Answer to Question #144818 in Mechanics | Relativity 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

(ii) The amplitude, period and frequency of oscillation after a long time has elapsed

Answer 1

$$\frac{d²x}{dt²} + 8x = 20\cos2t.$$

The solution

$$x=A\cos2t+B\sin2t.$$

We obtain

$$-4A\cos2t-4B\sin2t - 8A\sin2t+8B\cos2t$$ $$+ 8A\cos2t+8B\sin2t = 20\cos2t.$$

Hence

$$4A+8B=20,\quad -8A+4B=0.$$ $$B=2A, \; A=1.$$ $$x=\cos2t+2\sin2t=\sqrt{5}\cos(2t+\arctan2).$$ $$x=\cos2t+2\sin2t=\sqrt{5}\cos(2t+\arctan2).$$

Therefore:

amplitude $a=\sqrt{5};$

$a=\sqrt{5};$

period $T=\frac{2\pi}{\omega}=\pi;$

frequency $f=1/T=1/\pi.$