MAT2680 Differential Equations, FA2013

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  • mass
  • November 2, 2013 at 11:24 pm #13566

    Afzal
    Participant

    .hello professor,
    can please look at the solution of number 5 in the attach file. i am little confuse how they get m=mg/g were mg=2 and g=9.8 so how they got 1/16.

    November 2, 2013 at 11:39 pm #17339

    Keven Deng
    Participant

    I think you have to convert g=9.8m/s to g=32 ft/s since its using inches so it becomes m=2lb/32 ft/s =1/16

    November 2, 2013 at 11:51 pm #17340

    Afzal
    Participant

    Thank you keven :)

    November 3, 2013 at 12:31 am #17341

    Afzal
    Participant

    hi i have another question. how they come up with the initial condensation u(0) =1/4, u′(0) = 0? ….thank you.

    November 3, 2013 at 3:35 am #17342

    Ezra Halleck
    Participant

    From the equilibrium position, the mass was pulled down 3 inches, which is 1/4 ft.

    Also, it was just released so that the initial velocity is 0.

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