my\'\' + y\' + ky = 0, where m is the mass of the object, ? is the damping coeff

Question

my'' + y' + ky = 0,
where m is the mass of the object, ? is the damping coefficient and k is the spring constant. At rest position we found the relation Y k - mg = 0, where g = 9.8 is approximately the acceleration due to gravity and Y is the distance the spring is stretched passed its natural spring length. Suppose an object with a mass of .5 kilograms is attached to a spring and stretches it .1 meters. Suppose that the magnitude of the damping force was measured to be 3 Newtons when the object is moving 3 meters per second. (Reminder: the damping force was assumed to be proportional to the velocity)

a) Formulate the DE.

b) Suppose I pull on the object hanging from the spring an additional distance of .2 meters and release. Solve the DE in a) using these initial values. Describe the behavior as time goes to infinity.

c) Suppose now that the damping force is proportional to the square of the velocity. Refor- mulate the DE in a), but do not solve it.

Explanation / Answer

a) Given that = 3 when y' = 3, => = 1

again yk = mg

k = mg/y,

= .5*9.8/0.1 = 49

so, k = 49

The DE is 0.5y'' + y' + 49y = 0

=> y'' + 2y' + 98y =0

b) given y(t=0) = 0.2

so, the Characteristic equations are :

m^2 + 2m + 98 = 0

m = - ± (-97)

so, y(t) = e-t[c1.cos 97 t + c2.sin97t]

now, y(0) = c1 = 0.2

=> y(t) = e-t[0.2 cos 97 t + c2.sin97t]

clearly as t - > , y(t) -> 0

c)  y' = a.(y')2

= > = ay'

so, the D.E is 0.5y'' + a(y')2 + 49y =0

or, y'' + 2ay'2 + 98 y =0

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