Page 483 - Dynamics of Mechanical Systems
P. 483

0593_C13_fm  Page 464  Monday, May 6, 2002  3:21 PM





                       464                                                 Dynamics of Mechanical Systems


                       where A is an amplitude to be determined and where ψ is defined as:


                                                            D
                                                          ψ = pt + φ                          (13.10.3)
                       where φ is a phase angle to be determined.
                        Next, let dA/dt and dφ/dt be given by:

                                                           2π
                                                 dA  =−     ∫  Fcos ψψ  = α  A ()             (13.10.4)
                                                                   d
                                                 dt    2π p
                                                           0
                       and


                                                           2 π
                                                 dφ  =     ∫     ψψ =  β()
                                                 dt   2 π pA  Fsin  d    A                    (13.10.5)
                                                           0

                        If Eqs. (13.10.4) and (13.10.5) can be solved for A and φ, then an approximate solution
                       to Eq. (13.10.1) may be expressed in the form:

                                                       x =  Asin ( pt + ) φ                   (13.10.6)


                                         ˙ x
                       with the derivative   given by:
                                                                (
                                                       ˙ x =  Ap cos pt + ) φ                 (13.10.7)

                        To illustrate the procedure, consider again the nonlinear pendulum equation:

                                                        ˙˙
                                                       θ +( ) l  sinθ= 0                      (13.10.8)
                                                           g
                       Suppose we approximate sinθ as:

                                                            ≈−
                                                                  3
                                                        sinθθ θ 6                             (13.10.9)
                       Then, Eq. (13.10.8) becomes:

                                                    ˙˙      θ )  3  +(g l  θ ) = 0           (13.10.10)
                                                    θ −(g 6l

                       Hence, by comparison with Eq. (13.10.1), we can identify  , f, and p as:

                                                             = g 6l                           (13.10.11)


                                                         f θθ ( ) =− θ  3                    (13.10.12)
                                                             ˙
                                                            ,
   478   479   480   481   482   483   484   485   486   487   488