Answer to Question #144399 in Classical Mechanics for Daniel Kanyugo

Question

The position of a particle moving along the x-axis is determined by the

equation

d^2x/dt^2+8x=20cos2t

. If the particle starts from rest at x = 0,

determine

(i) The position x as a function of time

Answer 1

Solution

This is second order differential equation

$\frac{d^2x}{dt^2}+8x=20\cos 2t$

Firstly found it's

CF and PI

$x(t) =CF+PI$

CF is

(m²+8) =0

it's solution become

$m=-2\sqrt{2}i \\m=+2\sqrt{2}i$

So CF=$(A\cos2\sqrt{2}t+B\sin2\sqrt{2}t)$

And

PI=$\frac{20\cos2t}{(D^2+8)}=\frac{20\cos2t}{(-4+8)}=5\cos2t$

Therefore

Position become

$x(t) =(A\cos2\sqrt{2}t+B\sin2\sqrt{2}t)+5\cos 2t$