(1 point) A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m . The mass is started in motion with initial position x0=5m and initial velocity v0=7m/s . Determine the position function x(t) in meters.

The function is x = e^(-t/2) * (0.792*sin12t + 5cos12t)

Explanation:

we have to:

m = mass = 4 kg

k = spring constant = 577 N/m

c = damping constant = 4 N*s/m

The differential equation of motion is equal to:

m(d^2x/dt^2) + c(dx/dt) + k*x = 0

Replacing values:

4(d^2x/dt^2) + 4(dx/dt) + 577*x = 0

Thus, we have:

4*x^2 + 4*x + 577 = 0

we will use the quadratic equation to solve the expression:

x = (-4 ± (4^2 - (4*4*577))^1/2)/(2*4) = (-4 ± (-9216))/8 = (1/2)  ± 12i

The solution is equal to:

x = e^(1/2) * (c1*sin12t + c2*cos12t)

x´ = (-1/2)*e^(1/2) * (c1*sin12t + c2*cos12t) + e^(-t/2) * (12*c1*cos12t - 12*c2*sin12t)

We have the follow:

x(0) = 5

e^0(0*c1 + c2) = 5

c2 = 5

x´(0) = 7

(-1/2)*e^0 * (0*c1 + c2) + e^0 * (12*c1 - 0*c2) = 7

(-1/2)*(5) + 12*c1 = 7

Clearing c1:

c1 = 0.792

The function is equal to:

x = e^(-t/2) * (0.792*sin12t + 5cos12t)

magnified images possible images can be real = projected onto a screen

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west