Integrating a Gaussian Pulse for Kinetic Energy Calculation

Click For Summary
SUMMARY

The discussion focuses on evaluating the kinetic energy integral of a Gaussian pulse defined as \(\psi(y,t) = Ae^{-\left(\frac{y-ct}{a}\right)^2}\) traveling along an infinite string with linear mass density \(\rho\) and tension \(T\). The kinetic energy is expressed as \(\frac{\rho}{2} \int_{-\infty}^{\infty} \left(\frac{\partial \psi}{\partial t}\right)^2 dy\), which simplifies to \(\frac{\rho}{2} \int_{-\infty}^{\infty} \left(\frac{2c(y-ct)}{a^2} \psi\right)^2 dy\). The user initially struggled with integration techniques but successfully identified the use of integration by parts and the error function to solve the integral.

PREREQUISITES
  • Understanding of Gaussian functions and their properties
  • Familiarity with kinetic energy concepts in physics
  • Knowledge of integration techniques, including integration by parts
  • Basic understanding of error functions and their applications
NEXT STEPS
  • Study the properties and applications of Gaussian functions in physics
  • Learn about the error function and its role in integrals involving exponential decay
  • Practice integration by parts with various functions, focusing on exponential and polynomial combinations
  • Explore the derivation and applications of kinetic energy in wave mechanics
USEFUL FOR

Students and professionals in physics, particularly those studying wave mechanics and kinetic energy calculations, as well as anyone interested in advanced integration techniques.

atqamar
Messages
55
Reaction score
0
I'm having a hard time evaluating this integral.

A Gaussian pulse [tex]\psi (y,t) = Ae^{-( \frac{y-ct}{a} )^2}[/tex] is traveling in an infinite string of linear mass density [tex]\rho[/tex], under tension [tex]T[/tex].

I know the Kinetic Energy is the integral of the partial: [tex]\frac{\rho}{2} \int_{-\infty}^{\infty} (\frac{\partial \psi}{\partial t})^2 dy[/tex]. I evaluate the partial, and this simplifies to [tex]\frac{\rho}{2} \int_{-\infty}^{\infty} (\frac{2c(y-ct)}{a^2} \psi)^2 dy[/tex].

I don't know where to proceed from here. I tried u-substitution, and integration by parts, with no success. I think the error function is useful in this, but we haven't covered this in the physics course yet.
 
Physics news on Phys.org
Integral is of the form [itex]\int x^2 e^{-x^2}dx[/itex] notice that this is the same as [itex]\int x (x e^{-x^2})dx[/itex] where the term between brackets is easy to integrate. Now use partial integration.
 
Oh stupid me. That wasn't tough at all. Thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K