A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The

Question

A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 16 ft/s. Determine the time at which the mass passes through the equilibrium position. (Use

g = 32 ft/s2

for the acceleration due to gravity.)

B) Find the time after the mass passes through the equilibrium position at which the mass attains its extreme displacement from the equilibrium position.

Explanation / Answer

Weight ,W = mg
4 = m*32 => m = 1/8

Here We'll first Write the differential equation

m* (d^2x / dt^2) = -kx - b*(dx/dt) ; where b = 1 is damping constant and m is the mass mass and k is spring constant

1/8*(d^2x / dt^2) = -2x - 1*(dx/dt)

d^2x/dt^2 + 8dx/dt + 16x = 0 ,------------(1)

noW We'll solve equation (1)

r^2 + 8r + 16 = 0

(r + 4)^2 = 0

=> r1 = r2 = -4,

Hence the system is critically damped, and the general solution is

x(t) = C1 e^(-4t) + C2 t*e^(-4t)

NoW We'll apply the initial conditions that is , at t = 0, x(0) = -1 (initially 1 foot above equilibrium)

and initial velocity x'(0) = 16 ft/sec

=> -1 = C1 e^(0) + C2 *0*e^(0) => C1 = -1
and

x'(t) = C1 e^(-4t)*d/dt(-4t) + C2 [t*e^(-4t)*d/dt(-4t) + e^(-4t)*1]
16 = -1*e^(0)*(-4) + C2 [0*e^(0)*(-4) + e^(0)*1]
16 = 4 + C2 [0 + 1]
C2 = 12

b>
Hence x'(t) = 4e^(-4t) + 12*e^(-4t) [1 - 4t]

we see that x'(t) = 0 = 4 e^(-4t) + 12e^(-4t) [1 - 4t]

=>16e^(-4t) = 24te^(-4t)
=> t = 2/3 seconds

c>
and the corresponding extreme displacement can be found from
x(t) = - e^(-4t) + 12t*e^(-4t)

x(t) = 12t*e^(-4t) - e^(-4t)

x(2/3) = 12*(2/3)*e^(-4*(2/3)) - e^(-4*(2/3))

x = 8*e^(-8/3) - e^(-8/3) = .4863 ft

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