Answer to Question #94079 in Physics for Puspedu

Question

For a damped harmonic oscillation, the equation of motion is
m (d2x/dt2) + γ (dx/dt) + kx = 0
with m = 0.25 kg, γ = 0.07 kg s−1 and k = 85 N m−1. Calculate (i) the period of motion,
(ii) number of oscillations in which its amplitude will become half of its initial value,
(iii) the number of oscillations in which its mechanical energy will drop to half of its
initial value, (iv) its relaxation time, and (v) quality factor.

Answer 1

(i)

$$T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.25}{85}}=0.34\ s$$

(ii)

$$N=\frac{t}{T}=\frac{2m\ln{2}}{γT}=\frac{2(0.25)\ln{2}}{(0.07)(0.34)}=15$$

(iii)

$$\tau=\frac{2m}{γ}=\frac{2(0.25)}{(0.07)}=7.1\ s$$

(iv)

$$Q=\frac{2m\pi}{γT}=\frac{2(0.25)\pi}{(0.07)(0.34)}=66$$