How Do You Solve the Damped Oscillator Equation Using Newton's Second Law?

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Using Newton's 2nd Law for a damped oscillator:

[tex]ma = -kx - \alpha x[/tex]

which is a second order linear DE. To solve it we use the trial integrating factor [itex]e^{\lambda x} [/tex] to come to the root equation<br /> <br /> [tex]mx^2 + \alpha x + k = 0[/tex] where we can find our two solutions to be <br /> <br /> [tex]r_{1} \ and \ r_{2} \frac{-\alpha}{2m} +/- \frac{1}{2m} \sqrt{\alpha^2 - 4km}[/tex]<br /> <br /> And the function [itex]x(t)[/itex] is supposed to be represented by<br /> [tex]x(t) = Ae^{r_1 t } + Be^{r_2 t}[/tex] which can be solved with th given initial conditions [itex]x(0) = 0 \ and \ v(0) = v_0[/itex].<br /> <br /> I'm supposed to end up with the product of two exponentials which dissolve to a cosine and sin function. <br /> Four lines up is where I'm lost.. I'm not sure how to jump to the two exponential functions. What I have is:<br /> <br /> [tex]x(t) = Ae^{r_1 t}(\cos(x(\alpha^2 - 4km)) +/- i \sin(x(\alpha^2 - 4km))[/tex] which just smells really wrong. I can solve the particular equations once I can get to the general solution, but I'm stuck there. I know how the damping is determined and everything.. I just can't get the general equation :([/itex]
 
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The differential equation modelling this system is given by:

[tex]m\Ddot{x} + 2\beta \dot{x} + \omega^2 x = 0[/tex]

With [itex]\beta \ and \ \omega[/itex] defined as

[tex]\beta = \frac{b}{2m} \ and \ \omega = \sqrt{\frac{k}{m}}[/tex]

with m being mass, k the characteristic constant of the physical system, and b the damping coefficient. The equation is solved via the auxilliary equation which simplifies to

[tex]r_{1,2} = -\beta \pm \sqrt{\beta^2 - \omega^2}[/tex]

The damping coefficient is less than the frequency of the system in underdamped harmonic motion, thus [itex]\beta < \omega[/itex] and imaginary numbers appear. We introduce another simplification, [itex]\omega_1[/itex] defined by

[tex]\omega_1 = \sqrt{\omega^2 - \beta^2}[/tex]

Thus [tex]r_{1,2} = -\beta \pm \imath \omega_1[/tex] and [itex]x(t)[/itex] can be expressed by the exponentials:

[tex]x(t) = \exp{\left((-\beta + \imath \omega_1)t}\right) + \exp{\left((-\beta - \imath \omega_1)t}\right) }[/tex] and by Euler's formula we simplify to the general solution to underdamped harmonic motion:

Equation A
[tex]x(t) = e^{-\beta t}\left(A\cos(\omega_1 t) + B\sin(\omega_1 t) )[/tex]
 
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[tex]\frac{T}{2} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_n a_m\cos(\omega _n t) \cos(\omega_m t) {\left(\frac{2\pi}{L} x\right)}^2 n m \int_0^L \cos \left(\frac{2\pi n}{L} x\right) \cos\left(\frac{2\pi m}{L} x\right) dx[/tex]
 
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This is not really introductory physics, but anywho, have you taken differential equations yet? This is a classic example probably found in your book if you have one on O.D.E. That post number 3 looks scary as hell to me, what is that?
 
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what is that?

iii) (challenging) now consider a wave made from a superposition of modes:

[tex]\sum_{n=1}^{\infty} a_n\cos(\omega_n t)\sin(\frac{2\pi n}{L} x)[/tex]

calculate the total energy of this wave and comment on your answer.
I get the KE and PE each to be 0 by the expression in post 3. We have arguments for it being 0 and arguments against it being 0.
 
[tex]\alpha ^{x+n} 2 dy + \left( \alpha ^3 x^2 - 3 \alpha ^n y) dx = 0[/tex]
 
[tex]f = \frac{1}{2L}\sqrt{\frac{mg}{\mu}}[/tex]
 
Prove

[tex]\left(\frac{ia - 1}{ia+1}\right)^{ib} = exp(-2b cot^{-1} a)[/tex] and

[tex]\sum_{n=1}^{N} \cos(2n-1) x = \frac{\sin 2n x}{2\sin x}}[/tex]
 
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Whozum, are those questions with which you need help?

If so, for the second one, you could multiply numerator and denominator by 2sin(x) and write 2SinACosB as Sin(A+B) + Sin(A-B).
 
siddharth said:
Whozum, are those questions with which you need help?

If so, for the second one, you could multiply numerator and denominator by 2sin(x) and write 2SinACosB as Sin(A+B) + Sin(A-B).

No, I'm just using this as a tex notepad since no one was helping when I needed it. I'm good with all these.

[tex]v = \lambda f[/tex]

[tex]\lambda = 4L[/tex]

[tex]v = 4L f[/tex]

[tex]f = \frac{v}{4L}[/tex]