# Laplace transform of Diff. Eq containing sine Function

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In summary, the homework statement is that the laplace transform of the pendulum equation is s^{2} + \alpha s + g \frac{\omega}{s^{2}+\omega^{2}} = 0. I think the problem is that the \sigma term is dropping out, and I'm not sure what to do about it.

## Homework Statement

Find the Laplace transform of the pendulum equation.

## Homework Equations

The pendulum equation:
$$\frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin(\theta) = 0$$

$$s = \sigma + i \omega$$

## The Attempt at a Solution

Taking the laplace transform I get

$$s^{2} + \alpha s + g \frac{\omega}{s^{2}+\omega^{2}} = 0$$

I usually drop the $$\sigma$$ term in $$s = \sigma + i \omega$$, which leaves

$$s^{2} + \alpha s + g \frac{\omega}{-\omega^{2}+\omega^{2}} = 0$$

The $$\frac{\omega}{-\omega^{2}+\omega^{2}}$$ is problematic, but I'm not sure what to do. I don't see the usefulness of the $$\sigma$$ variable, and I'm not sure that I'm doing this correctly to begin with

You're not doing the transform correctly. For laplace you're just taking the real part of z in $$exp^{-zt}, z = \sigma + i \omega$$.

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I'm not sure what you're saying but I had a thought. I think the Laplace of the sine function is
$$\mathcal{L} \left\{\sin(t)\right\} = \frac{1}{s^{2}+1}$$

which would make the transform

$$s^{2} + \alpha s + g \frac{1}{s^{2}+1} = 0$$

This is more sensible, but I still need to verify it some how.

What do you get if you take the Laplace transform of just a second deriavative?

I think it is something like

$$\mathcal{L} \left\{y(t)\right\} = s^{2} Y(s)$$

presuming that
$$\dot y(0) = 0$$
and
$$\ddot y(0) = 0$$

I think the sine term would make this a nonlinear differential equation, and I'm now I'm thinking that maybe the Laplace transform doesn't work on nonlinear differential equations.

Sure it does!

Why do you keep dropping the Y(s) terms? That's what you need to take the inverse laplace transform of! Solve for Y(s) and take the inverse transform.

I'm trying to get a transfer function of:

$$\frac{d^{2}\theta}{dt^{2}} + \alpha \frac{d \theta}{dt} + g \sin( \theta ) = f(t)$$

With the Y(s) terms I get:

$$s^{2} Y(s) + \alpha s Y(s) + g \frac{1}{s^{2}+1}} = F(s)$$

For the transfer function, I want to Y(s) and F(s) on the same side like this:

$$\frac{Y(s)}{F(s)}$$

but I don't see and way to factor things out that way.

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