Calculating the Total Energy of a Transverse Wave on a String

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Homework Help Overview

The problem involves calculating the total energy of a transverse wave on a string characterized by tension T and mass density μ. The wave function is given as ψ(x,t) = f(x±vt), where f(x) is an even function that approaches zero as x approaches ±∞. The task is to show that the total energy can be expressed as an integral involving the derivative of f.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the kinetic and potential energy contributions to the total energy, with attempts to integrate these over the string. Questions arise regarding the role of the function f and its derivatives in the energy expression.

Discussion Status

Some participants have provided insights into the relationships between the derivatives of f and the energy components, while others have pointed out potential errors in reasoning and dimensional inconsistencies. There is ongoing exploration of the correct formulation of the total energy integral.

Contextual Notes

Participants are navigating through the implications of the wave function and its derivatives, with some noting the need to clarify the integration variable and the limits of integration. There is also mention of the need to ensure dimensional consistency in the expressions used.

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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