(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A certain oscillator satisfies the equation

x''+4x=0

x=[tex]\sqrt{3}[/tex]

|x'|= 2

show that , in subsequent motion,

x=[tex]\sqrt{3}[/tex] 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([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)

x(t)=C*cos([tex]\Omega[/tex]*t-[tex]\gamma[/tex])

x''+[tex]\Omega[/tex]^{2}*x=0

3. The attempt at a solution

x(t)=A*cos([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)

x'(t)= A[tex]\Omega[/tex]*sin([tex]\Omega[/tex]*t) + B*[tex]\Omega[/tex]*cos([tex]\Omega[/tex]*t)

x''(t)=A[tex]\Omega[/tex]^{2}*cos([tex]\Omega[/tex]*t)-B*[tex]\Omega[/tex]^{2}*cos([tex]\Omega[/tex]*t)

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

[tex]\Omega[/tex]=2

x' = [tex]\sqrt{3}[/tex]*2*sin(2t)-2*B*cos(2t)

x''=[tex]\sqrt{3}[/tex]*4*cos(2t)+4*B*sin(2t)

x''= -4[tex]\sqrt{3}[/tex]

[tex]\sqrt{3}[/tex]*4*cos(2t)+4*B*sin(2t)=-4[tex]\sqrt{3}[/tex]

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

[tex]\sqrt{3}[/tex]*cos(2t)+B*sin(2t)=-[tex]\sqrt{3}[/tex]

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

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# Homework Help: Simple harmonic motion problem

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