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

In summary, the conversation discusses solving a differential equation to show simple harmonic motion starting from Hooke's Law. The steps involve setting up the equation, solving it using integration, and finding the general solution using the auxiliary equation. The solution is found to be x(t)=e^{-i\sqrt{\frac{k}{2m}}t}.
  • #1
Bill Foster
338
0

Homework Statement



Show simple harmonic motion starting from Hooke's Law.


The Attempt at a Solution



[tex]F=-kx[/tex]

[tex]=m\frac{d^2x}{dt^2}=-kx[/tex]

[tex]\frac{1}{x}\frac{d^2x}{dt^2}=-\frac{k}{m}[/tex]

[tex]=\frac{1}{x}\frac{d}{dt}\frac{dx}{dt}=-\frac{k}{m}[/tex]

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

[tex]\int\frac{1}{x}dx=-\frac{k}{2m}t^2[/tex]

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

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

But it should be:

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

I thought I knew how to do this. :redface:
 
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  • #2
Well normally you could just put F=ma and then put it the form [itex]a= - \omega^2 x[/itex] 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
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
Nobody knows?
 
  • #5
I don't see how to use this to get to the right answer, but:

[tex]\int\frac{1}{x}dx \neq \ln (x)[/tex]

[tex]\int\frac{1}{x}dx = \ln |x| + C[/tex]

So the derivation would continue:

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

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

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 [itex]a cos(\omega t)[/itex], and go from there.

I'm going to look at this more tomorrow, I can't believe I don't know either.
 
  • #6
dotman said:
I'm going to look at this more tomorrow, I can't believe I don't know either.


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


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?
 

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

What is a simple harmonic oscillator?

A simple harmonic oscillator is a system that exhibits periodic motion around an equilibrium point. It is characterized by a restoring force that is proportional to the displacement from the equilibrium point and a constant frequency of oscillation.

What are some real-life examples of a simple harmonic oscillator?

Some common examples of simple harmonic oscillators include a mass on a spring, a pendulum, and a guitar string.

How is the period of a simple harmonic oscillator determined?

The period of a simple harmonic oscillator is determined by the mass of the object, the strength of the restoring force, and the initial conditions of the system.

What is the difference between simple harmonic motion and simple harmonic oscillator?

Simple harmonic motion refers to the repetitive back and forth motion of an object, while a simple harmonic oscillator is a physical system that exhibits this type of motion. Simple harmonic motion can occur in other systems, such as a damped oscillator, while a simple harmonic oscillator is a specific type of system.

How does amplitude affect a simple harmonic oscillator?

The amplitude, or maximum displacement, of a simple harmonic oscillator affects the energy and velocity of the system. A larger amplitude will result in a higher energy and faster oscillations, while a smaller amplitude will result in lower energy and slower oscillations.

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