A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring feet. The ball is started in motion from the equilibrium position with a downward velocity of 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.

The solution when after t seconds the ball is y feet below its rest position is y (t) =3 et ^⁻16

Step-by-step explanation:

Solution

Given that:

The hollow steel ball weight (W) = 4 pounds

We take acceleration, g. = 32 ft/sec

The mass m = W/g = 4/32

=1/8 slug

The damping constant β = 4

The starting displacement , y (0) = 0

The initial velocity y' (0) = 3

The ball stretches the spring  = 1/8 feet

Now

From Hooke's Law, F =ks

here,

k = constant spring

so,

4 =k (1/8)

k= 32

Now from Newton's law,

my'' = -ky -βy'

= (1/8) y'' = -32y - 4y'

y'' = -256 y - 32 y'

y'' + 32y' + 256 = 0 this is the equation (1)

Thus,

we solve for DE (1)

The equation (auxiliary) is m² + 32m + 256 = 0

so,

m² + 16m + 16 m _ 256 =0

m (m + 16 ) + 16 (m +16) = 0

(m +16) (m + 16 ) = 0

m₁ = - 16,

m₂ = - 16

In this case, the root is known as repeated roots

Now,

The solution is y (t) = c₁e^⁻16 t + c₂ te^⁻16 t

Now, we make use of the initial condition y (0) =0

y(0) = c₁e^⁻16 (0) + c₂ (0)e^⁻16(0)

0 = c₁ e^0 + 0

c₁ = 0

Again, we apply the initial condition y' (0) = 3

y' (t)c₁e^⁻16 t (-16( + c₂ (te^⁻16 t (-16) + e^-16)

y' (t)= - 16c₁e^⁻16 t  + c₂ ( -16te^ ⁻16 + e^-16)

y' (0) =- 16c₁e^⁻16 (0)  + c₂ ( -16(0)e^ ⁻16 (0) + e^-16 (0))

3  = 16c₁ + c₂ (1)

So,

3  = -16(0) + c₂ (1)

c₂ = 3

Therefore, the required solution is y(t) = (0) e^⁻16 + 3 et ^⁻16

y (t) =3 et ^⁻16

x=4

step-by-step explanation: