The motion of a particle is given by x(t)=(25cm)cos(13t), where t is in s.?

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find the first time at which the kinetic energy is twice the potential energy.

2 Answers

  • Motion Of A Particle

  • EDIT – added alternative solution at the bottom.

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    If you are allowed to use standard equations, just consider a mass (m) on a spring (spring constant k):

    x = Acos(ωt) where ω = √(k/m)

    Note, since x(t)=(25cm)cos(13t), we can match coefficients so:

    A = 25cm (=0.25m) and

    ω = 13 (rad/s)

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    v = dx/dt = -Aωsin(ωt)

    Eₚ = ½kx²

    Eₖ =½mv²

    We want Eₖ = 2Eₚ

    ½mv² = 2(½kx²)

    mv² = 2kx²

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    Since x = Acos(ωt) and v = -Aωsin(ωt)

    m(-Aωsin(ωt))² = 2k(Acos(ωt))²

    mA²ω²sin²(ωt) = 2kA²cos²(ωt)

    mω²sin²(ωt) = 2kcos²(ωt)

    sin²(ωt)/cos²(ωt) = 2k/(mω²)

    ω = √(k/m) so ω² = (k/m), and sin()/cos() = tan() so;

    tan²(ωt) = 2k/(m(k/m)

    tan²(ωt) = 2

    tan(ωt) = √2

    ωt = tan⁻¹√2

    Remember to switch calculator to radians:

    ωt = 0.9553

    t = 0.9553/ω = 0.9553/13 = 0.0735s

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    Alternatively, you can find where the kinetic energy is 2/3 of its maximum value because then 1/3 of the total energy is potential energy.

    v max (when sn()=1) =

    KE = ½mv² = ½m(-Aωsin(ωt))² = ½mAω²sin²(ωt)

    The max. value odof sin() is 1, so the max value of KE = is when sin() = 1:

    KEmax = ½mAω²1² = ½mAω²

    When KE = (2/3)KEmax

    ½mAω²sin²(ωt) = (2/3)½mAω²

    ω²sin²(ωt) = (2/3)ω²

    sin²(ωt) = (2/3)

    sin(ωt) = √(2/3)

    ωt = sin⁻¹√(2/3)

    . . .= 0.9553

    Then the working is the same as in the first method.

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