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. (Note: the general solution of the equation of motion is given by: x 

= A Cos 2t + B Sin 2t, where A and B are arbitrary constants)

Answer 1

$$x(t)=A \cos 2t + B \sin 2t\\x(0)=\sqrt{3}=A\\v(t)=2B\cos 2t -2A\sin 2t\\v(0)=-2=2B$$

Thus,

$$x(t)=\sqrt{3} \cos 2t - \sin 2t$$