--- title: "Damping from logarithmic decrement" subject: "Theory of Mechanism and Machine II" chapter: 6 key: "semester_7/theory_of_mechanism_and_machine_ii:logarithmic-decrement-damping-numerical" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/logarithmic-decrement-damping-numerical site: BCT Engineering --- # Damping from logarithmic decrement Theory of Mechanism and Machine II · Chapter 6: Vibration of Single Degree of Freedom Systems - Asked in: year not given, 6 marks - Video: [Mechanical Vibration Lecture 11 || Logarithmic Decrement || SDOF damped vibration](https://www.youtube.com/watch?v=YQwy6cJJkiI) — Mechanical Engineering by Ashish Purohit - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/logarithmic-decrement-damping-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-6 ## Question A vibrating system has a mass of 25 kg and a spring of stiffness 20 kN/m with a viscous damper. It is found that the amplitude of free vibration reduces to 25% of its initial value after two complete cycles. Find the logarithmic decrement, the damping ratio, the damping coefficient, and the damped natural frequency. How many cycles are needed for the amplitude to fall to 10% of the initial value? ## Short answer Log decrement δ = (1/2) ln(1/0.25) = 0.6931. Damping ratio ζ = δ/√(4π²+δ²) = 0.1097. Natural frequency ω_n = 28.28 rad/s, damped f_d = ω_n√(1-ζ²)/(2π) = 4.474 Hz. Damping coefficient c = ζ·2√(km) = 155.1 N·s/m. Cycles to 10%: n = ln(10)/δ ≈ 3.32. ## Answer **Data:** $m = 25$ kg, $k = 20000$ N/m, $x_2/x_0 = 0.25$ after $n = 2$ cycles. #### Logarithmic decrement $$ \delta = \frac{1}{n}\ln\frac{x_0}{x_n} = \frac{1}{2}\ln\frac{1}{0.25} = \frac{\ln 4}{2} = 0.6931 $$ #### Damping ratio $$ \delta = \frac{2\pi\zeta}{\sqrt{1-\zeta^2}} \Rightarrow \zeta = \frac{\delta}{\sqrt{4\pi^2 + \delta^2}} = \frac{0.6931}{\sqrt{39.478 + 0.4805}} = 0.1097 $$ #### Natural and damped frequency $$ \omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{20000}{25}} = 28.28\ \text{rad/s} $$ $$ \omega_d = \omega_n\sqrt{1-\zeta^2} = 28.28\sqrt{1 - 0.1097^2} = 28.11\ \text{rad/s}, \qquad f_d = \frac{\omega_d}{2\pi} = 4.474\ \text{Hz} $$ #### Damping coefficient $$ c_c = 2\sqrt{km} = 2\sqrt{20000 \times 25} = 1414.2\ \text{N·s/m} $$ $$ c = \zeta c_c = 0.1097 \times 1414.2 = 155.1\ \text{N·s/m} $$ #### Cycles to reach 10% $$ \frac{x_0}{x_n} = e^{n\delta} = 10 \Rightarrow n = \frac{\ln 10}{\delta} = \frac{2.3026}{0.6931} = 3.32\ \text{cycles} $$ **Answer:** $\delta = 0.6931$; $\zeta = 0.1097$; $c = 155.1$ N·s/m; $f_d = 4.474$ Hz; about 3.32 cycles to fall to 10%. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Amplitude falls to 25% after two cycles. Which expression gives the logarithmic decrement? A. delta = 2 ln(4) B. delta = ln(0.25)/2 C. delta = (1/2) ln(1/0.25) = 0.693 D. delta = ln(4)/4 Answer: C. delta = (1/n) ln(x0/xn) with n = 2 and x0/xn = 4, giving ln4/2 = 0.6931. (Key point: Logarithmic decrement over n cycles) 2. Using delta = 2 pi zeta / sqrt(1 - zeta^2) with delta = 0.6931, the damping ratio is about: A. 0.693 B. 0.110 C. 1.10 D. 0.0110 Answer: B. zeta = delta / sqrt(4 pi^2 + delta^2) = 0.6931/6.32 = 0.1097. (Key point: Damping ratio from delta) 3. How is the damping coefficient c obtained here? A. c = zeta x c_c with c_c = 2 sqrt(km) = 1414 N s/m, so c is about 155 N s/m B. c = zeta x k C. c = 2 m wn with zeta ignored, about 1414 N s/m D. c = delta x m Answer: A. c = zeta c_c = 0.1097 x 1414.2 = 155.1 N s/m. (Key point: Damping coefficient) 4. How many cycles are needed for the amplitude to fall to 10% of the initial value? A. 2.50 B. 4.00 C. 3.32 D. 10 Answer: C. x0/xn = e^(n delta) = 10, so n = ln10/0.6931 = 3.32 cycles. 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