A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true. (a) the mass is initially released from rest from a point 1 meter below the equilibrium position

A) y(t) = - 0.286e^(-9) + 1.287e^(-2)

Step-by-step explanation:

A) M = 1kg

K = 18 N/m

by' = 11y'

So, using Newton's second law for the system, we obtain;

my" + by' + Ky = 0

So;

y" + 11y' + 18y = 0

The auxiliary equation of this is;

m² + 11m + 18 = 0

Solving for the roots to obtain, m = -9 and -2

So the general Solution is;

y(t) = C1 e^(-9) + C2 e^(-2)

and y'(t) = -9C1 e^(-2) - 2C2 e^(-2)

Now, from the question,

y(0) = 1 and y'(0) = 0

And so;

C1 + C2 = 1

And -9C1 - 2C2 = 0

Making C2 the subject in the second equation, we get;

C2 = - (9/2) = - 4.5C1

Putting - 4.5C1 for C2 in first equation to obtain;

C1 - 4.5C1 = 1

- 3.5C1 = 1

C1 = -(1/3.5) = -0.286

C2 = -4.5 x (-0.286) = 1.287

Thus equation of motion here is;

y(t) = - 0.286e^(-9) + 1.287e^(-2)