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Lecture-X
Small oscillation
Stability:
The result F = -dU/dx is useful not only for computing the force but also for visualizing
the stability of a system from the potential energy plot.
Suppose there is a force on the particle is F = -dU/dx, and the system is in equilibrium
where there is no force i.e. dU/dx = 0. If this occurs at a minimum of U it is a stable
equilibrium whereas if it is at a maximum of U, the equilibrium is unstable. Say, dU/dx
= 0 occurs at some point xo. To test for stability we must determine whether U has a
minimum or a maximum at xo. One needs to examine d2U/dx2 at xo.
If the second derivative is positive, the equilibrium is stable; if it is negative, the system
is unstable. If d2U/dx2 = 0, one must look at higher derivatives. If all derivatives vanish
so that U is constant in a region about xo, the system is said to be in a condition of
neutral.
Energy and Stability-The Teeter Toy
The teeter toy consists of two identical weights which hang from a peg on drooping
arms, as shown. Find the condition for stability of the toy with respect to rocking.
The solution is θ =0, as we expect from symmetry.
At equilibrium,
For the second derivative to be positive,
Bounded and unbounded motion:
Since kinetic energy K=E-U can never be negative, the motion
of the system is constrained to regions where U < E.
Harmonic oscillator: total energy is constant, E is
represented by a horizontal line. Motion is limited to the
shaded region where E > U; the limits of the motion, x1
and x2, are called the turning points. The motion is
bounded.
U=A/r: There is a distance of closest approach, rmin, but
the motion is not bounded for large r since U decreases
with distance. Unbounded motion.
A commonly used potential energy function to describe
the interaction between two atoms is the Lennard-Jones
potential
For E > 0, the motion is unbounded, and the atoms are free to fly apart.
For E < 0, the motion is bounded, and the atoms never approach closer
than ra or move farther apart than rb.
Time period of bounded motion
2
Since,
1  dx 
K = m   = E −U ,
2  dt 
2( E −U )
dx
=
or
dt
m
x2
t=
∫
x1
dx
2( E −U )
m
where x1 and x2 are the turning points. The time period is then T=2t.
Consider the potential shown in the figure. Find the period of oscillation.
b
t=∫
0
2b
dx
+∫
2E b
m
dx
2( E −U0 )
m
1  m

= b 1 +

3  U0

1  m

Thus, T = 2b 1 +

3  U0

=b
m
m
+b
3U 0
U0
E=3U0/2
U=2U0
U=U0
U=0
b
b
Small oscillation
Nearly every bound system oscillates like a harmonic oscillator if it is slightly perturbed
from its equilibrium position.
Expand U(r) about r0, the position of the potential
minimum.
where
Frequency of small oscillation:
∂U
mx = F ( x) = −
= −kx,
∂x
the effective spring constant.
2
x + ω x = 0,
k
1 d 2U
ω=
=
m
m dr 2
r0
Molecular Vibrations
Suppose that two atoms of masses m1 and m2 are bound together in a molecule with energy so
low that their separation is always close to the equilibrium value ro. With the parabola
approximation, the effective spring constant is k=(d2U/dr2)Iro. How can we find the vibration
frequency of the molecule?
This vibrational motion, characteristic of all molecules,
can be identified by the light the molecule radiates. The
vibrational frequencies typically lie in the near infrared
(3X1013 Hz), and by measuring the frequency, the value of
d2U/dr2 at the potential energy minimum. For the HCI
molecule, the effective spring constant turns out to be
500 N/m.
Frequency from energy pre-factors:
If the potential and kinetic energies are given by:
where q is a generalized displacement.
The total energy can be obtained as:
Since the total energy is constant:
The equation of motion can be obtained as:
or
2
q + ω q = 0
where
ω=
A
B
Example: Small oscillation of teeter toy
Hence,
s 2 = L2 + l 2 − 2 Ll cos α
Formal solution of Simple harmonic oscillator:
Consider a spring mass system:
If at time t = 0 the position of the mass is x(0) and
its velocity v(0),
The amplitude A and the phase φ are given by
The time period of motion
Energy Considerations
For a harmonic oscillator:
and
&
The total energy is constant as expected.
Let us calculate the time average values of the potential and kinetic energies over one
period of oscillation
Thus,
The time average kinetic and potential energies are equal. When friction is present,
this is no longer exactly true.
Damped harmonic oscillation:
Consider a spring mass system with a damping force
Net force acting on the system is
Equation of motion:
Or,
Trial solution:
with
γ = b / m,
x = x0 eα t
Then the general solution can be written as
x = x Aeα1t + xB eα 2t
where xA and xB are constants.
Consider
Then, x = Ae −(γ
2 )t
cos (ω1t + φ )
The motion is similar to the undamped case except
that the amplitude decreases exponentially in time
and the frequency of oscillation ω1 is less than the
undamped frequency ω0 .
From the work energy theorem:
Physically, E(t) decreases with time because the friction force continually dissipates
energy.
How E(t) depends on time can be found by calculating the kinetic and potential
energies K(t) and U(t). The velocity v is given by
x = Ae
−( γ 2 )t
cos (ω1t + φ )
If the motion is only lightly damped, 'Y / WI « 1, and the coefficient of the second term
in the bracket is small. Then
and
and
Since the damping is assumed to be small,
and taking
,
,
The energy decreases exponentially in time.
The Q of an Oscillator
The degree of damping of an oscillator is often specified by a dimensionless parameter
Q, the quality factor, defined by
The energy dissipated in a short time ∆t is
The time to oscillate through one radian is
Hence, one radian of oscillation requires time
A lightly damped oscillator has Q »1.