Page 182 - Schaum's Outline of Theory and Problems of Applied Physics
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CHAP. 14]                        SIMPLE HARMONIC MOTION                               167



        SOLVED PROBLEM 14.2
              A spring with a force constant of 200 N/m is compressed by 8 cm. A 250-g ball is placed against the end
              of the spring, which is then released. What is the ball’s velocity when it leaves the spring?
                  The kinetic energy of the ball equals the elastic potential energy of the compressed spring. Hence
                                            KE = PE
                                           1 mv = ks 2
                                              2
                                                 1
                                           2     2


                                            k      200 N/m
                                       v =    s =        (0.08 m) = 2.26 m/s
                                            m      0.25 kg
        SOLVED PROBLEM 14.3
              Two springs, one of force constant k 1 and the other of force constant k 2 , are connected end-to-end. (a) Find
              the force constant k of the combination. (b)If k 1 = 5 N/m and k 2 = 10 N/m, find k.
              (a) When a force F is applied to the combination, each spring is acted on by this force. Hence the respective
                  elongations of the springs are
                                                      F         F
                                                  s 1 =    s 2 =
                                                      k 1      k 2
                  and the total elongation of the combination is
                                                       F   F   F(k 1 + k 2 )
                                            s = s 1 + s 2 =  +  =
                                                      k 1  k 2    k 1 k 2
                  Since F = ks for the combination,
                                                       F    k 1 k 2
                                                   k =   =
                                                       s   k 1 + k 2
                                               (5 N/m)(10 N/m)
              (b)                           k =              = 3.33 N/m
                                                5 N/m + 10 N/m
                  The force constant of the combination is smaller than either of the individual force constants.


        SIMPLE HARMONIC MOTION
        In periodic motion, a body repeats a certain motion indefinitely, always returning to its starting point after a
        constant time interval and then starting a new cycle. Simple harmonic motion is periodic motion that occurs when
        the restoring force on a body displaced from an equilibrium position is proportional to the displacement and in the
        opposite direction. A mass m attached to a spring executes simple harmonic motion when the spring is pulled out
        and released. The spring’s PE becomes KE as the mass begins to move, and the KE of the mass becomes PE again
        as its momentum causes the spring to overshoot the equilibrium position and become compressed (Fig. 14-1).
            The amplitude A of a body undergoing simple harmonic motion is the maximum value of its displacement
        on either side of the equilibrium position.


        PERIOD AND FREQUENCY
        The period T of a body undergoing simple harmonic motion is the time needed for one complete cycle; T is
        independent of the amplitude A. If the acceleration of the body is a when its displacement is s,
                                                      s

                                            T = 2π −
                                                      a

                                                      displacement
                                        Period = 2π −
                                                       acceleration
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