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

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.