# Simple harmonic oscillator

1. Nov 6, 2009

### Bill Foster

1. The problem statement, all variables and given/known data

Show simple harmonic motion starting from Hooke's Law.

3. 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.

2. Nov 6, 2009

### 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)

3. Nov 7, 2009

### 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?

4. Nov 7, 2009

### Bill Foster

Nobody knows?

5. Nov 7, 2009

### 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.

6. Nov 7, 2009

### rock.freak667

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

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?