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in Laws of motion by (15 points)
edited by

Derive an expression for the logarithmic decrement of a damped oscillator

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1 Answer

+1 vote
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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 (ωdt+ ϕ)

x2\(Xe^{-ζ\omega_nt_2}\) sin (ωdt+ 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 = \(\frac{1}{n}\)\(log_e(\frac{x_0}{{x_n}})\)

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