Page 135 - Schaum's Outline of Differential Equations
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118                   SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS                 [CHAP.  14




         motions, while forced  damped  systems (assuming the external force to be  sinusoidal) yield both transient and
         steady-state motions.
            Free undamped motion defined by Eq. (14.11) with a 1 = 0 and/(?) = 0 always has solutions of the form



         which  defines simple harmonic motion. Here  c 1, c 2, and ft) are  constants  with ft) often referred  to  as circular
         frequency.  The natural frequency j'is





         and it represents  the number  of complete  oscillations  per  time unit undertaken  by  the  solution. The period of
         the system of the time required  to complete  one oscillation  is





            Equation  (14.12)  has the alternate  form



         where the amplitude         the phase angle  (j) = arctan (c 2lcj),  and k is zero when c 1 is positive and unity
         when c 1 is negative.




                                           Solved Problems





         14.1.  A steel ball weighing  128 Ib is suspended from  a spring, whereupon the spring is stretched 2 ft from  its
               natural length. The ball is started in motion with no initial  velocity by displacing it 6 in above the equi-
               librium  position. Assuming no  air resistance,  find  (a) an  expression  for  the position  of  the ball  at any
               time t, and  (b)  the position  of the ball att=nl 12 sec.
               (a)  The  equation of motion is governed by  Eq.  (14.1).  There is no externally applied force,  so (t)  = 0, and  no
                                                                                      F
                                                                                                 2
                   resistance  from  the  surrounding  medium,  so  a = 0. The  motion is  free  and  undamped. Here g = 32  ft/sec ,
                   m = 128/32 = 4 slugs, and it follows  from  Example 14.1 that k = 64 Ib/ft.  Equation (14.1) becomes x + 16x = 0.
                   The roots of its characteristic equation are  X = ±4i, so its solution is


                   At t = 0, the position of the ball is  x 0  = -  j  ft  (the minus sign is required because the ball is initially displaced
                   above  the equilibrium position, which is in the negative direction). Applying this initial  condition to  (_/),  we
                   find that



                   so (1) becomes


                   The  initial velocity is given as v 0 = 0 ft/sec.  Differentiating  (2), we obtain

                                          v(t)  = x(i)  = 2sin4t + 4c 2 cos4f
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