1. A mass on a long hanging spring oscillates vertically with no damping Choose

Question

1. A mass on a long hanging spring oscillates vertically with no damping Choose the equilibrium position to be y 0 and up to be the positive time t = 0, the velocity is 3.0 m/s downwards. The amplitude is 1.5 m. did not solve Part A, in Parts B and C you may give your answer in te direction. At (Note: If you rms of T.) A. What is the period T of the oscillations? B. Write expressions for position, velocity and acceleration as a function of time? C. Suppose damping is added to the system such that the damping coefficient is 0.80 times that of critical damping. What is the period T" of the damped oscillations?

Explanation / Answer

1. given
mass = m
no damping, c = 0
yeq = 0
upward is the positive direction
at t = 0, v(t) = 3 m/s
A = 1.5 m

A. Period of osscilation = T
  T = 2*pi*sqroot(m/k)
  now,

  w = sqrt(k/m)
  and
  equation of motion can be written as
  y = Asin(wt + phi)
  hence
  v = Awcos(wt + phi)
  v(0) = -3 = 1.5*w*cos(phi)
  -2 = w*cos(phi)
  hence
  for phi = 180 deg
  w = 2 rad/s
  and
  y = 1.5*sin(2t + 180) = -1.5*sin(2t)

B. y = -1.5*sin(2t)
  we can also write T = 2*pi/w
  hence
  w = 2*pi/T
  hence
  y = -1.5*sin(2*pi*t/T)
  for w = 2
  2*pi/T = 2
  T = pi
  hece
  y = -1.5*sin(2t)

similiarly
  v = -(3*pi/T)*cos(2*pi*t/T)
   for w = 2
   v = -3cos(2t)

  a = (6*pi^2/T^2)(sin(2*pi*t/T))
  for w = 2
  a = 6sin(2t)

  C. c = 0.8c'
   c' = 2*sqrt(mk)
   c = 1.6*sqrt(mk)
   w' = sqroot(k/m - c^2/4m^2) = 0.6sqroot(k/m) = 0.6*w
   hence
   T' = 2*pi/w' = 2*pi*T/0.6*2*pi = 1.6667T
   hence
   for T = pi
   T' = 5.235987755 s

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