ANSWER TO CHAPTER 14 HOMEWORK

1.  (a)  When we compare x = (3.8m)cos[(7p/4s^-1)t +p/6] to a general displacement, x = Acos(wt + f), we see that

                          w = 7p/4s^-1, so the period is

                          T = 2p/w = 2p/(7p/4s^-1) = 8/7s.

               the frequency is f = 1/T = 7/8Hz = 0.875Hz.

          (b)  The velocity is

                          v = dx/dt = -(3.8m)(7p/4s^-1)sin[(7p/4s^-1)t + p/6] = -(20.9m/s)sin[(7p/4s^-1)t + p/6].

                 At t = 0, we have

                          x_o = (3.8m)cos(p/6) = 3.3m;

                          v_o = -(20.9m/s)sin(p/6) = -10.4m/s.

            (c)  The acceleration is

                          a = dv/dt = -(3.8m)(7p/4s^-1)²cos[(7p/4s^-1)t + p/6] = -(115m/s²)cos[(7p/4s^-1)t + p/6].

                   At t = 2.0s, we have

                          v = dx/dt = -(20.9m/s)sin[7p/4s^-1)(2.0s) + p/6] = +18m/s;

                          a =  -(115m/s²)cos[(7p/4s^-1)(2.0s) + p/6] =  -57m/s².

2.  (a)  The period of the motion is independent of amplitude:

                          T =2p(m/k)^1/2 = 2p[(0.650kg)/(184N/m)]^1/2 = 0.373s.

               The frequency is f = 1/T = 1/(0.373s) = 2.68Hz.

         (b)  Because the mass is struck aat the equilibrium position, the initial speed is the maximum speed.

               WE find the amplitude from

                          v_o = Aw = 2pfA;

                         22.6m/s = 2p(2.68Hz)A, which gives A = 0.134m.

          (c)  The maximum acceleration is

                          a = w²A =(2pf)²A = [2p(2.68Hz)]²(0.134m) = 38.0m/s².

          (d)  Because the mass starts at the equilibrium position, we have a sine function.

                 If we take the position x-direction in the direction of the initial velocity,         

                we have

                          x = Asin(wt) = Asin(2pft) = (0.134m)sin[2p(2.68Hz)t] =   (0.134m)sin[(16.8s^-1)t]

          (e)  We find the total energy from the maximum kinetic energy:

                         E =K_max = 1/2mv_o² = 1/2(0.650kg)(2.26m/s)² =  1.66J.

          (f)  We find the kinetic energy from the toal energy:

                         E = K + 1/2kx²;

                         1.66J = K + 1/2(184N/m)[(0.40)(0.134m)]²; which gives K = 1.40J.

3.  (a)  For the frequency we have                         f = (g/L)1/2/2p =[(9.80m/s2)/(0.68m)]1/2/2p = 0.60Hz.          (b)  We use energy conservation between the release point and the lowest point:                       Ki + Ui = Kf + Uf;                       0 + mgh = 1/2mvo2 + 0;                       (9.80m/s)(0.68m)(1- cos12°) = 1/2vo2,                which gives vo =  0.54m/s.