# Homework Help: Simple harmonic motion problem

1. Sep 29, 2008

### Benzoate

1. The problem statement, all variables and given/known data
A certain oscillator satisfies the equation

x''+4x=0

x=$$\sqrt{3}$$
|x'|= 2
show that , in subsequent motion,

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

Deduce the amplitude of the oscillations. How long does it take for the particle reach its first origin?

2. Relevant equations

General equation: x(t)=A*cos($$\Omega$$*t) + B*sin($$\Omega$$*t)
x(t)=C*cos($$\Omega$$*t-$$\gamma$$)
x''+$$\Omega$$2*x=0
3. The attempt at a solution
x(t)=A*cos($$\Omega$$*t) + B*sin($$\Omega$$*t)
x'(t)= A$$\Omega$$*sin($$\Omega$$*t) + B*$$\Omega$$*cos($$\Omega$$*t)

x''(t)=A$$\Omega$$2*cos($$\Omega$$*t)-B*$$\Omega$$2*cos($$\Omega$$*t)

t=0 at x=$$\sqrt{3}[/tex} , therefore , A=[tex]\sqrt{3}$$
$$\Omega$$=2

x' = $$\sqrt{3}$$*2*sin(2t)-2*B*cos(2t)
x''=$$\sqrt{3}$$*4*cos(2t)+4*B*sin(2t)

x''= -4$$\sqrt{3}$$

$$\sqrt{3}$$*4*cos(2t)+4*B*sin(2t)=-4$$\sqrt{3}$$

dividing 4 out on both sides of the equation I am left with:

$$\sqrt{3}$$*cos(2t)+B*sin(2t)=-$$\sqrt{3}$$

Not sure how to obtain B . Wait a minute, doesn't t=0 at |x'| so B= -1 since particle is projected towards origin. In addition , finding the period seems so trivial to me: tau=2*pi/omega and omega = 2 , therefore the period is tau= pi right?

Now isn't this the equation for amplitude: C=sqrt(B^2+A^2)

My book says the period is pi/6 , rather than pi. I know from the principal root and Euler Formula's I know that 2(cos(pi/6)-i*sin(pi/6))= sqrt(3)-i. not sure what that has to do with the time since the equation for period is : tau= 2*pi/omega