Energy Diagrams

Education
CSIR NET-JRF, GATE, IIT-JAM, JEST, TIFR and GRE for Physics
H.N. 28 A/1, Jia Sarai, Near IIT-Delhi, Hauz Khas, New Delhi-110016
Contact: +91-89207-59559, 8076563184
Website: www.pravegaa.com | Email: pravegaaeducation@gmail.com
Chapter 1
Stability Analysis and
Phase Diagram
1. Energy Diagrams
We can often find the most interesting features of the motion of a one dimensional system by
using an energy diagram, in which the total energy E and the potential energy U are plotted as
functions of position. The kinetic energy K  E  U is easily found by inspection. Since kinetic
energy can never be negative, the motion of the system is constrained to regions where U  E .
Energy Diagram of Bounded Motion
Energy
K  E U
x2 x
x1
#: +91-89207-59559, 8076563184
1
Education
Contact: +91-89207-59559, 8076563184
Here is the energy diagram for a harmonic oscillator. The potential energy U  kx 2 / 2 is a
parabola centered at the origin. Since the total energy is constant for a conservative system, E
is represented by a horizontal straight line. Motion is limited to the shaded region where E  U ;
the limits of the motion, x1 and x2 in the sketch, are sometimes called the turning points.
Here is what the diagram tells us. The kinetic energy, K  E  U is greatest at the origin. As the
particle flies past the origin in either direction, it is slowed by the spring and comes to a complete
rest at one of the turning points x1 , x2 . The particle then moves toward the origin with increasing
kinetic energy and the cycle is repeated.

The harmonic oscillator provides a good example of bounded motion. As E increases, the turning
points move farther and farther off, but the particle can never move away freely. If E decreased,
the amplitude of motion decreases, until finally for E  0 the particle lies at rest at x  0 .
Energy Diagram of Unbounded Motion
The pot U  A / r , where A is positive. There is a distance of closest approach rmin , as shown in
the diagram, but the motion is not bounded for large r since U decreases with distance. If the
particle is shot toward the origin, it gradually losses kinetic energy until it comes momentarily to
rest at rmin . The motion then reverses and the particle moves back towards infinity. The final and
initial speeds at any point are identical; the collision merely reverses the velocity.
Energy
U
E
K  E U
rmin r
For positive energy, E  0 , the motion is unbounded, and the atoms are free to fly apart. As the
diagram indicates, the distance of closest approach, rmin , does not change appreciably as E is
increased. The kinetic energy will be zero at rmin and as r increases the potential energy
decreases, but kinetic energy K  E  U will increase sharply.
#: +91-89207-59559, 8076563184
2
Education
Contact: +91-89207-59559, 8076563184
Example:
E1
E2 V1 V3
V2
e x
a b c d
V4
In the above figure, potential energy in different regions are given, where
V  8 J , a  x  b V  6 J , cxd
V  x   1 and V  x    3
V2  3 J , b  x  c V4  4 J , d  x  e
The potential is assumed to be zero in all other regions.
(a) What will be kinetic energy in all region if total energy E is 10J ?
Solution: If ‘ T ’ is kinetic energy and ‘ V ’ is potential energy, then total energy E  T  V , so
kinetic energy is T  E  V .
For total energy E  E1 all regions are classical allowed region.
So, in region x  a , V ( x)  0 , so T  10  0  10 J
In region a  x  b , V ( x)  V1  8 J so T  10  8  2 J
In region b  x  c , V ( x)  V2  3 J so T  10  3  7 J
In region c  x  d , V ( x)  V3  6 J so T  10  6  4 J
In region d  x  e , V ( x)  V4  4 so T  10  ( 4)  14 J
In region e  x , V ( x)  0 so T  10  0  10 J
(b) What will be kinetic energy in all regions, if total energy E is 5J ?
Solution: If T is kinetic energy and V is potential energy then total energy E  T  V , so kinetic
energy is T  E  V
So, in region x  a , V ( x)  0 so, T  5  0  5 J
In region a  x  b , V ( x)  V1  8 J hence V1  E2 so, T  0 (classical forbidden region)
In region b  x  c , V ( x)  V2  3 J so, T  5  3  2 J
In region c  x  d , V ( x)  V3  6 J  V3  E2 so, T  0 (classical forbidden region)
In region d  x  e , V ( x)  V4  4 so, T  5  (4)  9 J

In region e  x , V ( x)  0 so, T  5  0  5 J
#: +91-89207-59559, 8076563184
3

Energy Diagrams

Uploaded by

Copyright:

Available Formats

Energy Diagrams

Uploaded by

Document Information

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Energy Diagrams

Uploaded by

Copyright:

Available Formats

Education

kinetic energy and the cycle is repeated.

decreases, but kinetic energy K  E  U will increase sharply.

In region c  x  d , V ( x)  V3  6 J  V3  E2 so, T  0 (classical forbidden region)

In region d  x  e , V ( x)  V4  4 so, T  5  (4)  9 J

You might also like