Question

An object of mass 2kg is attached to a spring. A force of 5nt is applied to move the object 0.5m from its equilibrium position. The damping force of the object sliding on the table is int when the velocity is 0.25m/second. The object is pulled to the left until the spring is stretched lm and then released with the initial velocity of 2m/second to the right.

Set up an I.V.P. to describe the motion of the object and solve it, then state the amplitude function of the motion.

Answer 1

`\mathbf{x(t) = ( \sqrt 5)/(2)e^(-t) }`

Explanation:

Given that:

mass of the object = 2 kg

A force of 5nt is applied to move the object 0.5m from its equilibrium position.

i.e

Force = 5 newton

Stretchin (x) =0.5 m

Damping force = 1 newton

Velocity = 0.25 m/second

The object is pulled to the left until the spring is stretched lm and then released with the initial velocity of 2m/second to the right

SOLUTION:

If F = kx

Then :

5 N = k(0.5 m)

where ;

k = spring constant.

k = 5 N/0.5 m

k = 10 N/m

the damping force of the object sliding on the table is 1 newton when the velocity is 0.25m/second.

SO;

`C (dx)/(dt)= F_d`

`C* 0.25 = 1`

C =
`(1 \ N )/(0.25 \ m/s)`

C = 4 Ns/m

NOW;

`m (d^2x)/(dt^2)+ C (dx)/(dt)+ kx = 0`

Divide through by m; we have;

`(m)/(m)(d^2x)/(dt^2)+ (C)/(m) (dx)/(dt)+ (k)/(m) x= 0`

`(d^2x)/(dt^2)+ (C)/(m) (dx)/(dt)+(k)/(m)x= 0`

`(d^2x)/(dt^2)+ (4)/(2) (dx)/(dt)+(10)/(2)x= 0`

`(d^2x)/(dt^2)+ 2 (dx)/(dt)+5x= 0`

we all know that:

`x(t) = Ae^((\alpha \ t))` ------ (1)

SO;

`\alpha ^2 + 2\alpha + 5 = 0`

`\alpha = (-2 \pm √((4)-(4*5)))/(2)`

`\alpha = -1 \pm 2i`

Thus ;

`x(t) = e^(-t)[A \sin (2t)+ B cos (2t)]` ------------ (1)

However;

`(dx)/(dt) = e^(-t)[A \sin (2t)+ B cos (2t)]+ 2e ^(-t) [A \cos (2t)- B \ Sin (2t)]` ------- (2)

From the question ; we are being told that;

The object is pulled to the left until the spring is stretched 1 m and then released with the initial velocity of 2m/second to the right.

So ;

`x(0) = -1 \ m`

`(dx)/(dt)|_(t=0) = 2 \ m/s`

x(0) ⇒ B = -1

`(dx)/(dt)|_(t=0) =- B +2A`

`=- 1 +2A`

1 = 2A

A =
`(1)/(2)`

From (1)

`x(t) = e^(-t)[A \sin (2t)+ B cos (2t)]`

`x(t) = e^(-t)[(1)/(2) \sin (2t)+ (-1) cos (2t)]`

`x(t) = e^(-t)[(1)/(2) \sin (2t)-cos (2t)]`

Assuming;

`A cos \ \phi = (1)/(2)`

`A sin \ \phi = 1`

Therefore:

`A = \sqrt{(1)/(4)+1}`

`A = \sqrt{(1+4)/(4)}`

`A = \sqrt{(5)/(4)}`

`A ={ ( \sqrt5)/( \sqrt4)}`

`A ={ ( \sqrt5)/( 2)}`

where;

`\phi = tan ^(-1) (2)`

Therefore;

`x(t) = (\sqrt 5)/(2)e^(-t) \ sin (2 t - \phi)`

From above ; the amplitude is ;

`\mathbf{x(t) = ( \sqrt 5)/(2)e^(-t) }`