A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 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

x(t) = 7/4(e^-7t) - ¾(e^-3t)

Explanation:

Given

Let m represents the mass attached to the spring. m = 1kg

Represent the spring constant with k; k = 21N/m

Represent the positive damping constant with β; β = 10

From Newton second law for the system, we have

m.d²x/dt² = -kx - β.dx/dt --- divided through by m

d²x/dt² = -(k/m)x - (β/m)dx/dt

Suppose the system is in equilibrium

d²x/dt² + (k/m)x + (β/m)dx/dh = 0

Substitute in the values of m,k and β.

This gives

d²x/dt² + (21/1)x + (10/1)dx/dt = 0

d²x/dt² + 21x + 10dx/dt = 0

The auxiliary equation is

m² + 21m + 10 = 0

And the solution is

m = -7 and m = -3.

The general solution is then

x(t) = c1e^-7t + c2e^-3t

Given that

x(0) = 1 and x'(0) = 0

c1 + c2 = 1

-3c1 - 7c2 = 0

c2 = -¾

c1 = 7/4

x(t) = 7/4(e^-7t) - ¾(e^-3t)

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.