Question

A force of 15 lb stretches a spring 3 ft. A 8-lb weight is attached to the spring and the system is immersed in a medium that imparts a damping force equal to its instantaneous velocity. (a) Find the equation of motion if the weight is released from rest 15 inches above equilibrium position x(t) Preview (b) the weight is released 15 inches above the equilibrium position with an upward velocity of 2 ft/s x(t) Preview Using the equation from part b, (c) Find the the first time that the weight passes through the equilibrium point: t = Preview (d) Find the time when the object is at maximum displacement below equilibrium: t = Preview Preview (d) Find the maximum displacement below equilibrium in inches: in

Please show all work! Thank you!

P.S. 5.2 Q2

Answer 1

a) is spring constant

F 15 - 5 3

W 8 m = g = 1 32 4

Equation of motion is

ma 0 "x 5 0 4

So is the characteristic equation so

4 20r 0 D2+4D 20

(D 2)2 16 D -2+ 4

So is general solution

2t x(t)ce cos(4t) e2t sin(4t)

Initial condition implies $x(0)=c_1=\frac{15}{12},\,x'(0)=-2c_1+4c_2=0\Rightarrow c_2=\frac{c_1}{2}=\frac{15}{24}$

So that is the equation of motion

15 15 -2t cos(4t) e-2 sin(4t) 24 (t) 12

b) If

15 27 r(0)C1 z'(0)2c14c2 2 4c2 2(1c1) 6 _

So
27 C2 = 24 15 C1 12
and so
15 -2t cos(4t) 2t sin(4t) -e 24 (t) 12

Whose graph is given below:

1:5 27 2 2t sin (4 f() cos (4t)+ 24 12 1 0.5 (0.576, 0 25 15 0.5 1 (0.853, -0.2733) -0.5

c) Using the above solution, we have is the time it touches the equilibrium position

(t 0t0.576

d) The time it is at maximum displacement below the equilibrium point is $t\approx 0.853$

e) The maximum displacement is $x(0.853)\approx 3.28$ inches (0.273 feet in inches)

$\blacksquare$

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