Page 102 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)
P. 102

1.11  harmoniC analysis   99
                                                  x(t)
                                                 (a) (b)  (c)

                                               A
                                    (d)        A                        t
                                    (e)
                                                     O t   t
                                                       2
                                                      (i)
                                                   x (t)
                                                    1
                                                     A
                                                                        t
                                                   O    t   t
                                                  A     2
                                                 (ii) Odd function


                                                 x 2 (t)
                                                     A
                                                                        t
                                                  O      t
                                                  A
                                                      t
                                                      2
                                                (iii) Even function
                                    FiGure 1.58  Even and odd functions.




                                    we find from Eq. (1.91),

                                                  t     4A       1       2p12n - 12     t
                                            x ¢t +  ≤  =              sin          ¢t +  ≤
                                                           a
                                             1
                                                  4     p  n = 1 12n - 12    t          4
                                                        4A       1        2p12n - 12t   2p12n - 12
                                                       =   a          sin b           +            r   (1.94)
                                                        p  n = 1 12n - 12      t             4

                                    Using the relation sin1A + B2 = sin A cos B + cos A sin B, Eq. (1.94) can be expressed
                                    as

                                                   t     4A        1       2p12n - 12t   2p12n - 12
                                              x ¢t +  ≤ =     b         sin           cos
                                                           a
                                             1
                                                   4     p  n = 1 12n - 12     t             4
                                                              2p12n - 12t   2p12n - 12
                                                          + cos          sin          r                (1.95)
                                                                   t            4
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