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## Homework Statement

Derive the energy of the standing wave.

## Homework Equations

Standing wave equation: $$y(x,t)=2Asin(kx)cos(ωt)$$

## The Attempt at a Solution

I am trying to derive the energy of the standing wave. But I am kind of stuck at the approach.

Standing wave has no translational kinetic energy. But there's some energy of the amplitude which it moves between two nodes and makes some kind of an oscillation motion.

I tried many approaches but I couldn't figure it out exactly.

Lets start with the equation of the standing wave.

$$y(x,t) = 2Asin(kx)cos(ωt)$$ or equally $$y(x,t) = A(x)cos(ωt)$$

where $$A(x)=2Asin(kx)$$

Here I tried 2 approaches. First we can try to think the wave as a string. And there's force action on the string and then we can use the

$$W = \int {\vec{F}d\vec{x}}$$

Second approach was to think little pieces from the string and call them ##dm##

and $$dm=qdx$$ where ##q## is the density of the string and ##dx## is the small length on the string. Then try to calculate the energy of the every ##dm## and take an integral.

Lets try the first approach,

Take the time derivative of the ##y(x,t)##.

$$ v = \frac {\partial y(x,t)}{\partial t}=-A(x)ωsin(ωt)$$

$$ a = \frac {\partial^2 y(x,t)} {\partial t^2}=-A(x)ω^2cos(ωt)$$

Now we have

$$W = \int\int{dma\vec{j}dx\vec{j}}$$

$$W = \int\int {dm-A(x)ω^2cos(ωt)\vec {j}dx\vec{j}}$$

and it becomes

$$W = \int\int {-A(x)ω^2cos(ωt)dmdx}$$

$$W = 2Aω^2cos(ωt)\int\int {-sin(kx)dmdx}$$

Its wrong I guess. But I am not sure where.

For the second approach I simply have

$$KE = \int {\frac {1} {2} dmv^2}$$

$$KE = \int {\frac {1} {2} dm(A(x))^2ω^2sin^2(ωt)}$$

$$KE = 2A^2qω^2sin^2(ωt)\int { sin^2(kx)dx}$$

Here I am confused about the boundries of the integral I am thinking it can be ##0## to ##2\pi## or ##0## to ##\lambda##.

And I don't think I need potential energy for the calculations. Or I can try to use a point ##x## where the potential energy is zero and total energy is equal to kinetic energy.