# Homework Help: Isn't there anybody that can help?

1. Nov 27, 2007

### Rubidium

1. The problem statement, all variables and given/known data
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)
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

### Astronuc

Staff Emeritus
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.