3. After a mass weighing 10 pounds is attached to a 5-foot spring, the spring me

Question

3. After a mass weighing 10 pounds is attached to a 5-foot spring, the spring measures 7 feet. This mass is removed and replaced with another mass that weighs 8 pounds. The entire system is placed in a medium that offers a damping force that is numerically equal to the instantaneous velocity Find the equation of motion if the mass is initially released from a point foot below the equilibrium position with a downward velocity of 1 ft/s. Express the equation of motion in a form only involving a sine function (no cosine) Find the times at which the mass passes through the equilibrium position heading downward. a. b. c.

Explanation / Answer

3. weight of mass , W = 10 pounds

length of spring when this mass is attached, l = 7 ft

Lo = 5ft

so spring constant k = W/(l - Lo) = 5 lb per ft

now m = 8 pounds

c = 1

hence

a. equation of motion = mx" + cx' + kx = 0

8x" + x' + 5x = 0

initial conditions

x'(t = 0) = 1 ft/s

x(t = 0) = 1 ft

b. so the characeteristic equation of this equation gives us

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

c^2 - 4mk = 1 - 4*8*5 < 0

hence

underdamped osscilator

so the solution is of the form

x = e^(-gamma*t)*A*sin(w1t - alpha)

w1 = sqroot(w^2 - gamma^2)

w = sqroot(k/m) = 0.7905

gamma = c/2m = 1/2*8 = 0.0625

x = e^(-0.0625t)Asin(0.788t - phi)

x(t = 0) = -Asin(phi) = 1

sin(phi) = -1/A

cos(phi) = sqroot(1 - 1/A^2) = sqroot(A^2 - 1)/A

x'(t = 0) = -0.0625e^(-0.0625t)Asin(0.788t - phi) + 0.788e^(-0.0625t)Acos(0.788t - phi) = 1

A*0.0625sin(phi) + A*0.788cos(phi) = 1

-0.0625 + 0.788*sqroot(A^2 - 1) = 1

A = 1.678

phi = -36.562 deg = -0.6381 rad

so

x = 1.678*e^(-0.0625*t)*sin(0.788t + 0.6381)

c. x = 1.678*e^(-0.0625*t)*sin(0.788t + 0.6381) = 0

0.788t + 0.6381 = n*pi

t =

x' > 0

cos(0.788t + 0.6381) > 0

hence n = 2,4,6 ..

t = 7.163s, 15.1373 s ..

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