Mathematics for Physics 3: Dynamics and Differential Equations
8
52
Lecture 8: Periodic motion about an equilibrium point
In the last lecture we have seen a simple example of periodic motion: that of the pendulum in the small-angle
approximation. In this particular example the potential was quadratic, allowing us to write the exact solution for
the motion right away. This is no more the case for large oscillations. However, certain features of the analysis do
carry over to more general cases, where the potential is not quadratic in the coordinate.
Figure 9: Three examples of potential functions: quadratic as in Hooke’s law or pendulum in the small-angle
approximation, a cosine-shape potential as in the pendulum without the small-angle approximation, and finally one
with a rather different, asymmetric shape. The three have some common features: a local minimum and restoring
force acting towards this minimum from both directions.
Let us now analyse the motion in such potentials V (x) without making specific assumptions on their functional
form. We have seen that for any one-dimensional conservative force the total mechanical energy
m 2
ẋ + V (x) = E
2
is conserved. E is constant throughout the motion. Assume now that the potential rises both to the right and to the
left of where the particle is located initially, so that there is a finite range in x such that V (x) < E. Let us denote
the two points on the boundary of this range, where V (x) = E, by xa and xb , respectively. These points are the
turning points, as the kinetic energy, and therefore the velocity ẋ vanishes there. Since the kinetic energy is never
negative, regions that fall beyond these points can never be reached. The motion is confined to the range:
xa ≤ x ≤ xb
where
xa , xb admit V (xa ) = V (xb ) = E
even if the potential drops down again below the level V (x) = E (in quantum mechanics the situation is qualitatively
different, as there is a probability of tunnelling through regions where the potential is higher than the energy; in
classical mechanics this is impossible).
Potentials with two turning points necessarily lead to periodic motion. To see this recall that the motion from
xa to xb is uniquely determined by the following first order differential equation
� �
�
2
ẋ =
E − V (x)
(143)
m
and the initial condition x(ta ) = xa ; similarly the motion from xb to xa is uniquely determined by
� �
�
2
ẋ = −
E − V (x)
m
(144)
and the initial conditions x(tb ) = xb . Note that the initial conditions on the velocity ẋ(ta ) = ẋ(tb ) = 0 are already
encapsulated in (143) and (144) through the fact that V (xa ) = V (xb ) = E.
Using (143) one can compute the time of the motion between the two turning points:
� � xb
m
dx
�
∆ta→b =
(145)
2 xa
E − V (x)
and in the opposite direction:
∆tb→a = −
�
m
2
�
xa
xb
dx
�
= ∆ta→b
E − V (x)
(146)
Mathematics for Physics 3: Dynamics and Differential Equations
Therefore the period of the periodic motion is 2∆ta→b , namely
� xb
√
�
∆ta→b→a = 2m
xa
53
dx
E − V (x)
.
(147)
The other special point besides the turning points, is the equilibrium point. This is the point where no force
is active, namely there the potential attains its minimum:
�
dV ��
F (x0 ) = −
=0
(148)
dx �x=x0
It is useful to expand the potential around its minimum:
1
V (x) = V (x0 ) + (x − x0 ) V � (x0 ) + (x − x0 )2 V �� (x0 ) + · · ·
� �� � 2
=0
We often redefine the reference level for the potential such that V (x0 ) = 0, leading to the familiar Harmonic potential,
which is quadratic in the coordinate, up to O((x − x0 )3 ) corrections:
1
(x − x0 )2 V �� (x0 ) + · · ·
2
(149)
dV
= −V �� (x0 )(x − x0 ) + O((x − x0 )2 )
dx
(150)
V (x) =
The corresponding equation of motion is
mẍ = −
We emphasize that the quadratic form of the potential (149) provides a first approximation around equilibrium for
any potential. This approximation is sometimes referred to as linearising the equation of motion, as is indeed clear
from (150).
Example: Computing the oscillation period for large and small oscillations
Let us consider the specific case of a particle that is subject to a conservative force of the form:
F (x) =
2a
− 2bx
x3
where a and b are positive constants. The corresponding potential is:
V (x) =
a
+ bx2
x2
The equilibrium point, where the force vanishes (and the potential reaches its minimum) is at x = xmin =
(151)
� a �1/4
b
.
