# Isn't there anybody that can help?

1. Nov 27, 2007

### Rubidium

1. The problem statement, all variables and given/known data
Power transported by a wave problem...Please Help!!!
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-t), where A(x)=Ae. What is the power transported by the wave as a function of x, where x>0?

2. P=vcos(kx-t)
=
v=
=2f
v=f-->f=

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)=vAcos(kx-t)
so, by substituting to "simplify": P(x)=(2)(A(x))cos(kx-t)
P(x)=4v(A(x))cos(kx-t)
P(x)=(4)(A(x))cos(kx-t)
P(x)=16FAecos[/tex](kx-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=Aesin(kx-t)
v_{y}=\frac{d}{dt}\left[Aesin(kx-t)\right]=-\omegaAecos(kx-t)
P=F_{Ty}v_{y}\approxF_{T}v_{y}tan\theta=-F_{T}\frac{\gamma}{dt}\frac{\gamma}{dx}
P=-F_{T}\left[-\omegaAecos(kx-t)\right]\left[kAecos(kx-t)\right]=F_{T}\omegakA^{2}ecos^{2}(kx-t)
So those are my two answers worked out completely showing all steps. Please help and thank you.
Also note that the greek letters look like superscripts but they aren't supposed to be.

2. Relevant equations

3. The attempt at a solution

2. Nov 27, 2007

### Staff: Mentor

Is this supposed to be a decaying exponential? If not, then the maximum amplitude is constant as a function of position.

3. Nov 28, 2007

### Shooting Star

I am assuming that A(x) is of the form A*exp(-x/b), because you have mentioned dissipation of energy. What I want to know is whether you know how to find the power of a wave when there is no dissipation? What has been taught in the class so far? Don't use formula blindly.

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