Suppose that a mass of 2 kg is attached a spring whose spring constant is 50. Th

Question

Suppose that a mass of 2 kg is attached a spring whose spring constant is 50. The system is damped such that b = 2. The mass is set in motion with an initial velocity of -8 m/s at a position 0 meters from equilibrium. Set up and solve a differential equation that models this motion. Write your solution in the form A cos (wt - alpha) where alpha is a positive number. Use your solution to fill in the information below: What is the amplitude of the motion? What is the value of w? _ Preview What is the phase shift? Preview

Explanation / Answer

for b = 12
k = 50
m = 2 kg
vo = -8 m/s
xo = 0 m

from newtons' secondlaw

ma + bv + kx = 0
let x = e^lambda*t

then m*lambda^2 + b*lambda + k = 0

lambda = (-b+-sqroot(b^2 - 4mk))/2m

now, b^2 - 4mk < 0 so underdamped osscilation

so the solution is of the form
x = e^(-gamma)t * a * cos(wt - alpha)
w = sqroot(wo^2 - gamma^2)
gamma = b/2m = 3
wo = sqroot(k/m) = sqroot(25) = 5 rad/s
w = 4 rad/s
x = ae^(-3t)cos(4t - alpha)
xo = 0 = ae^(-3*0)cos(-alpha) = acos(alpha) = 0
=> alpha = pi/2

v = a[e^(-3t)(-sin(4t - alpha)*4) - 3e^(-3t)(cos(4t - alpha))]
vo = -8 = 4a
a = -2

so, x = -2e^(-3t)cos(4t - pi/2)

so, a = -2
w = 4
alpha = pi/2

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