Critically damped oscillator: Classical mechanics help

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SUMMARY

The discussion focuses on determining the maximum initial speed of a critically damped oscillator that starts at a position \( x_0 > 0 \) without crossing the origin. The relevant equations include the position function \( x(t) = e^{-\gamma t}(A + Bt) \) and the velocity function \( v(t) = e^{-\gamma t}(B - Bt\gamma - A\gamma) \), where \( \gamma = \omega = \sqrt{k/m} \). The key conclusion is that to find the maximum initial speed \( v_0 \), one must solve for \( x(t) = 0 \) and derive the velocity function while incorporating initial conditions.

PREREQUISITES
  • Understanding of critically damped oscillators in classical mechanics
  • Familiarity with differential equations and their applications
  • Knowledge of the natural frequency \( \omega \) and damping coefficient \( \gamma \)
  • Proficiency in using LaTeX for mathematical expressions
NEXT STEPS
  • Study the derivation of the critically damped oscillator equations
  • Learn how to apply initial conditions in differential equations
  • Explore the concept of limits in the context of velocity functions
  • Investigate the relationship between damping ratios and system behavior
USEFUL FOR

This discussion is beneficial for students and educators in physics, particularly those studying classical mechanics, as well as engineers and researchers working with oscillatory systems and damping effects.

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Homework Statement



A critically damped oscillator with natural frequency [tex]\omega[/tex] starts out at position [tex]x_0>0[/tex]. What is the maximum initial speed (directed towards the origin) it can have and not cross the origin?

Homework Equations


For the case of critical damping,
[tex]x(t)=e^{(-\gammat)}(A+Bt) where \gamma=\omega=\sqrt{k/m}[/tex]


The Attempt at a Solution


Well first I derived the above equation (and verified it with my textbook). I then evaluated the initial position:
[tex]x(0)=A[/tex].
I then took the derivative of the position function to get velocity:
[tex]v(t)=(e^{-\gamma t}) ( B-Bt\gamma -A\gamma)[/tex]
Setting t=0 I obtained
[tex]v(0)=B-A\gamma[/tex] recalling that x0=A, [tex]v(0)= B-x_0\gamma[/tex]. I then solved for Beta: [tex]B=v_0 +\gamma x_0.[/tex]
I then tried substituting this back into the position equation, and solving for the initial velocity, the program is there is always time dependency that i can't get rid of... what am i doing wrong? any advice would be much appreciated!
Thanks
 
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umm its supposed to be e-tgamma(B-Bt(gamma)-A(gamma)) for the velocity and v(0)=B-A(gamma)=B-x_0(gamma) for the initial velocity
and x(t)=e-tgamma(A+Bt) for the position but for some reason I all the sudden fail at latex
 
I think I solved it... you're on the right track. Solve for x(t) in terms of gamma, w, Xo and Vo, then make x(t)=0 and solve for Vo. Next, derive V(t) by taking dx/dt (remember to include initial conditions!). Take the limit as t goes to inf of V(t), and substitute terms so that you're left with w and Xo, and you should get Vmax.
 

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