Question

A mass of slug, when attached to a spring, stretches it feet and then comes to rest in the equilibrium position. Starting at , an external force equal to is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to times the instantaneous velocity.

Answer 1

Answer:

Equation of motion is x(t) = $-te^{-4t}$ + $\frac{1}{4}$ sin(4t)

Explanation:

P.S - The exact question is -

Given - A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at t = 0, an external force equal to  $f(t) = 8 cos(4t)$ is applied to the system.

To find - Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to 8 times the instantaneous velocity.

Proof -

Given that,

Mass = 1 slug

We know that, 1 slug = 32 lb

Now,

Force, f = kx

⇒32 = k(2)

⇒k = 16

Now,

Given that, C = 8 ( 8 times the instantaneous velocity)

Now,

The differential equation of motion is equals to

mx'' + Cx' + kx = 8 cos(4t)

⇒x'' + 8x' + 16x = 8 cos(4t)               ...........(1)

Let the General solution of equation (1) be

x(t) = x(c) + x(p)

Now,

The auxiliary equation is

m² + 8m + 16 = 0

m² + 4m + 4m + 16 = 0

m (m+4) + 4 (m+4) = 0

⇒(m+4)(m+4) = 0

⇒m = -4, -4

So,

The Complimentary equation becomes

x(c) = $Ae^{-4t} + Bte^{-4t}$                  ...........(2)

Now,

Let the particular solution be

x(p) = C cos(4t) + D sin(4t)

x'(p) = -4C sin(4t) + 4D cos(4t)

x''(p) = -16C cos(4t) - 16D sin(4t)

It also satisfy equation (1)

Equation (1) becomes

-16C cos(4t) - 16D sin(4t) + 8 [ -4C sin(4t) + 4D cos(4t) ] + 16 [ C cos(4t) + D sin(4t) ] = 8 cos(4t)

⇒-16C cos(4t) - 16D sin(4t) - 32C sin(4t) + 32D cos(4t) ] + 16C cos(4t) + 16D sin(4t) ] = 8 cos(4t)

⇒-4C sin(4t) + 4B cos(4t) = cos(4t)

By comparing, we get

4B = 1 , A = 0

⇒ B = $\frac{1}{4}$ , A = 0

So, The particular solution becomes

x(p) =  $\frac{1}{4}$ sin(4t)

Now,

The General solution becomes

x(t) = $Ae^{-4t} + Bte^{-4t}$ + $\frac{1}{4}$ sin(4t)       .......(3)

Now,

Given that, At t = 0, initial velocity is zero and the system starts equilibrium

⇒x(0) = 0, x'(0) = 0

By putting t = 0 in equation (3) , we get

A = 0

Now,

Differentiate equation (3), we get

x'(t) = $-4Ae^{-4t} + Be^{-4t} - 4Bte^{-4t}$ + $\frac{1}{4}$ *4 cos(4t)

Put t = 0, we get

0 = -4A + B + 1

⇒B = -1

∴ we get

The general solution becomes

x(t) = $-te^{-4t}$ + $\frac{1}{4}$ sin(4t)

Equation of motion is x(t) = $-te^{-4t}$ + $\frac{1}{4}$ sin(4t)