Question

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to 2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 3 ft/s. (Use g

Answer 1

`\mathbf{x = 3te ^(-2 √(2t))}`

Step-by-step explanation:

From the given information concerning the spring-mass system:

Let us apply Hooke law.

Then, we have:

mg = ks

8 = k4

k = 8/4

k = 2

Provided that the mass weighing 8 lbs is attached to a spring.

Then, we can divide it by gravity 32 ft/s².

∴

m = 8/32

m = 1/4 slugs

The medium that offers the damping force
`\beta = √(2)`

Now, let us set up a differential equation that explains the motion of the spring-mass system.

The general equation is:

`mx '' + \beta x' + kx = 0`

where;

`m = (1)/(4)`

k = 2, and

`\beta = √(2)`

Then;

`(1)/(4)x'' + √(2) x' + 2x = 0`

By solving the above equation, the auxiliary equation is:

`m^2 + 4 √(2) m + 8n = 0`

Using quadratic formula:

`(-b \pm √(b^2 -4ac))/(2a)`

`m = \frac{-4 √(2) \pm \sqrt{(4 √(2))^2 -4(1)(8) }}{2}`

`m = (-4 √(2) \pm √(32 -32 ))/(2)`

`m = -2 √(2)`

Since this is a repeated root, the solution to their differential equation took the form.

`x_c = c_1 e^(mt) + c_2 t e^(mt)`

`x_c = c_1 e^(-2√(2) t) + c_2 t e^(-2√(2) t)`

From the initial condition.

At equilibrium position where the mass is being from:

x(0) = 0

Also, at the downward velocity of 3 ft/s

x'(0) = 3

Then, at the first initial condition:

`x_c (0) = c_1 e^(-2√(2) *0) + c_2 (0) e^(-2√(2) *0)`

`0= c_1 e^(0) + 0`

`0= c_1`

At the second initial condition;

`x' = -2 √(2) c_1 e^(-2 √(2) t ) -2 √(2) c_2 t e^(-2 √(2) t ) + c_2 e^(-2 √(2) t)`

where;

x'(0) = 3

`x' (0) = -2 √(2) c_1 e^(-2 √(2)* 0 ) -2 √(2) c_2 (0) e^(-2 √(2) *0 ) + c_2 e^(-2 √(2) *0)`

`3 = -2 √(2) * 0 *e^0 - 0 + c_2 e^0`

`3 = 0 + c_2`

`3 = c_2`

Replacing in the constraints, the equation of the motion is:

`\mathbf{x = 3te ^(-2 √(2t))}`