How Does Energy Transfer in a Triangular Wave Pulse?

[applestrudle]

Homework Statement

one end of a fixed string is moved transversly at a constant speed u for a time τ and is moved back to it's starting point with velocity -u during the next time interval. as a result a triangular pulse is set up and moves along the string with speed v. Calculate the kinetic and potential energies associated with the pulse, and show that their sum is equal to the work done by the transverse force that has to be applied at the end of the string.

Homework Equations

Work = F x displacement
Power = Work x time

PE density = T/2 (dy/dx)^2
KE density = μ/2 (dy/dt)^2

v = (T/μ)^0.5 (T = tension)

The Attempt at a Solution

the triangle pulse:

y = A f(x-vt)

KE density = μ/2 A^2[-vAf'(x-vt)]^2
PE density = t/2 A^2 [f'(x-vt)]^2

using v = (T/mu)^0.5 you get that PE = KE therefore

total energy density = μ A^2 v^2 [f'(x-vg)]^2

work = power x time

W = T x u x 2τ

-TvA[f'(x-vt)] x 2τ

I can't seem to rearrange them to get them the same. I tried using v = λ/(2τ)

thank you

 

[mfb]

total energy density = μ A^2 v^2 [f'(x-vg)]^2

I guess "g" is a typo and you mean t?
If you replace v^2 by T/μ there, the equations get more similar.

W = T x u x 2τ

I don't understand where this comes from. The force required to move the line is not T.

 

[applestrudle]

I guess "g" is a typo and you mean t?
If you replace v^2 by T/μ there, the equations get more similar.

I don't understand where this comes from. The force required to move the line is not T.

Isn't the force your putting on the end of the string the tension T?

 

[mfb]

The motion is transversal, the sidewards force (=relevant for the wave generation) is different.

 

[paranormal]

for the question and the equations provided. It seems like you are on the right track in terms of using the equations for work, power, and energy density. However, there are a few errors in your calculations.

First, in your attempt at a solution, you wrote that PE = KE, but this is not necessarily true for all cases. In this problem, we are dealing with a wave pulse, so the potential energy will depend on the displacement of the string, while the kinetic energy will depend on the velocity of the string. Therefore, we cannot equate them.

Second, when calculating the potential energy density, you wrote t/2 instead of T/2. This is a minor error, but it will affect your final answer.

Third, when calculating the work done by the transverse force, you need to use the total displacement of the string, which is 2A. So the work should be W = T x u x 2A.

Finally, in order to show that the sum of kinetic and potential energies is equal to the work done by the transverse force, you need to integrate the energy densities over the entire length of the string. This will give you the total energy associated with the pulse. Then, you can compare it to the work done by the transverse force.

I hope this helps. Keep up the good work!