Simple Harmonic Oscillator: From Hooke's Law to Harmonious Motion

[Bill Foster]

Homework Statement

Show simple harmonic motion starting from Hooke's Law.

The Attempt at a Solution

F=-kx

=m\frac{d^2x}{dt^2}=-kx

\frac{1}{x}\frac{d^2x}{dt^2}=-\frac{k}{m}

=\frac{1}{x}\frac{d}{dt}\frac{dx}{dt}=-\frac{k}{m}

\int\int\frac{1}{x}d\left(dx\right)=-\frac{k}{m}\int dt\int dt

\int\frac{1}{x}dx=-\frac{k}{2m}t^2

\ln\left(x\right)=-\frac{k}{2m}t^2

x\left(t\right)=e^{-\frac{k}{2m}t^2}

But it should be:

x\left(t\right)=e^{-i\sqrt{\frac{k}{2m}}t}

I thought I knew how to do this.

 

[rock.freak667]

Well normally you could just put F=ma and then put it the form a= - \omega^2 x and that would be sufficient to show SHM.


Else to solve the DE, you would need to know that the equation y''+ky=0 has solutions y₁=cos(kt) and y₂=sin(kt)

 

[Bill Foster]

I want to solve the differential equation. But as you can see, somewhere I need to take the square root of the exponential argument.

Where, in my steps, did I miss that?

 

[Bill Foster]

Nobody knows?

 

[dotman]

I don't see how to use this to get to the right answer, but:

\int\frac{1}{x}dx \neq \ln (x)

\int\frac{1}{x}dx = \ln |x| + C

So the derivation would continue:

\ln |x| + C =-\frac{k}{2m}t^2

|x| = e^{-\frac{k}{2m}t^2 + C}

Like I said, not sure how to use this, or if it helps. I could see squaring x, then taking the square root, in lieu of the absolute value signs, but, then, I don't see where the i comes out. Most of the derivations I've seen 'guess' at the solution of the 2nd order DE to be of the form a cos(\omega t), and go from there.

I'm going to look at this more tomorrow, I can't believe I don't know either.

 

[rock.freak667]

I'm going to look at this more tomorrow, I can't believe I don't know either.

m\frac{d^2x}{dt^2}=-kx \Rightarrow \frac{d^2x}{dt^2}+\frac{k}{m}x=0


the auxiliary equation is r²+(k/m)=0 so r=±√(k/m)i

when you have roots in the form r=λ±μi what is the general solution x(t) equal to?