Figure 10: The potential V (x) of (151), here plotted with a = 2 and b = 2, in two different scales: the plot on the
left extends over a large range in both x and energies, where large oscillations occur, while the one on the right is an
enlargement of the region near the minimum, where small oscillations occur. It is clear that the latter is quadratic
to a very good approximation, while the former is not.
Mathematics for Physics 3: Dynamics and Differential Equations
54
√
For any energy E > 2 ab there will be oscillations between the two turning points where
E = V (x) =
The solution are:
a
+ bx2
x2
√
E 2 − 4ab
E + E 2 − 4ab
2
y1 ≡
=
,
y2 ≡ x2 =
2b
2b
2
where we defined y = x . Let us now compute the period of the oscillations according to (147)
� x2
� x2
√
√
dx
dx
�
�
∆tx1 →x2 →x1 = 2m
= 2m
a
E − V (x)
x1
x1
E − 2 − bx2
x
√
� x2
�
√
xdx
2m y2
dy
�
�
= 2m
=
2
4
2
Ey
−
a − by 2
x1
y
Ex − a − bx
1
�
�
�
� y2
� y2
m
dy
m
dy
m
�
�
=
=
=π
2b y1
2b
2b
(y2 − y)(y − y1 )
E
a
y1
y − − y2
b
b
E−
x21
√
(152)
The final step was done using the substitution:
y = y1 cos2 (φ) + y2 sin2 (φ)
which gives:
dy = 2(y2 − y1 ) sin(φ) cos(φ)dφ
and
and therefore
�
(y2 − y)(y − y1 ) = (y2 − y1 ) sin(φ) cos(φ)
�
y2
y1
�
dy
(y2 − y)(y − y1 )
=
�
π/2
2dφ = π .
0
The conclusion is�that with the particular potential (151) the period of oscillations is as given in (152), namely,
m
∆tx1 →x2 →x1 = π 2b
. Remarkably, in this case the period of oscillations does not depend at all on the energy E nor
on the parameter a (which determines the steepness of the potential at small x). It only depends on the mass and
on b. Thus, it must be the same result as one would obtain in the approximation of small oscillations, as we shall
indeed confirm below. N. B. The independence of the period on the energy is not a general result; it is special to
the potential of (151). In the general case, the period of large oscillations does depend on the energy and on the
details of the potential away from the minimum. You will see this explicitly in the tutorial and homework questions.
Let us now analyse the problem in the harmonic approximation, namely considering small oscillations right from
the start. As already discussed this amounts to expanding the potential up to the second power in the coordinate,
see (149) above.
� a �1/4
it is useful to expand in δ = x − xmin
Considering the potential (151) near the minimum xmin =
b
√
a
V (x) =
+ b(δ + xmin )2 = 2 ab + 4bδ 2 + O(δ 3 )
(153)
2
(δ + xmin )
Dropping the subleading terms, O(δ 3 ), we return to the familiar case of the quadratic potential which we analysed
in the previous lecture. It is no else than the familiar harmonic motion case (simple pendulum at small oscillation
or Hooke’s law), where
�
m
k
m� 2
E = T + V = ẋ2 + x2 =
ẋ + ω 2 x
(154)
2
2
2
�
with ω = k/m. Clearly the potential in eq. (151) is just the same as in (154) with k = 8b. Here we already know
the result for the period:
√
√
√
2π m
2π
2π m
π m
∆tx1 →x2 →x1 =
= √
= √
= √
(155)
ω
k
8b
2b
in agreement with our previous calculation in (152).
Alternatively, one may compute the period in the harmonic approximation from the general formula of eq. (147).
Now the turning points δ1,2 are the roots of E − V = 0,
√
E − 2 ab − 4bδ 2 = 0
Mathematics for Physics 3: Dynamics and Differential Equations
namely
δ1 = −
�
√
E − 2 ab
,
4b
55
δ2 = +
�
√
E − 2 ab
4b
We thus obtain:
∆tx1 →x2 →x1 =
=
√
2m
�
m
2b
�
�
x2
x1
δ2
δ1
�
�
in agreement with the previous calculations.
dx
E
4b
dδ
−
√
ab
2b
√
�
δ2
dδ
√
E − 2 ab − 4bδ 2
�
�
� δ2
m
dδ
m
�
=
=π
2b
2b
(δ2 − δ)(δ − δ1 )
δ1
− δ2
E − V (x)
=
2m
δ1
�
(156)