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Math Help - Shm again!

  1. #1
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    Shm again!

    A particle in SHM about O has period T. It passes through P with V moving away from O. Prove that the time taken before it returns to P is given by T/pi * arctan[vT/2pix] where x is the distance OP.
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    Quote Originally Posted by Sunyata View Post
    A particle in SHM about O has period T. It passes through P with V moving away from O. Prove that the time taken before it returns to P is given by T/pi * arctan[vT/2pix] where x is the distance OP.
    In the problem, the required time is double the time taken by the particle to move from the extreme position to the point P.
    The displacement x is given as
    x = A*cos(ωt)
    The velocity v is given by
    V = \omega\sqrt{A^2 - x^2}

    V = \omega\sqrt{A^2 - {Acos{\{omegat}^2}
    V = ωΑ*sin(ωt)
    V/x = ω*tan(ωt)
    ωt = arctan(V/ωx)
    t = 1/ω*arctan(V/ωx)
    T' = 2t = 2/ω*arctan(V/ωx)
    Put ω = 2π/T and find T'
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  3. #3
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    Quote Originally Posted by Sunyata View Post
    A particle in SHM about O has period T. It passes through P with V moving away from O. Prove that the time taken before it returns to P is given by T/pi * arctan[vT/2pix] where x is the distance OP.
    In the problem, the required time is double the time taken by the particle to move from the extreme position to the point P.
    The displacement x is given as
    x = A*cos(ωt)
    The velocity v is given by
    V = ωsqrt{A^2 - x^2}
    V = ωsqrt{A^2 - [Acos(ωt)]^2}
    V = ωΑ*sin(ωt)
    V/x = ω*tan(ωt)
    ωt = arctan(V/ωx)
    t = 1/ω*arctan(V/ωx)
    T' = 2t = 2/ω*arctan(V/ωx)
    Put ω = 2π/T and find T'
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