Calculating the Total Energy of a Transverse Wave on a String

In summary, the conversation discusses the calculation of the total energy in a string with tension T and mass density μ that has a transverse wave with a function ψ(x,t) = f(x±vt), where f(x) is an even function that goes to zero as x approaches ±∞. It is shown that the total energy is given by the integral ∫dw*T*((f'(w))2 with limits of integration at ±∞, and that the infinitesimal energy is equal to T*((dψ/dx)2)dx. However, there is a mistake in the original attempt at a solution and the correct form for the total energy is μ/2[(f'2(x+vt)*dx+v2
  • #1
gimak
56
1

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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  • #2
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
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
gimak said:
dψ/dx = dψ/dt
That is not even dimensionally correct.
Differentiate f(x+vt) wrt each of x, t.
 
  • #5
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
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
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?
 

1. What is the energy of a transverse wave?

The energy of a transverse wave is the amount of mechanical energy that is transferred as the wave propagates through a medium. It is a measure of the wave's amplitude and frequency.

2. How is the energy of a transverse wave related to its amplitude?

The energy of a transverse wave is directly proportional to its amplitude. This means that as the amplitude increases, the energy of the wave also increases.

3. Can the energy of a transverse wave be negative?

No, the energy of a transverse wave cannot be negative. It is always a positive quantity as it represents the amount of energy being transferred by the wave.

4. What factors affect the energy of a transverse wave?

The energy of a transverse wave is affected by its amplitude, frequency, and the properties of the medium through which it propagates. For example, a wave with a higher amplitude or frequency will have more energy, and a denser medium will have a higher energy transfer compared to a less dense medium.

5. How is the energy of a transverse wave measured?

The energy of a transverse wave can be measured using an energy meter or by calculating it using the wave's amplitude and frequency. It is typically measured in joules (J) or ergs (erg).

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