This problem has 3 parts. The next part is revealed after the previous part is c

Question

This problem has 3 parts. The next part is revealed after the previous part is correctly answered. A body of mass 4 gr attached to a vertical spring stretches it 1000/9 cm to reach an equilibrium position. When the body moves in the air it experiences a resistance opposite to its movement, proportional to its speed, with a damping constant d = 40 gr/sec. Approximate the acceleration of gravity by g = 1000 cm/sec. (3/10) Find the spring constant k and the body-spring natural frequency omega_0. (2/10) Find a differential equation of the form y" = F(y, if) for the displacement function y from the equilibrium position, with y positive downwards. Use yp to denote de derivative if. F(y, y') = (1/10) Find, r+, r-, roots of the characteristic polynomial of the differential equation in part (b). Convention: If the roots are real, then r + r-; if the roots are complex, then r plusminus = alpha plusminus beta i, with alpha, beta reals and beta > 0.

Explanation / Answer

mass=m=4 gram=0.004 kg

elongation=1000/9 cm=10/9 m

as force=k*x

==>0.004*10=k*(10/9)

==>k=0.036 N/m

natural angular frequency=sqrt(k/m)=3 rad/sec

part b:

force balancing equation for the spring will be:

F=-k*y-d*(dy/dt)+m*g

using the values for different parameters,

0.004*y''=-0.036*y-0.04*y'+0.004*10

==>y''+10*y'+9*y=0.04...(1)

characterstics polynomial:

m^2+10*m+9=0

==>m=-1 and m=-9

hence r+=-1

and r-=-9

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