## Monday, April 09, 2012

### Math Bait II: The Spring-Mass System with Laplace Transforms

In my last post, we solved the differential equation

with traditional methods.  Today, we will examine the same problem using Laplace Transforms.

Let’s briefly review the definition of the Laplace Transform and a few of its properties:

$F(s) = L\left \{x(t)\right \} = \int_{-\infty}^{\infty}x(t)e^{-st}dt \\ L\{tx$$t$$\}=-\frac{d}{ds}X(s) \\ L\left \{mx'' + kx\right \} = mL\left \{x''\right \} + kL\left \{x\right \} \\ L\{x''\} = s^2X(s) - sx(0) - x'(0) \\ L\{A\cos\omega t\} = A\frac{s}{s^2+\omega^2} \\ L\{A\sin\omega t\} = A\frac{\omega}{s^2+\omega^2}$

Returning to our equation, we can apply the properties to rewrite it in the s domain:

and solve for X(s):

In order to find the inverse transform of X(s) and solve the problem, we must decompose the denominator of the first term using partial fraction expansion:

Much as we matched the coefficients for the sine and consine terms in the method of undetermined coefficients in the previous problem, we here must match the coefficients for the powers of s, yielding a system of equations and their solution:

Armed with these coefficients, we can now rewrite X(s) in terms of our expanded fractions:

In order to calculate the inverse Laplace transform, it would be easier if our terms matched those of the properties provided above. Rewriting:

It is now a trivial matter to find the final solution:

Resonant Frequency

Finding the result for the resonant frequency, while algebraically intense, is relatively straightforward with Laplace transforms. We begin by replacing ω with √(k/m) to obtain:

$\left.X(s)\right|_{\omega = \sqrt\frac{k}{m}} = \frac{As}{\left(s^2 + \frac{k}{m}\right)(ms^2 + k)} + \frac{x_0ms}{ms^2 + k}$

Before proceeding, and since we are expecting the same answer we obtained using traditional methods, let's apply the multiply-by-t property above to the sine function:

Now let's rewrite X(s) above so that it's terms match the formula:

Here again, it is a trivial matter to find the solution:

$x(t) = L^{-1}\{X(s)\} \\= \frac{A}{2\sqrt{km}}t\sin\left(\sqrt{\frac{k}{m}}t\right) + x_0\cos\left(\sqrt\frac{k}{m}t\right)$

#### 1 comment:

rajput said...

I am here to discuss a simple definition of Laplace,Laplace transform is a integral transform with many applications in mathematics. Termed as , it is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms it to a function F(s) with a complex argument s.
calculus law of exponential change