# Oscillation with damping

by songoku
Tags: damping, oscillation
 P: 767 1. The problem statement, all variables and given/known data Spring – mass system with spring constant k = 40 N/m and mass 10 kg. a. Find the angular speed and period. Draw the response X versus time t b. Linear damping is added with ζ = 4 %. Find the angular speed and period. Draw the response c. Viscous damping is added with c1 = 0.03. Find the angular speed and period. Draw the response d. Another viscous damping with c2 = 0.015 added parallel to c1. Find the angular speed and period e. If a force F = F0 cos (2.3t) is applied to the system, will it be closer to the resonance, undamped system or system with viscous damping? f. The spring is divided into 4 parts of equal length and arranged as follows. Find the angular speed and period of the system. g. 2 viscous damper are added into the above system in series between point B and C with damping coefficient c1 = 0.03 and c2 = 0.02. Find the period and angular speed of this system 2. Relevant equations kseries = k1 + k2 1/kparallel = 1/k1 + 1/k2 k' = (1 - ζ )k $$ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}$$ ω = √(k/m) ω = 2π/T 3. The attempt at a solution OK actually my teacher didn't teach anything in class. Only gave formula and homework, so I am just trying to use the formula without understanding the concept here because the test is tomorrow. Please excuse my poor understanding and for now I haven't drawn the response graph because I really don't understand how a. ω = √(k/m) = √(40/10) = 2 rad/s ω = 2π/T T = π s b. k' = (1 - ζ )k = (1 - 0.04) . 40 = 38.4 N/m ω = √(k'/m) = 1.96 rad/s T = 2π/ω = 3.2 s c. $$ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}$$ ω' = 2 √(1-(0.03/20)2) ≈ 1.99 rad/s T = 2π/ω = 3.14 s d. c = c1 + c2 = 0.04 $$ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}$$ ω' ≈ 1.99 rad/s T = 2π/ω = 3.14 s e. no clue at all f. because k is inversely proportional to length and the spring is divided into 4 parts, so the value of k for each part is 160 N/m After some calculation, ktotal = 400 N/m ω = √(k'/m) = 2√10 rad/s T = 2π/ω = 0.99 s g. do not understand at all I am not sure whether my work right or wrong... Thanks

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