Amplitude of harmonic oscillator

In summary, the conversation discusses finding the general solution and amplitude of the steady state solution for a simple harmonic oscillator with a sinusoidal driving force. The solution is of the form (sin(wt)*(1-w^2)-cos(wt)*w)/f(w), where f(w) is equal to (\omega^2-\omega_0^2)^2+(\lambda\omega)^2. The amplitude of the steady state solution is found to be 1/sqrt(f(w)). The conversation also includes a discussion on using trigonometric identities and Maple to simplify the equation and confirm the solution. Ultimately, the solution is derived by considering a right triangle with lengths \omega^2-\omega_0^2 and \lambda\omega
  • #1
longrob
52
0

Homework Statement


I have a simple harmonic oscillator system with the driving force a sinusoidal term. The question is to find the general solution and the amplitude of the steady state solution


Homework Equations



I found the steady state part of the solution. It is of the form:
(sin(wt)*(1-w^2)-cos(wt)*w)/f(w)



The Attempt at a Solution



I'm told that the amplitude of the system in steady state should be 1/sqrt(f(w)) but I can't derive it. Can anyone help ?

Thanks
 
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  • #2
use trig identity and reduce the equation to sin(A-B), amplitude is the factor by which the stated term is raised
 
  • #3
But the coefficients of each sinusoidal term are different...
 
  • #4
is your steady state solution correct? it gets crazy if i reduce it to some trig identity
and what is f(w) equal to?
 
  • #5
Yes I'm sure it's correct. This is homework so I'm loathed to post the whole details. However, take a look here:
http://nuweb.neu.edu/dheiman/U600/DHO.pdf
Look at the section "Driven Harmonic Oscillator" on page 4. The solution given is exactly what I want (with lambda, w_0 and F_0/m all equal to 1). I just can't derive it from my own solution. By hand I just used a trial solution of p.cos(wt) + q.sin(wt) and equated coefficients, which lead to the form I posted above. I also used Maple to check this, and it agrees.
 
  • #6
The (1-w^2) factor isn't under a square root?
 
  • #7
no, it's not...
 
  • #8
Another look at this in Maple:
[Sorry, fiddling with latex...]
 
Last edited:
  • #9
In Maple:
[tex]{\it s2s}\, := \,{\frac { \left( -{\omega_{{0}}}^{2}+{\omega}^{2} \right) \sin \left( \omega\,t \right) +\cos \left( \omega\,t \right) \omega\,\lambda}{{\omega}^{4}+ \left( {\lambda}^{2}-2\,{\omega_{{0}}}^{2} \right) {\omega}^{2}\\
\mbox{}+{\omega_{{0}}}^{4}}}\]}[/tex]

which simplifies to:

[tex]{\frac {\cos \left( \omega\,t-\arctan \left( -{\omega_{{0}}}^{2}+{\omega}^{2},\omega\,\lambda \right) \right) }{ \sqrt{{\lambda}^{2}{\omega}^{2}+{\omega_{{0}}}^{4}-2\,{\omega_{{0}}}^{2}{\omega}^{2}\\
\mbox{}+{\omega}^{4}}}}\]}[/tex]

And that's what I'm trying to achieve by hand...

Maple code is
s2s:=((-omega[0]^2+omega^2)*sin(omega*t)+cos(omega*t)*omega*lambda)/(omega^4+(lambda^2-2*omega[0]^2)*omega^2+omega[0]^4);
simplify(convert(s2s,phaseamp,t));

Thanks
 
  • #10
Oh, that helps. Note that the denominator is [itex](\omega^2-\omega_0^2)^2+(\lambda\omega)^2[/itex]. Consider a right triangle with one leg of length [itex]\omega^2-\omega_0^2[/itex] and the other of length [itex]\lambda\omega[/itex]. Does that help?
 
  • #11
Indeed, yes. So easy now. Thanks !
 

What is the amplitude of a harmonic oscillator?

The amplitude of a harmonic oscillator represents the maximum displacement from the equilibrium position of the oscillating object.

How is the amplitude of a harmonic oscillator related to its energy?

The amplitude of a harmonic oscillator is directly proportional to its energy. As the amplitude increases, so does the energy of the oscillating object.

What factors affect the amplitude of a harmonic oscillator?

The amplitude of a harmonic oscillator is affected by the initial displacement, the mass of the object, and the stiffness of the restoring force.

What is the formula for calculating the amplitude of a harmonic oscillator?

The formula for calculating the amplitude of a harmonic oscillator is A = xmax - xeq, where A is the amplitude, xmax is the maximum displacement, and xeq is the equilibrium position.

How does the amplitude of a harmonic oscillator change over time?

The amplitude of a harmonic oscillator decreases over time due to the dissipation of energy through friction and other external forces. This process is known as damping.

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