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.