or below $Q=1/2$, a damped period and a sequence of same-side oscillation peaks no longer exist.