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

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.