How Do You Calculate the Energy of a Standing Wave?

[Arman777]

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.

 

[Orodruin]

And I don't think I need potential energy for the calculations.

The total energy includes a potential term. It is given by $U = T(L-L_0)$, where $T$ is the tension in the string, $L$ the string length and $L_0$ the rest length of the string.

Or I can try to use a point x where the potential energy is zero and total energy is equal to kinetic energy.

You are correct that you can try to find a situation where all of the energy is in the form of kinetic energy. However, this will be a time $t$, not a position $x$.

It is very unclear to me what you are trying to do with your first approach. Your second approach for the kinetic energy is correct.

 

[Arman777]

The total energy includes a potential term. It is given by U=T(L−L0)U=T(L−L0)U = T(L-L_0), where TTT is the tension in the string, LLL the string length and L0L0L_0 the rest length of the string.

Hmm I didn't know that. I was thinking to about to add a potential but I didnt know how to express it. But this seems nice.

. Your second approach for the kinetic energy is correct.

But then my integral is wrong ? I mean How can I introduce the time component to the equation to find the total energy as mentioned ?
If the integral is correct then I ll take the boundries as $0$ to $\lambda$.

It is very unclear to me what you are trying to do with your first approach

Well since $dm$ on the string will encounter with a force of F due to the its motion. And I tried to calculate that force on each dm and then multiply it with dx. Thats not much satisfactory I guess.

Edit: I also didnt quite understand why time would be important. I can kind of see why but x is also important right. Since for example in node points there's no kinetic energy.

 

[Orodruin]

I mean How can I introduce the time component to the equation to find the total energy as mentioned ?

You need to find the moment in time when all the energy is in the form of kinetic energy in the string.

I also didnt quite understand why time would be important.

Because the total energy in the string will change from being in the form of kinetic energy to being in the form of potential energy. When the string reaches its maximal amplitude, all of the string changes direction and is therefore not moving. In other words, at that moment the kinetic energy is zero. The question you need to ask yourself is the opposite one, i.e., when is the potential energy zero?

 

[Arman777]

Maybe I can express it as

$$TE = T+U $$
$$TE = 2A^2qω^2sin^2(ωt)\int { sin^2(kx)dx} + T(L-L_0)$$
Also wait $$U=T(L-L_0)$$ but this will be negative then right since $L_0$ is always bigger then $$L_0>L$$

 

[Arman777]

You need to find the moment in time when all the energy is in the form of kinetic energy in the string.Because the total energy in the string will change from being in the form of kinetic energy to being in the form of potential energy. When the string reaches its maximal amplitude, all of the string changes direction and is therefore not moving. In other words, at that moment the kinetic energy is zero. The question you need to ask yourself is the opposite one, i.e., when is the potential energy zero?

Well I guess from general equation we can say that $$cos(wt)=0$$ so $$t=nπ/2$$ I mean the string should be like straigth line. We can see that from
$$U=T(L-L_0)$$
When $L=L_0$, $U =0$

 

[Orodruin]

Also wait $$U=T(L-L_0)$$ but this will be negative then right since $L_0$ is always bigger then $$L_0>L$$

This is not correct. In fact, it is the exact opposite. The equilibrium length is always shorter than $L$. As soon as you stretch the string out, it becomes longer.

 

[Arman777]

As soon as you stretch the string out, it becomes longer.

Yes that's what I am thinking. $L_0$ is the streched length (0 wavelength) and $L$ is the position where non-zero wavelength ?

So KE in max value will be,
$$KE = 2A^2qω^2sin^2(ωnπ/2)\int_0^λ sin^2(kx) \, dx$$
$$TE=KE_{max}=A^2qω^2sin^2(ωnπ/2)λ$$

That seems wrong...or is it True ?

 

[Orodruin]

Yes that's what I am thinking. $L_0$ is the streched length (0 wavelength) and $L$ is the position where non-zero wavelength ?

No. This is not what I was saying. $L_0$ is the length of the string at rest. $L$ is the length of the string when it is oscillating. Clearly $L \geq L_0$.

So KE in max value will be,
$$KE = 2A^2qω^2sin^2(ωnπ/2)\int_0^λ sin^2(kx) \, dx$$
$$KE_{max}=A^2qω^2sin^2(ωnπ/2)λ$$

That seems wrong...or is it True ?

Why do you think it seems wrong?

 

[Arman777]

No. This is not what I was saying. $L_0$ is the length of the string at rest. $L$ is the length of the string when it is oscillating. Clearly $L \geq L_0$.Why do you think it seems wrong?

Okay I see now.

Well its just visiually not appealing. $sin^2(ωnπ/2)$ term bothers me. Also normally, when wavelength increases energy decrease. Like gamma rays have more energy than radio waves.

 

[Orodruin]

Yes, your assumption on $t$ is not correct. What is the value of $\sin^2(\alpha)$ when $\cos(\alpha) = 0$?

 

[Arman777]

Yes, your assumption on $t$ is not correct. What is the value of $\sin^2(\alpha)$ when $\cos(\alpha) = 0$?

I should say just 1 ?

 

[Orodruin]

I should say just 1 ?

And hence ...

 

[Arman777]

And hence ...

Umm
$$TE=KE_{max}=A^2qω^2λ$$ ?

 

[Orodruin]

Does that feel more comfortable?

 

[Arman777]

Does that feel more comfortable?

Yeah

I guess its true. Thanks a lot for your help

 

[Arman777]

I want to ask something how wavelength can be proportional to Total Energy ?

 

[Orodruin]

It is not. You have integrated only one wavelength so what you are getting is energy per wavelength of the wave. This is not the same as average energy per string length since the wavelength depends on the wavelength. Also note that you have the frequency $\omega$ squared in your expression.

 

[Arman777]

I guess I understand. thanks again.

 

[Arman777]

@Orodruin In my equation there's no 1/2 is that normal ?

 

[Orodruin]

That depends on how you normalize the amplitude.

 

[Arman777]

That depends on how you normalize the amplitude.

I see your point now. I was discussing with a friend and normally, If we write down the equation as $$y(x,t)=Asin(kx)cos(wt)$$ we are getting 1/4 in the end.