Sects. 2.6 & 2.7
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Transcript Sects. 2.6 & 2.7
Energy Methods: Sect. 2.6
• Consider a point particle under the influence of a
conservative force in 1d (for simplicity).
Conservation of mechanical energy:
E = T+U = (½)mv2 + U(x) = constant
• Rewrite this, using v = (dx/dt) as:
v as a function of x!
v(t) = (dx/dt) = [2(E - U(x))/m]½ v(x)
Clearly, this requires (for real v): E U(x)
• For general U(x), & given initial position xo at t = to,
formally solve the problem by integrating (get t(x),
rather than x(t)): (limits: xo x)
t - to = ∫dx[2(E - U(x))/m]-½
t as a function of x!
t(x) - to = ∫dx[2(E - U(x))/m]-½
• Given U(x), (in principle) integrate this to get
t(x) & (in principle) algebraically invert t(x) to
get x(t), which is what we want!
– In future chapters, we will do (in detail):
1. The harmonic oscillator:
U(x) = (½)kx2
2. Gravitation:
U(x) = -(k/x)
• General U(x): Learn a lot about particle motion by
analyzing plot of U(x) vs. x for different energies E:
T = (½)mv20
Any real physical situation requires
E = T + U(x) U(x)
T = (½)mv2 0 E = (½)mv2 + U(x) U(x)
• Consider motion for different E.
E = E1: Bounded & periodic
between turning points xa & xb.
Bounded Particle never gets out of the region xa x xb.
Periodic: Moving to left, will stop at xa, turn around & move
to right until stops at xb, & turn around again, repeating forever.
xa & xb are called Turning Points for obvious reasons.
Turning Points: v = 0, T= 0, E1 = U(x). Gives xa & xb
E = T + U = constant, but T & U change throughout motion.
T = (½)mv2 0 E = (½)mv2 + U(x) U(x)
• Consider motion for different E.
E = E2: Bounded & periodic
between turning points
xc & xd and separately between
turning points xe & xf .
Bounded The particle never
gets out of region xc x xd or out of region xe x xf.
Periodic: Goes from one turning point to another, turns
around & moves until stops at another turning point. Repeats
forever. Particle is in one valley or another. Can’t jump from
one to another without getting extra energy > E2 (but, in QM:
Tunneling!) Turning points: v = 0, T= 0, E2 = U(x). Gives
xc , xd , xe & xf
T = (½)mv2 0 E = (½)mv2 + U(x) U(x)
• Motion for different E:
E = E0: Since E = U(x),
T= 0 & v = 0. Particle
doesn’t move. Stays at x0 forever.
x0 is determined by E0 = U(x0)
E = E3: If the particle is initially
moving to the left, it comes in from infinity to turning point
xg, stops, turns around, & goes back to infinity. Turning point
xg determined by v = 0, T=0, E3 =U(x).
E = E4: Unbounded motion. The particle can be at any
position. Its speed changes as E - U(x) = T = (½)mv2 changes.
• Motion of particle at energy E1: This is similar to the massspring system.
• Approximate potential for
xa x xb is a parabola: U(x) (½)k(x-x0)2
x0 = equilibrium point (Ch. 3!)
• For the motion of a particle at energy E: If there are 2
turning points, xa & xb, the situation looks like the figure.
The Approximate Potential
for xa x xb is a parabola:
U(x) (½)k(x-x0)2
where x0 is the
Stable Equilibrium Point
xb
xa
x0
Equilibrium Points
• Equilibrium Point Point where the particle will stay
& remain motionless.
– Stable Equilibrium Point If the particle
is displaced slightly away from that point, it
will tend to return to it. (Like the bottom of parabolic potential well).
– Unstable Equilibrium Point If the
particle is displaced slightly away from that
point, it will tend to move even further away from it. (Like the
top of an upside down parabolic barrier.)
– Neutral Equilibrium Point If the particle is
displaced slightly away from that point, it will
tend to stay at new point. (Like a flat potential).
• Assume that the equilibrium point is at x = 0.
In general, expand U(x) in a Taylor’s series
about the equilibrium point [(dU/dx)0 (dU/dx)x0]
U(x) U0 + x(dU/dx)0 + (x2/2!)(d2U/dx2)0
+ (x3/3!)(d3U/dx3)0 + ...
• By definition, if x = 0 is an equilibrium point,
the force = 0 at that point: F0 - (dU/dx)0 = 0
We can choose U0 = 0 since the zero of the potential is arbitrary.
So:
U(x) (x2/2!)(d2U/dx2)0 + (x3/3!)(d3U/dx3)0 + ...
• For a general potential U(x), not far from an
equilibrium point, keep the lowest order term only:
U(x) (x2/2!)(d2U/dx2)0 Or:
U(x) (½)kx2
where k (d2U/dx2)0
• Equilibrium conditions:
1. k = (d2U/dx2)0 > 0 : Stable equilibrium.
U(x) Simple harmonic oscillator potential
2. k = (d2U/dx2)0 <0: Unstable equilibrium.
3. k = (d2U/dx2)0 = 0: May be neutral
equilibrium, but must look at higher order terms.
Example 2.12
A string, length b, attached at A, passes over a pulley at B, 2d
away, & attaches to mass m1. Another pulley, with mass m2
attached passes over string, pulls it down between A & B.
Calculate the distance x1 when system is in equilibrium. Is the
equilibrium stable or unstable? Work on the board!
Example 2.13
Potential: U(x) = -Wd2(x2+d2)/(x4+8d4) Sketch this potential
& discuss motion at
various x. Is it bounded
or unbounded? Where
are the equilibrium
positions? Are these
stable or unstable?
Find the turning
points for E = -W/8.
Limitations of Newtonian Mechanics Sect. 2.7
• Implied assumptions of Newtonian Mechanics:
– r, v, t, p, E are all measurable (simultaneously!)
– All can be specified with desired accuracy, depending
only on the sophistication of our measuring instruments.
True for MACROSCOPIC objects!
Not true for MICROSCOPIC (atomic & smaller) objects!
– Quantum mechanics is needed for these! Heisenberg
uncertainty, for example tells us that ΔxΔp (½)ħ
We cannot precisely know the x & p for a particle
simultaneously!
– Quantum mechanics Newtonian mechanics as
size of the object increases.
• Newtonian mechanics also breaks down when
the speed v of a particle approaches a
significant fraction of the speed of light c.
– Need Special Relativity for these cases (Ch. 14)
– The is no concept of absolute time.
– Simultaneous events depend on the reference
frame.
– There is time dilation.
– There is length contraction.
– Light speed c is limitation on speed of objects.
• Practical limitation to Newtonian mechanics:
– It is impractical when dealing with systems of
huge numbers of particles 1023.
– Even with the most sophisticated & powerful
computers, we cannot simultaneously solve this
many coupled differential equations!
– For such problems we need the methods of
Statistical Mechanics (Physics 4302).
– This uses the methods of probability & statistics to
compute average properties of the system.