Question

A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 1515 feet. The ball is started in motion from the equilibrium position with a downward velocity of 88 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.

Answer 1

Answer:

$\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}$

Explanation:

Given that:

The weight of ball = 4 pounds

The spring stretch  x = 1/15 feet

Using the relation of weight on an object:

W = mg

m = W/g

m = 4 / 32

m = 1/8

Now, from Hooke's law:

F = kx

4 =k(1/5)

k = 20 lb/ft

However, since the air resistance is 4 times the velocity;

Then, we can say:

C = 4

Now, for the damped vibration in the spring-mass system, we have:

$m\dfrac{d^2 y}{dx^2}+ c\dfrac{dy}{dt}+ky = 0$

$(\dfrac{1}{8})\dfrac{d^2 y}{dx^2}+ 4\dfrac{dy}{dt}+20y = 0$

$\dfrac{d^2 y}{dx^2}+ 32\dfrac{dy}{dt}+160y = 0$

Solving the differential equation:

m² + 32m + 160 = 0

Solving the equation:

m = -25.80 or m = -6.20

So, the general solution for the equation is:

$y (t)= c_1 e^{-6.20t}+c_2e^{-25.80t}$

$y '(t)=-6.20 c_1 e^{-6.20t}-25.80c_2e^{-25.80t}$

y(0) = 0   ;  y'(0) = 8

$y (0)= c_1 e^{-6.20(0)}+c_2e^{-25.80(0)}$

$c_1 +c_2 = 0 ---(1)$

At y'(0) = 8

$y '(0)=-6.20 c_1 e^{-6.20(0)}-25.80c_2e^{-25.80(0)} \\ \\ 8=-6.20 c_1 e^{0}-25.80c_2e^{0} \\ \\ 8=-6.20 c_1 -25.80c_2--- (2)$

From (1), let $c_1 = -c_2$, then replace the value of c_1 into equation (2)

$8=-6.20 (-c_2)-25.80c_2$

$8=6.20c_2-25.80c_2$

$8=-19.60c_2$

$c_2=\dfrac{ 8}{-19.60}$

$c_2 = -0.408$

From $c_1 = -c_2$

$c_1 = -(-0.408)$

$c_1 = 0.408$

∴

The required solution in terms of t is:

$\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}$