Show wave kinetic energy and potential energy are the same

[JayKo]

Homework Statement

Consider a traveling wave y=f(x-vt) on a string. Show that the kinetic and potential energies are always equal to each other. Remember that the potential energy of a wave on a string (over one wavelength) is given by

Homework Equations

E(potential)= 1/2 * F * \int (\partialy/\partialx)^2 dx
E(kinetic) = 1/2 * \mu * \int (\partialy/\partialt)^2 dx

F=\muv^2
ps:the integrand is integrate from 0 to lamba.
3. The Attempt at a Solution (\partial^{2}y/\partialt^{2})=-\omega^{2}A sin(\omegat-kx)

(\partial^{2}y/\partialx^{2})=-k^{2}A sin(\omegat-kx)

so my question is, is it i just need to integrate both the 2nd order partial derivative that i will be show that both potential and kinetic are the same? thanks.

 

[G01]

What work have you done on this problem so far?

Remember that you MUST show some work or thought on a problem before we can help you. We help people with homework problems. We do not do their homework for them.

https://www.physicsforums.com/showthread.php?t=94379

 

[JayKo]

hi G01, thanks for the prompt reply. i am typing my working now.

 

[JayKo]

ok, just updated my question, hope some can answer my question.thanks

i m not asking for working, i just need clarification of my question. thanks.

 

[G01]

Homework Statement

Consider a traveling wave y=f(x-vt) on a string. Show that the kinetic and potential energies are always equal to each other. Remember that the potential energy of a wave on a string (over one wavelength) is given by

Homework Equations

E(potential)= 1/2 * F * \int (\partialy/\partialx)^2 dx
E(kinetic) = 1/2 * \mu * \int (\partialy/\partialt)^2 dx

F=\muv^2
ps:the integrand is integrate from 0 to lamba.
3. The Attempt at a Solution (\partial^{2}y/\partialt^{2})=-\omega^{2}A sin(\omegat-kx)

(\partial^{2}y/\partialx^{2})=-k^{2}A sin(\omegat-kx)

so my question is, is it i just need to integrate both the 2nd order partial derivative that i will be show that both potential and kinetic are the same? thanks.

Yes, your goal is to show that the potential energy integral and kinetic energy integral must be equal. Now, I just want to point out one possible error. In your relevant formulas for potential and kinetic energy, you have the expressions:

(\frac{\partial y}{\partial x})^2 and

(\frac{\partial y}{\partial t})^2

By these do you mean, "the second derivative" or "the first derivative squared?"

 

[JayKo]

Yes, that is how I would go about this problem. Now, I just want to point out one possible error. In your relevant formulas for potential and kinetic energy, you have the expressions:

(\frac{\partial y}{\partial x})^2 and

(\frac{\partial y}{\partial t})^2

By these do you mean, "the second derivative" or "the first derivative squared?"

the question is written as (\frac{\partial y}{\partial x})^2 and

(\frac{\partial y}{\partial t})^2, actually that is my question, is first order partial derivative squared=second order partial derivative ?

 

[G01]

actually that is my question, is first partial derivative squared=second partial derivative ?

No, the two quantities are not generally equal to one another. So, you are going to want to work with first derivatives here and square them, since that is what each expression involves.

 

[G01]

JayKo,

You seem to have a good idea of where you need to go with the problem at this point. I have to head out, but when I get home later I'll check back on this thread to see how you are doing. Try the calculations and see if you can show the equality. f you have any troubles post them, and I'll see if I can help later. Good Luck!

 

[JayKo]

No, the two quantities are not generally equal to one another. So, you are going to want to work with first derivatives here and square them, since that is what each expression involves.

i see, got it now, in that case, i just substitute

(\frac{\partial y}{\partial x})^2=(k * A cos (omega*t-kx))^2

and the same for (\frac{\partial y}{\partial t})^2, with power of omega drop to 1. right? thanks

 

[JayKo]

JayKo,

You seem to have a good idea of where you need to go with the problem at this point. I have to head out, but when I get home later I'll check back on this thread to see how you are doing. Try the calculations and see if you can show the equality. f you have any troubles post them, and I'll see if I can help later. Good Luck!

Thanks G01, no problem. i guess we are of different time zone, here is 1:13 am 03 Feb 08.is time to go to bed. will work out this after wake up. thanks and appreciated your help. when you are free then check out my working here. thanks again.

 

[G01]

i see, got it now, in that case, i just substitute

(\frac{\partial y}{\partial x})^2=(k * A cos (omega*t-kx))^2

and the same for (\frac{\partial y}{\partial t})^2, with power of omega drop to 1. right? thanks

Looks like you got the idea Jay. Good luck! If you have any more trouble feel free to ask.

 

[JayKo]

Hi G01, will show the working later. rushing with other assignment

 

[JayKo]

taking \partialy/\partialt=-\omegaA*cos(kx-\omegat)

\partialy/\partialx=kA*cos(kx-\omegat)

integrate from 0 to \lambda (\partialy/\partialt)^2=
\omega^{2}A^{2}\intcos^{2}(kx-\omegat)
=\omega^{2}A^{2}[x/2+(1/4(sin(2kx-2\omegat))]
=\omega^{2}A^{2}[\lambda/2+1/4(sin(2(2\pi\lambda/\lambda-2\omegat)+sin(2\omegat))]
=\omega^{2}A^{2}[\lambda/2+1/4(sin(4\pi-2\omegat)+sin(2\omegat))]
=\omega^{2}A^{2}[\lambda/2+1/4(-2\omegat)+sin(2\omegat))]
=\omega^{2}A^{2}[\lambda/2.
subst =\omega^{2}A^{2}[\lambda/2]. into

E(kinetic)
and get E = 1/4 *\mu \omega^{2}A^{2}[\lambda].do the same for E (potential)

integrate partial derivative of y/t from 0 to lamba will get k^{2}A^{2}[\lambda]/2

and subst it into E(potential) and subst F=\muv^2 to get 1/4 *\mu \omega^{2}A^{2}[\lambda][proved]oh boy oh boy, finally i got it. thanks G01 :D