Energy of a transverse wave

1. Jul 20, 2017

gimak

1. The problem statement, all variables and given/known data
There's a string with tension T & mass density μ that has a transverse wave with ψ(x,t) = f(x±vt). f(x) is an even function & goes to zero as x→±∞
Show that the total energy in the string is given by ∫dw*T*((f'(w))2; limits of integration are ±∞

2. Relevant equations
The kinetic energy of an infinitesimal part of string is KE = 0.5*μ*dx*(dψ/dt)2.

Its potential energy is 0.5*T*dx*(dψ/dx)2.

Note that w=ck. c = √(T/μ)

3. The attempt at a solution
Total energy = KE + PE = μ/2[(dψ/dt)2 + v2*(dψ/dx)2]. I add them & integrate over ±∞. However, where does the f(w) come from. What does it have to do with f(x±vt)?

2. Jul 20, 2017

haruspex

You have gone a bit wrong by eliminating T and keeping v. The form of answer has T but not v.

Also, you have omitted the dx terms from the total energy. Please post your working for the integration step.

3. Jul 20, 2017

gimak

Ok, so I figured out that dψ/dx = dψ/dt. So infinitesimal energy = μ/2(2V2*(dψ/dx)2)dx. Since v2 = T/μ, infinitesimal energy = T*((dψ/dx)2)dx. Now I integrate to get total energy for whole string: total energy = ∫T*((dψ/dx)2)*dx. Limits of integration are ±∞. I don't see how they go from this to the answer.

4. Jul 20, 2017

haruspex

That is not even dimensionally correct.
Differentiate f(x+vt) wrt each of x, t.

5. Jul 21, 2017

gimak

df/dx = f(x+vt)*dx;
df/dt = f(x+vt)*v*dt
infinitesimal energy = KE + PE = μ/2[(f2(x+vt)*dx+v2*f2(x+vt)*dt] = 0.5*μ*(1+v2)*f2(x+vt)*(dx+dt)

Total energy = ∫0.5*μ*(1+v2)*f2(x+vt)*dx I'm guessing dt goes to zero since we're integrating with respect to space/over entire spring & not time. Is this right?

6. Jul 21, 2017

gimak

Oh I forgot something:

df/dx = f'(x+vt)*dx;
df/dt = f'(x+vt)*v*dt

So KE + PE = μ/2[(f'2(x+vt)*dx+v2*f'2(x+vt)*dt] = 0.5*μ*(1+v2)*f'2(x+vt)*(dx+dt)

Forgot to add primes in front of f's to indicate derivative.

7. Jul 21, 2017

haruspex

You do not want dx in there twice. Delete one of them. Similarly dt in the next line.
For the integral, you are interested in the total energy at any instant, so what is the variable of integration?