# Homework Help: Harmonic wave equation

1. Jun 28, 2015

### SuperSpasm

1. The problem statement, all variables and given/known data
I'm using a couple different textbooks on waves, and it seems they've possibly contradicted one another.
I think the problem may be that one deduced the equation for displacement in a transverse harmonic wave
y(x,t) through the equation for displacement at t=0 [y(x,0)] and the other for displacement at x=0 [y(0,t)]
they both use the example of a rope shaken in set time intervals to make a harmonic wave, with an x axis set at the position of the rope "at rest", and seem to agree when the wave is going in the negative x direction.
I'm trying to cram this for a test tomorrow and don't really fully understand it, so I'll try to put whatever equations they used to derive this in so maybe you can understand the difference, sorry if this gets long or redundant.

2. Relevant equations
1st textbook (this one I was able to follow):

$t=0:$
$y = Asin(\frac{2\pi x}{\lambda} + \phi)$

$t\neq 0:$
$y = Asin(\frac{2\pi (x \pm vt)}{\lambda} + \phi)$
$y = Asin(\frac{2\pi x}{\lambda} \pm \frac{2\pi vt}{\lambda} +\phi)$

$v = \lambda f,\ \frac{v}{\lambda} = f$
$\omega = 2\pi f,\ k = \frac{2\pi}{\lambda}$

finally:
(1) $y = Asin(kx\pm \omega t + \phi)$

2nd textbook (in the beginning they refer only to a wave moving in the positive x direction):

$x=0$
$y(x,t) = y_{0} (t-\frac{x}{v})$ <- this is where I lost them.

then they reference a previous chapter, siting that harmonic motion is given by this:
$y_0(t) = Acos(\omega t + \phi)$
$y(x,t) = Asin[\omega (t - \frac{x}{v})]$
(2) $y(x,t) = Asin(\omega t - kx)$

and for waves moving in the negative x direction:
$y(x,t) = Asin(\omega t + kx)$

the discrepency here is that (1) = -(2) [since $\omega t - kx = -(kx - \omega t)$ and $sin(-a) = -sin(a)$]

as I write this I'm starting to think it maybe has to do with the second book omitting the phase constant $\phi$? Nonetheless I'm kinda lost with the second book's reasoning. Can anyone explain this to me?

2. Jun 28, 2015

### PeroK

The second equation clearly assumes that $y(0,0) = 0$; whereas, the first equation allows $y(0,0)$ to be non-zero. The two equations are, however, equivalent apart from this. What you're missing is that $A$ is an arbitrary constant. So, for a particular solution, you would have $A$ in one case and $-A$ in the other.

3. Jul 6, 2015

### SuperSpasm

I think I get it now, thanks!