# Deriving SHM equations

1. Jan 24, 2006

### whozum

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

$$ma = -kx - \alpha x$$

which is a second order linear DE. To solve it we use the trial integrating factor $e^{\lambda x} [/tex] to come to the root equation $$mx^2 + \alpha x + k = 0$$ where we can find our two solutions to be $$r_{1} \ and \ r_{2} \frac{-\alpha}{2m} +/- \frac{1}{2m} \sqrt{\alpha^2 - 4km}$$ And the function [itex] x(t)$ is supposed to be represented by
$$x(t) = Ae^{r_1 t } + Be^{r_2 t}$$ which can be solved with th given initial conditions $x(0) = 0 \ and \ v(0) = v_0$.

I'm supposed to end up with the product of two exponentials which dissolve to a cosine and sin function.
Four lines up is where I'm lost.. I'm not sure how to jump to the two exponential functions. What I have is:

$$x(t) = Ae^{r_1 t}(\cos(x(\alpha^2 - 4km)) +/- i \sin(x(\alpha^2 - 4km))$$ 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 cant get the general equation :(

Last edited: Jan 24, 2006
2. Jan 27, 2006

### whozum

The differential equation modelling this system is given by:

$$m\Ddot{x} + 2\beta \dot{x} + \omega^2 x = 0$$

With $\beta \ and \ \omega$ defined as

$$\beta = \frac{b}{2m} \ and \ \omega = \sqrt{\frac{k}{m}}$$

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

$$r_{1,2} = -\beta \pm \sqrt{\beta^2 - \omega^2}$$

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

$$\omega_1 = \sqrt{\omega^2 - \beta^2}$$

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

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

Equation A
$$x(t) = e^{-\beta t}\left(A\cos(\omega_1 t) + B\sin(\omega_1 t) )$$

Last edited: Jan 27, 2006
3. Jan 31, 2006

### whozum

$$\frac{T}{2} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a_n a_m\cos(\omega _n t) \cos(\omega_m t) {\left(\frac{2\pi}{L} x\right)}^2 n m \int_0^L \cos \left(\frac{2\pi n}{L} x\right) \cos\left(\frac{2\pi m}{L} x\right) dx$$

Last edited: Jan 31, 2006
4. Jan 31, 2006

### Cyrus

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?

Last edited: Jan 31, 2006
5. Jan 31, 2006

### whozum

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

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

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.

6. Feb 2, 2006

### whozum

$$\alpha ^{x+n} 2 dy + \left( \alpha ^3 x^2 - 3 \alpha ^n y) dx = 0$$

7. Feb 3, 2006

### whozum

$$f = \frac{1}{2L}\sqrt{\frac{mg}{\mu}}$$

8. Feb 7, 2006

### whozum

Prove

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

$$\sum_{n=1}^{N} \cos(2n-1) x = \frac{\sin 2n x}{2\sin x}}$$

Last edited: Feb 7, 2006
9. Feb 8, 2006

### siddharth

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).

10. Feb 9, 2006

### whozum

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.

$$v = \lambda f$$

$$\lambda = 4L$$

$$v = 4L f$$

$$f = \frac{v}{4L}$$