Question

Due Thu 06/06/2019 2:5 A force of 20 lb stretches a spring 2 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 18 inches above equilibrium position. z(t) Preview (b) the weight is released 18 inches above the equilibrium position with an upward velocity of 3 ft/s. r(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 (d) Find the maximum displacement below equilibrium in inches: in Preview

Answer 1

The equation of motion can be written as, (downwards is positive and upwards is negative)

mä= mg - bi - ka

It is given that 20lb stretches the string by 2ft, therefore k is 10 lb/ft. The damping force is equal to the instantaneous velocity hence b = 1. Substituting values we get,

8 (8 x 32.2) (1)-(10) 8 10 257.6

Solving the differential equation without any initial condition we get,

(te.0025 (c1sin(1.12)C2cos(1.12t))25.76

(Inches are converted to ft)

(a) In this case the initial condition is that at t = 0, x = -1.5 (negative since it is above), and the initial velocity is 0 at t = 0. Substituting t = 0, x = -1.5, we get,

1.5e.0625(0) (c1sin(1.12(0))c2cos(1.12(0)))25.76 C2 -1.5- 25.76 -27.26

Taking derivative of the above equation and substituting x'(0) = 0, we get,

C1-1.53

Hence the solution is,

(te0.025t (-1.53sin (1.12t) 27.26cos(1.12t)25.76

(b) Here the initial conditions are at t = 0, x = -1.5 and x' = -3 (upwards hence negative). Therefore on substituting the initial conditions and solving we get, (as done above),

(te0.025t (-4.21sin(1.12t) 27.26cos(1.12t)25.76

Plotting the graph we get, ( x-axis is time and y-axis is displacement )

-40 20 www 0 20 40 60 80

(c) From the graph we get the point as,

40 20 20 40 60 (0.398, 0)

Hence the time is 0.398 seconds.

(d) Maximum displacement below equilibrium means that we have to find the time when the graph has the maximum positive value, hence the point is, (2.892,48.746)

(2.892, 48.746) -40 20 0 60 40 20

Therefore the time is 2.892 seconds.

(e) From the graph point we get the maximum displacement below the equilibrium as, 48.746 ft, which is 584.94 in.