Ask Question
24 November, 22:39

A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then stretched an additional 1 inch and released. find the equation of motion, the amplitude, and the period. neglect friction. then

+3
Answers (1)
  1. 25 November, 01:59
    0
    Given:

    k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs

    Solution:

    F = - kx

    mx" = - kx

    x" + (k/m) x = 0

    characteristic equation:

    r^2 + k/m = 0

    r = i*sqrt (k/m)

    x = Asin (sqrt (k/m) t) + Bcos (sqrt (k/m) t)

    ω = sqrt (k/m)

    2π/T = sqrt (k/m)

    T = 2π*sqrt (m/k)

    T = 2π*sqrt (0.03106 slugs / 100 lb/ft)

    T = 0.1107 s (period)

    x (0) = 1/12 ft = 0.08333 ft

    x' (0) = 0

    1/12 = Asin (0) + Bcos (0)

    B = 1/12 = 0.08333 ft

    x' = Aω*cos (ωt) - Bω*sin (ωt)

    0 = Aω*cos (0) - (1/12) ω*sin (0)

    0 = Aω

    A = 0

    So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[ (2π/0.1107) t]
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then stretched an ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers