Oscillation with damping

  1. 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.
    [​IMG]

    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

    [tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]

    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. [tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]
    ω' = 2 √(1-(0.03/20)2) ≈ 1.99 rad/s
    T = 2π/ω = 3.14 s

    d. c = c1 + c2 = 0.04
    [tex]ω'=ω_o \sqrt{1-(\frac{c}{2m})^2}[/tex]

    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
     
  2. jcsd
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