Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I'm studying vibrations; so I'm using Beer-Johnston-Cornwell Dynamics book. I am worry about the equation for Underdamped Vibration, which in the book it is: $$x_{(t)}=x_0e^{-\lambda t}\sin(\omega_d+\phi)$$; where $$\omega_d=\sqrt{\omega_n^2+c^2/4m^2}$$.

I think that $x_0$ would be replaced by the result vector of constants $c_1$ and $c_2$ affected by the factor $e^{-\lambda t}$. It means, an $x_m$ or an amplitud, but not the initial position, because it could be 0, with an inicial velocity.

Also the graphic, it depicts the boundary equation with $x_0$. I attached a picture.

Can you help me and comment? How I should interpret $x_0$? Maybe I misunderstood this topic.

damp

share|improve this question

1 Answer 1

up vote 1 down vote accepted

$x_0$ is proportional to $\sqrt{E_{initial}}$ square root of initial energy in the system.

For example suppose we have dumped motion of a mass on a spring and $x$ is distance from equilibrium position. Initial energy of the system can be fully in the tension of the spring (zero initial velocity).

Initial energy is $E_{initial} = \frac12k\,x_{initial}^2 = \frac12k\,x_0^2$

Generaly for system in the example: $x_0=\sqrt{2\frac{E_{initial}}{k}}$

We can change distribution of initial energy between potential and kinetic energy (changing $\phi$) but if sum of initial energy is the same then $x_0$ is also the same.

share|improve this answer
    
I agree with you. But, it's clear that $x_0$ is not the position in $t=0$, isn't? –  Isai Oct 27 '13 at 21:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.