1. Nov 23, 2007

### Rubidium

1. On a real string, some of the energy of a wave dissipates as the wave travels down the string. Such a situation can be described by a wave function whose amplitude A(x) depends on x: y=A(x)sin(kx-$$\omega$$t), where A(x)=A$$_{o}$$e$$^{-bx}$$. What is the power transported by the wave as a function of x, where x>0?

2. P=$$\mu$$v$$\omega$$$$^{2}$$cos$$^{2}$$(kx-$$\omega$$t)
$$\mu$$=$$\frac{m}{L}$$
v=$$\sqrt{\frac{F_{T}}{\mu}}$$
$$\omega$$=2$$\pi$$f
v=f$$\lambda$$-->f=$$\frac{v}{\lambda}$$

3. I tried two approaches, neither of which I am sure if it is the right way to solve the problem...not sure what the question is really asking for.
P(x)=$$\mu$$v$$\omega$$$$^{2}$$A$$^{2}$$cos$$^{2}$$(kx-$$\omega$$t)
so, by substituting to "simplify": P(x)=$$\mu$$$$\frac{v}{\lambda}$$$$\lambda$$(2$$\pi$$$$\frac{v}{\lambda}$$)$$^{2}$$(A(x))$$^{2}$$cos$$^{2}$$(kx-$$\omega$$t)
P(x)=$$\mu$$4$$\pi$$$$^{2}$$v(A(x))$$^{2}$$cos$$^{2}$$(kx-$$\omega$$t)
P(x)=$$\mu$$$$^{2}$$(4$$\pi$$)$$^{2}$$$$\frac{F_{T}}{\mu}$$(A(x))$$^{4}$$cos$$^{4}$$(kx-$$\omega$$t)
P(x)=16$$\mu$$$$\pi$$$$^{4}$$F$$_{T}$$A$$^{2}_{o}$$e$$^{-4bx}$$cos$$^{4}$$[/tex](kx-$$\omega$$t)
and I'm not sure where to go from there...if it's simplified enough for the answer or if it's the completely wrong approach.

Or:
y=A$$_{o}$$e$$^{-bx}$$sin(kx-$$\omega$$t)
v_{y}=\frac{d}{dt}\left[A$$_{o}$$e$$^{-bx}$$sin(kx-$$\omega$$t)\right]=-\omegaA$$_{o}$$e$$^{-bx}$$cos(kx-$$\omega$$t)
P=F_{Ty}v_{y}\approxF_{T}v_{y}tan\theta=-F_{T}\frac{\gamma}{dt}\frac{\gamma}{dx}
P=-F_{T}\left[-\omegaA$$_{o}$$e$$^{-bx}$$cos(kx-$$\omega$$t)\right]\left[kA$$_{o}$$e$$^{-bx}$$cos(kx-$$\omega$$t)\right]=F_{T}\omegakA$$_{o}$$^{2}e$$^{-bx}$$cos^{2}(kx-$$\omega$$t)