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Wave function interpretation $y(x,t) = (0.35m)\sin(10\pi t-3\pi x + \frac\pi{4})$ I used to deal with function with one variable And now theres are two, how can I interpret them?
Is $10\pi$ still equal to $\omega$ ? how about $-3\pi x$ and $ \frac\pi{4}$?

I just want to know the form of the equation?

Just like $y = mx$ I know $m$ is the slope.

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This person is clearly a novice. It doesn't matter now since he has accepted an answer but he needed a very, very simple explanation. Compared to what he got. –  Bogo May 27 '13 at 22:14

2 Answers 2

up vote 0 down vote accepted

The general solution to the classical wave equation is

$$y(x,t)=f(\mathbf{k}\cdot\mathbf{x}-wt)$$ where $f$ is an arbitrary function. In this case, $f(x)=(0.35m)\sin(-x+\frac{pi}{4})$.

Basically, whatever multiplies $t$ is $-\omega$, and whatever multiplies (or is dotted into) $\mathbf{x}$ is $\mathbf{k}$. Anything else is part of the arbitrary function.

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For more information on solving partial differential equations this way, check out the method of characteristics –  Dan May 22 '13 at 2:25
    
There is one problem with that metod, at least in Wikipedia, it only applies to first order PDE's and the wave equation is second order. –  Angel Joaniquet Tukiainen May 27 '13 at 21:09
    
@Angel Joaniquet Tukiainen: Wikipedia's wrong on this one. Hyperbolic differential equations always have two characteristics. See e.g. Riley, Hobson and Bence. –  Dan May 28 '13 at 9:32
    
Thanks, I will. –  Angel Joaniquet Tukiainen May 28 '13 at 19:46

The wave equations are second order equations on time and space:

$$ \nabla^2u(\mathbf{x},t) = \frac{1}{v^2}\frac{\partial^2 u(\mathbf{x},t)}{\partial t^2} $$

Where $v$ is the speed of the wave. If we plug in Dan's functions we have:

$$ \nabla^2 y(\mathbf{x},t) = \nabla^2 f(\mathbf{k}\cdot\mathbf{x}-wt) = (k_x^2 +k_y^2 +k_z^2 ) f (\mathbf{k}\cdot\mathbf{x}-wt) $$

and from the other side:

$$ \frac{\partial^2 y(\mathbf{x},t)}{\partial t^2} = \frac{\partial^2 f(\mathbf{k}\cdot\mathbf{x}-wt)}{\partial t^2} = w^2f(\mathbf{k}\cdot\mathbf{x}-wt) $$

With $v^2 = \frac{w^2}{k_x^2 +k_y^2 +k_z^2}$ the equation is satisfied. So you already can compute the velocity of your wave, for example.

As for your graphing intuition, you can think of $w$ of how much it oscillates in time, and $\mathbf{k}$ as how much it oscillates in space.

And your $\pi/4$ is a phase factor related to the initial conditions (or boundary conditions), and it can be seen as a shift in the oscillations.

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