Problem: From damped oscillation measurements to Q

Problem

A damped harmonic oscillator has (undamped) frequency $f_0 = 5{,}0\;\text{Hz}$. After $N = 20$ oscillations the amplitude is observed to reduce by $50\%$. Calculate: (a) the constant $\gamma$, (b) the quality factor $Q = \omega_0/\gamma$.

Solution

(a) Period $T_0 = 1/f_0 = 0{,}20\;\text{s}$. After $20$ oscillations, a time $t = 20 \cdot 0{,}20 = 4{,}0\;\text{s}$ has elapsed. The envelope $A = A_0 e^{-(\gamma/2)t}$ halves:

$e^{-(\gamma/2)\cdot 4} = 0{,}5 \;\Rightarrow\; \gamma = \frac{2\ln 2}{4}$

$\ev{\gamma \approx 0{,}347\;\text{s}^{-1}}$

(b) With $\omega_0 = 2\pi f_0 = 2\pi \cdot 5{,}0 \approx 31{,}4\;\text{rad/s}$:

$Q = \frac{\omega_0}{\gamma} = \frac{31{,}4}{0{,}347}$

$\ev{Q \approx 90}$

Collegamenti

Argomenti: Oscillations and simple harmonic motion Concetti: Damped oscillations · Period and frequency Competenze: Reading graphs