CHAPTER 14

  1. a. From the graph we have T = 12 s,ω = 2π/T = 2π/(12 s) = 0.524 rad/s (π/6 rad/s).
    b. vmax = 60 cm/s = ωA, A = vmax /ω = (60 cm/s)/(0.524 rad/s) = 114.6 cm. 
    c. f0 = π/3.
  2. a. cos(π - θ)  = cosθ,
       As at t = 0.685 s, x(t) = 3.0 cm, and at t = 0.886 s, x(t) = -3.0 cm; T/2 = 0.886 s - 0.685 s = 0.201 s. T = 0.402 s
       ω = 2π/T = 2π/(0.402 s) = 15.63 rad/s.
    b. x(t) = Acos(ωt + f0), at t = 0, x(t) = Acosf0 = A,  f0 = π.
        x (0.685 s) = Acos[(15.63 rad/s)(0.685 s) + π] = 3.0 cm, A = 3.09 cm.
  3. The frequency of oscillation is independent of the amplitude, f = (1/2π)(m/k)1/2
  4. T = 2πÖg/L
    TMars/TEarth  = (gMars/gEarth)1/2
    gMars/gEarth = (TMars/TEarth)2;
    gMars = gEarth(TMars/TEarth)2 = (9.8 m/s2)(2.45 s/1.50 s)2 = 26.1 m/s2.
  5. Find the time constant of the oscillation first.
    0.6A =  Ae-(50 s)/2τ,
    ln(0.6) = -(50 s)/2τ, τ = 48.94 s.
    Find the period T = 1/f = 1/(2.0 Hz) = 0.5 s.
    0.3A = Ae-t/2τ = Ae-t/2(48.84 s)
    -t/2(48.94 s) = ln(0.3), t = 117.8 s.
    n = t/T = (117.8 s)/(0.5 s) = 236 oscillations.
  6. Ignoring  the masses of the two springs, the forces on the block, spring 1 and spring 2 would be same. F = F1 = F2.
    The  length changes of the two springs are different. F1= k1x1, F2= k2x2,and F = kx; where x = x1 + x2.
    k1x1 = k2x2, x2 = (k1/k2)x1 ; x = x1 + (k1/k2)x1 = [1 + (k1/k2)]x1.
    F = F1, kx = k[1 + (k1/k2)]x1 = k1x1.
    k[1 + (k1/k2)] = k1, k = k1/[1 + (k1/k2)] = (k1k2)/(k1 + k1)
    f = (1/2π)(k/m)1/2, f1 = (1/2π)(k1/m)1/2, and  f2 = (1/2π)(k2/m)1/2.
    f2 = (1/2π)2(k/m) = (1/2π)2[(k1k2)/(k1 + k2)]/m = (1/2π)2[1/(1/k1 + 1/k2)]/m
        = (1/2π)2[(m/k1) + (m/k2)]
        = (1/f12) + (1/f22)

    7.