Classical Mechanics Problem: Particle in a Square Potential Well

Click For Summary
The discussion focuses on solving a classical mechanics problem involving a particle in a square potential well. Participants are trying to determine the velocity of the particle and clarify the concept of bounded motion, which implies that the particle's energy must fall within specific limits. The key equation discussed is E = K + U, where E is total energy, K is kinetic energy, and U is potential energy. It is emphasized that kinetic energy (K) is always positive, leading to the conclusion that the total energy E must be less than zero for a bounded system. The conversation highlights the importance of correctly interpreting energy values and their implications for the particle's motion.
Jozefina Gramatikova
Messages
62
Reaction score
9

Homework Statement


CLASSICAL MECHANICS
upload_2019-1-20_12-29-4-png.png
[/B]

Homework Equations


E=U+K[/B]

The Attempt at a Solution


50288802_758179237899437_8167705243417575424_n-jpg-_nc_cat-106-_nc_ht-scontent-flhr4-1-jpg.jpg

Guys, can you please help me with part b) ? I am not sure how to find the velocity. Thanks
 

Attachments

  • upload_2019-1-20_12-29-4-png.png
    upload_2019-1-20_12-29-4-png.png
    20.7 KB · Views: 846
  • 50288802_758179237899437_8167705243417575424_n-jpg-_nc_cat-106-_nc_ht-scontent-flhr4-1-jpg.jpg
    50288802_758179237899437_8167705243417575424_n-jpg-_nc_cat-106-_nc_ht-scontent-flhr4-1-jpg.jpg
    37.3 KB · Views: 748
Physics news on Phys.org
First off, find the range of mechanical energies. What could it possibly be if the motion is bounded? What does "bounded" mean classically?
 
Is it -u<E<0 ? The motion of the system is bounded if it begins at a point situated between two forbidden regions, the system’s trajectory will never leave the allowed region where it started (otherwise it would have to cross one of the forbidden regions).
 
Any relevant equations linking kinetic energy to velocity ?
 
upload_2019-1-20_22-42-39.png
 

Attachments

  • upload_2019-1-20_22-42-39.png
    upload_2019-1-20_22-42-39.png
    1.2 KB · Views: 607
Could you tell me if this is true?

If U=-u , then E ≈ 0

if U=0, then E ≈ -u
 
No, I can tell you it is not true.
U is the potential energy
E is the total energy of the particle, an independent variable in this exercise.
As you almost say (well, guess) in #3, the given information tthat the particle is bounded leads to ##-u \le E < 0##.

With K the kinetic energy and, as you say, E = K + U you can write down the bounds of K and thereby the bounds of v. And derive the period of the motion.
 
7594_2276574659287142_7618848920235409408_n.jpg?_nc_cat=100&_nc_ad=z-m&_nc_cid=0&_nc_ht=scontent.jpg
 

Attachments

  • 7594_2276574659287142_7618848920235409408_n.jpg?_nc_cat=100&_nc_ad=z-m&_nc_cid=0&_nc_ht=scontent.jpg
    7594_2276574659287142_7618848920235409408_n.jpg?_nc_cat=100&_nc_ad=z-m&_nc_cid=0&_nc_ht=scontent.jpg
    10.7 KB · Views: 499
Is it -u<=K<0 ? and then v<0 ?
 
  • #10
PLS help
 
  • #11
Ah...the curse of negative numbers...
Please, remember K is always a positive number.
 
  • #12
Perhaps it might help if you considered a concrete example. Suppose ##K= \frac{2}{3}u## where ##u>0##. What is ##E##?
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
16
Views
2K
Replies
12
Views
2K
Replies
5
Views
2K