How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method θ 1 ″ = − g ( 2 m 1 + m 2 ) s i n θ 1 − m 2 g s i n ( θ 1 − 2 θ 2 ) − 2 s i n ( θ 1 − θ 2 ) m 2 ( θ 2 ′ 2 l 2 + θ 1 ′ 2 l 1 c o s ( θ 1 − θ 2 ) l 1 ( 2 m 1 + m 2 − m 2 c o s ( 2 c o s ( 2 θ 1 − 2 θ 2 ) ) ) θ 2 ″ = 2 s i n ( θ 1 − θ 2 ) ( θ 1 ′ l 1 ( m 1 + m 2 ) + g ( m 1 + m 2 ) c o s θ 1 + θ 2 ′ 2 l 2 m 2 c o s ( θ 1 − θ 2 ) ) l 2 ( 2 m 1 + m 2 − m 2 c o s ( 2 c o s ( 2 θ 1 − 2 θ 2 ) ) ) These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method? m 1 , m 2 ==masses of pendulum 1 and 2, θ 1 , θ 2 == angles formed by the pendulums, θ 1 ′ = ω 1 , θ 2 ′ = ω 2

Question

How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method      θ    1    ″    =            −      g      (      2              m        1            +              m        2            )      s      i      n              θ        1            −              m        2            g      s      i      n      (              θ        1            −      2              θ        2            )      −      2      s      i      n      (              θ        1            −              θ        2            )              m        2            (              θ        2                  ′          2                            l        2            +              θ        1                  ′          2                            l        1            c      o      s      (              θ        1            −              θ        2            )                      l        1            (      2              m        1            +              m        2            −              m        2            c      o      s      (      2      c      o      s      (      2              θ        1            −      2              θ        2            )      )      )            θ    2    ″    =            2      s      i      n      (              θ        1            −              θ        2            )      (              θ        1        ′                    l        1            (              m        1            +              m        2            )      +      g      (              m        1            +              m        2            )      c      o      s              θ        1            +              θ        2                  ′          2                            l        2                    m        2            c      o      s      (              θ        1            −              θ        2            )      )                      l        2            (      2              m        1            +              m        2            −              m        2            c      o      s      (      2      c      o      s      (      2              θ        1            −      2              θ        2            )      )      )      These are the equations, and How can I possibly solve this simultaneoouslt in RK4 method?      m    1    ,      m    2    ==masses of pendulum 1 and 2,       θ    1    ,      θ    2    == angles formed by the pendulums,       θ    1    ′    =      ω    1    ,      θ    2    ′    =      ω    2

poquetahr · Accepted Answer

You view your equations as a four-dimensional vector,   (      θ    1    ,      θ    2    ,      ω    1    ,      ω    2        )    T  . You have four first order differential equations, the two big ones you wrote are       ω    1    ′    ,      ω    2    ′   and your later       θ    1    ′    =      ω    1    ,      θ    2    ′    =      ω    2  . Now you are in the form       y    ′    =  f  (  y  ) so you can calculate the intermediate terms just fine.

How to solve this coupled 2nd order Differential equation of a double pendulum- Runge Kutta method

Answered question

Answer & Explanation

New Questions in Algebra I