Question

A certain oscillator satisfies the equation of motion: 𝑥••+ 4x = 0. Initially the particle is at

the point x = √3 when it is projected towards the origin with speed 2.

2.1. Show that the position, x, of the particle at any given time, t, is given by:

x = √3 cos 2t – sin 2t.

Answer 1

$\ddot x+4x=0,$

$x=c_1\cos 2t+c_2\sin 2t,$

$\dot x=2c_2\cos 2t-2c_1\sin 2t,$

$c_1=\sqrt 3,$

$c_2=1,$

$x=\sqrt 3\cos 2t+\sin 2t .$