Calculating the Total Energy of a Transverse Wave on a String

AI Thread Summary
The discussion focuses on calculating the total energy of a transverse wave on a string, defined by the wave function ψ(x,t) = f(x±vt). The kinetic and potential energy expressions are given, leading to the formulation of total energy as the integral of T*(dψ/dx)² over the limits of ±∞. Participants clarify the relationship between f(x±vt) and its derivatives, emphasizing the need to correctly apply the derivatives in the energy calculations. The conversation highlights the importance of maintaining dimensional consistency and correctly identifying the variable of integration for the total energy calculation. Ultimately, the total energy is confirmed to be represented by the integral of T*(f'(w))², integrating over the entire string.
gimak
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Homework Statement


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 ±∞

Homework 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)

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)?
 
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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.
 
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.
 
gimak said:
dψ/dx = dψ/dt
That is not even dimensionally correct.
Differentiate f(x+vt) wrt each of x, t.
 
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?
 
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.
 
gimak said:
df/dx = f'(x+vt)*dx;
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?
 
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