Show wave kinetic energy and potential energy are the same

In summary, the problem asked to show that the kinetic and potential energies for a traveling wave on a string are always equal. By integrating the first derivatives of the wave equation and substituting the values for F and \mu, it was shown that the two energies are equal to each other.
  • #1
JayKo
128
0

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 * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]x)^2 dx
E(kinetic) = 1/2 * [tex]\mu[/tex] * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]t)^2 dx

F=[tex]\mu[/tex]v^2
ps:the integrand is integrate from 0 to lamba.
3. The Attempt at a Solution ([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]t[tex]^{2}[/tex])=-[tex]\omega[/tex][tex]^{2}A[/tex] sin([tex]\omega[/tex]t-kx)

([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]x[tex]^{2}[/tex])=-k[tex]^{2}A[/tex] sin([tex]\omega[/tex]t-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.
 
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  • #2
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
 
  • #3
hi G01, thanks for the prompt reply. i am typing my working now.
 
  • #4
ok, just updated my question, hope some can answer my question.thanks :eek:

i m not asking for working, i just need clarification of my question. thanks.
 
  • #5
JayKo said:

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 * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]x)^2 dx
E(kinetic) = 1/2 * [tex]\mu[/tex] * [tex]\int[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]t)^2 dx

F=[tex]\mu[/tex]v^2
ps:the integrand is integrate from 0 to lamba.
3. The Attempt at a Solution ([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]t[tex]^{2}[/tex])=-[tex]\omega[/tex][tex]^{2}A[/tex] sin([tex]\omega[/tex]t-kx)

([tex]\partial[/tex][tex]^{2}[/tex]y/[tex]\partial[/tex]x[tex]^{2}[/tex])=-k[tex]^{2}A[/tex] sin([tex]\omega[/tex]t-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:

[tex](\frac{\partial y}{\partial x})^2[/tex] and

[tex](\frac{\partial y}{\partial t})^2[/tex]

By these do you mean, "the second derivative" or "the first derivative squared?"
 
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  • #6
G01 said:
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:

[tex](\frac{\partial y}{\partial x})^2[/tex] and

[tex](\frac{\partial y}{\partial t})^2[/tex]

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

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

[tex](\frac{\partial y}{\partial t})^2[/tex], actually that is my question, is first order partial derivative squared=second order partial derivative ?
 
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  • #7
JayKo said:
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.
 
  • #8
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!
 
  • #9
G01 said:
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

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

and the same for [tex](\frac{\partial y}{\partial t})^2[/tex], with power of omega drop to 1. right? thanks
 
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  • #10
G01 said:
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.
 
  • #11
JayKo said:
i see, got it now, in that case, i just substitute

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

and the same for [tex](\frac{\partial y}{\partial t})^2[/tex], 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.
 
  • #12
Hi G01, will show the working later. rushing with other assignment
 
  • #13
taking [tex]\partial[/tex]y/[tex]\partial[/tex]t=-[tex]\omega[/tex]A*cos(kx-[tex]\omega[/tex]t)

[tex]\partial[/tex]y/[tex]\partial[/tex]x=kA*cos(kx-[tex]\omega[/tex]t)

integrate from 0 to [tex]\lambda[/tex] ([tex]\partial[/tex]y/[tex]\partial[/tex]t)^2=
[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][tex]\int[/tex]cos[tex]^{2}[/tex](kx-[tex]\omega[/tex]t)
=[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][x/2+(1/4(sin(2kx-2[tex]\omega[/tex]t))]
=[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][[tex]\lambda[/tex]/2+1/4(sin(2(2[tex]\pi[/tex][tex]\lambda[/tex]/[tex]\lambda[/tex]-2[tex]\omega[/tex]t)+sin(2[tex]\omega[/tex]t))]
=[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][[tex]\lambda[/tex]/2+1/4(sin(4[tex]\pi[/tex]-2[tex]\omega[/tex]t)+sin(2[tex]\omega[/tex]t))]
=[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][[tex]\lambda[/tex]/2+1/4(-2[tex]\omega[/tex]t)+sin(2[tex]\omega[/tex]t))]
=[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][[tex]\lambda[/tex]/2.
subst =[tex]\omega[/tex][tex]^{2}[/tex]A[tex]^{2}[/tex][[tex]\lambda[/tex]/2]. into

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

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

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

Wave kinetic energy is the energy associated with the motion of a wave. It is a type of mechanical energy that is transferred through the movement of particles in a medium, such as water or air.

2. What is potential energy in relation to waves?

Potential energy in waves refers to the energy stored in a wave due to its position or shape. This energy is converted into kinetic energy as the wave moves through the medium.

3. How are wave kinetic energy and potential energy related?

In a wave, kinetic and potential energy are constantly exchanged as the wave propagates. When the wave is at its highest point, it has maximum potential energy, and when it is at its lowest point, it has maximum kinetic energy.

4. Why are wave kinetic energy and potential energy considered the same?

This is because, at any given point in time, the total energy of a wave is equal to the sum of its kinetic and potential energy. This concept is known as the principle of conservation of energy.

5. How does this concept apply to real-life situations?

Understanding the relationship between wave kinetic energy and potential energy is crucial in many fields, such as oceanography, meteorology, and engineering. It helps us predict and analyze the behavior of waves in different environments, such as predicting the energy output of ocean waves for renewable energy sources.

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