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Shock absorber

  1. Jun 4, 2014 #1
    The current of a regenerative shock absorber is modelled by
    I = -5e^(-0.5t) cos t - 10e^(-0.5t) sin t
    Given that the charge, q, in an electrical current is related to, I by I = dq/dt and that at t = 0 the charge of the regenerative shock absorber is q=80, find the charge when t = 5
    Hint= need to find the derivative of e^(-0.5t) cos t
    Part B
    Due to safety specifications the long term percentage charge of the circuit cannot exceed 92%. By first determining the lim as n approaches infinite, q(t). calculate whether the charge will exceed the safety limit
    discuss results
    please


    2. Relevant equations
    I = -5e^(-0.5t) cos t - 10e^(-0.5t) sin t
    e^(-0.5t) cos t


    3. The attempt at a solution

    I tried working it out by doing intergration by parts but I can't seem to get past that point

    Second Question:
    Develop a mathematical model that would lengthen the time until the shock stabilised by the given time. T= 2.51 show mathematical analysis of the situation

    d(t) = -5e^(-5t) cos (10t) is the original equation for a deflection of a rod in centimetres where t is time and d is deflection.

    More info:
    once a rod is released at time 0, it will spring back towards rest position where deflection is 0. It will go past rest which is called first rebound before rebounding again, going back through rest. the maximum distance of the rod below the rest position after this first rebound (dm) is used to measure the perfomance of the damper. dividing this rebound distance by the initial displacement (which is 5cm) gives the rebound ratio for that particular damper. if the ratio is below 1% the damper is working correctly
     
  2. jcsd
  3. Jun 4, 2014 #2

    Zondrina

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    Homework Helper

    Hint for part a):

    ##I = dq/dt##
    ##\int_{0}^{5} I dt = \int_{80}^{q} dq##

    EDIT: The OP has pmed me rather than posting so I'll make it a bit more clear by providing another hint. You need to use integration by parts to solve your problem.
     
    Last edited: Jun 4, 2014
  4. Jun 4, 2014 #3
    I can't seem to figure it out, I'm not very good with intergration by parts :(
     
  5. Jun 4, 2014 #4

    Zondrina

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    Homework Helper

    You want to compute this:

    ##\int_{0}^{5} -5e^{-0.5t} cos(t) - 10e^{-0.5t} sin(t) dt = \int_{80}^{q} dq##

    ##-5 \int_{0}^{5} e^{-0.5t}(cos(t) + 5sin(t)) dt = q - 80##

    For the integration by parts, suppose ##u = e^{-0.5t}## (you could also use ##u = cos(t) + 5sin(t)##).

    Also assume that ##dv = cos(t) + 5sin(t) dt## (you could also use ##dv = e^{-0.5t} dt##).

    Depending on your choice of substitution, I would say the one with ##u = e^{-0.5t}## is a bit easier, what are ##du## and ##v## respectively?

    Then break it down into ##uv|_{0}^{5} - \int_{0}^{5} v du##.
     
  6. Jun 4, 2014 #5
    Solved it thank you, I found the constant and worked from their, thank you
     
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