Calculating the Total Energy of a Transverse Wave on a String

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SUMMARY

The total energy of a transverse wave on a string with tension T and mass density μ is calculated using the formula ∫T*((dψ/dx)²)dx, with limits of integration from -∞ to +∞. The kinetic energy (KE) is defined as KE = 0.5*μ*dx*(dψ/dt)², while the potential energy (PE) is PE = 0.5*T*dx*(dψ/dx)². The integration process requires careful consideration of the derivatives of the wave function f(x±vt), leading to the conclusion that the total energy can be expressed as ∫0.5*μ*(1+v²)*f'²(x+vt)dx, where v² = T/μ.

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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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