Question

a block of mass 5 kg is attached to a vertical spring of spring constant 15.25 N/M and it executes an oscillatory motion. the damping coefficient due to air resistance is 4 Ns/m. calculate the percentage decrease in amplitude per cycle

Answer 1

the frequency if undamped would be ${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{15.25}{5}}=\sqrt{3.05}=1.75s^{-1}$
with damping $\omega =\sqrt{{\omega }^{{}_{0}2}-\frac{b^2}{4m^2}}=\sqrt{1.{75}^2-\frac{16}{4\times 5\times 5}}=\sqrt{3.06-0.16}=\sqrt{2.9}=1.70s^{-1}$
$x=A{e}^{-bt/2m}\cos (\omega t+\phi )overonecycle$, a time


$$\begin{gathered} T=\frac{2\pi }{\omega }theamplitudechangesfromAoto{A}_0{e}^{-b2\pi /2m\omega }forafractionaldecreaseof \\ \frac{A_0-A_0{e}^{-b2\pi /2m\omega }}{A_0}=1-e^{-\pi 4/5\times 1.70}=1-e^{-1.477} \\ taking\ln onbothsideweget \\ 1.477=147.7\% \end{gathered}$$