ωd = frequed of damped oscillation
Logarithmic decrement is defined of th natural logarthim of the ratio of successive amplitude on the same side of mean position.
We can write periodic time tp = \(\frac{2\pi}{\omega_d}\)
∵ ωd = \(\frac{2\pi}{\sqrt{(1-ζ^2)}\omega_n}\) (ζ = zeta)
ωd = \(\frac{2\pi}{(\sqrt{1-ζ^2)}\omega_n}\)
x1 = \(Xe^{-ζ\omega_nt_1}\) sin (ωdt1 + ϕ)
x2 = \(Xe^{-ζ\omega_nt_2}\) sin (ωdt2 + ϕ)
= \(Xe^{-ζ\omega_n(t_1+t_p)}\) sin (ωd(t1 + tp) + ϕ)
= \(Xe^{-ζ\omega_n(t_1+t_p)}\) sin (ωdt1 + ωdtp + ϕ)
= \(Xe^{-ζ\omega_n(t_1+t_p)}\) sin (ωdt1 + ωd.\(\frac{2\pi}{\omega_d}\) + ϕ)
= \(Xe^{-ζ\omega_n(t_1+t_p)}\) sin (2π + ωdt1 + ϕ)
We can write x2 = \(Xe^{-ζ\omega_n(t_1+t_p)}\) sin (ωdt1 + ϕ)
Now,
\(\frac{x_1}{x_2}\) = \(\frac{Xe^{-ζ\omega_nt_1}sin(\omega_dt_1+\phi)}{Xe^{-ζ\omega_n(t_1+t_p)sin(\omega_dt_1+\phi)}}\)
\(\frac{x_1}{x_2}\) = \(e^{-ζ\omega_n(t_1-t_1+t_p)}\)
\(\frac{x_1}{x_2}\) = \(e^{ζ\omega_nt_p}\)
Logarithmic decrement
δ = \(log_e(\frac{x_1}{{x_2}})\)
∵ \(\frac{x_1}{x_2}\) = \(e^{ζ\omega_nt_p}\)
δ = \(log_e(e^{ζ\omega_nt_p})\)
δ = \({ζ\omega_nt_p}\)
∵ tp = \(\frac{2\pi}{\omega_d}\)
δ = \({ζ\omega_n}\frac{2\pi}{\omega_d}\)
δ = \({ζ\omega_n}\frac{2\pi}{(\sqrt{1-ζ^2})\omega_n}\)
δ = \(\frac{ζ\pi}{\sqrt{1-ζ^2}}\)
This is expression for logarithmic decrement damped oscillatic another form
S = \(log_e(\frac{X_0}{{X_1}})\) = \(log_e(\frac{x_1}{{x_2}})\) = \(log_e(\frac{x_2}{{x_3}})\) = .... \(log_e(\frac{x_{n-1}}{{x_n}})\)
nδ = \(log_e(\frac{x_0}{{x_n}})\)
δ = \(\frac{1}{n}\)\(log_e(\frac{x_0}{{x_n}})\)