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small oscillations.pdf
1. SEMESTER-VI
PHYSICS-DSE: CLASSICAL DYNAMICS
Small Amplitude Oscillations
The theory of small oscillations was developed by Dโ Alembert, Joseph-Louis Lagrange and other
scientists. The study of the effect of all possible small perturbations to a dynamical system in mechanical
equilibrium is known as small oscillations. The theory of small oscillations is widely used in different
branches of physics viz. in acoustics, molecular spectra, coupled electric circuits etc.
1. Equilibrium and its types-
Equilibrium is the condition of a system in which all the forces- internal as well as external,
cancel out for some configuration of the system and unless the system is perturbed by an external
agency, it stays indefinitely in that state.
Equilibrium may be of following types-
a) Static Equilibrium-
Static equilibrium is a state of zero kinetic energy that continues indefinitely. In such
equilibrium the immediate surroundings of the system does not change with time. Example-
an object (e.g. book) lying still on a surface (e.g. table).
b) Dynamic Equilibrium-
Dynamic equilibrium is defined as the state when no net force acting on the system which
continues with zero kinetic energy. In such equilibrium the immediate surroundings of the
system change with time such that it exerts a balancing force on the system. Example- the
charge neutrality of atoms.
c) Stable Equilibrium-
In stable equilibrium, if a small displacement is given, the system tends to return to the
original equilibrium state. Example- the bob of a simple pendulum in equilibrium state.
d) Unstable Equilibrium-
In instable equilibrium, if a small displacement is given, the system does not return to the
original equilibrium state. Example- a large sized stone lying on the upper edge of a cliff.
e) Metastable Equilibrium-
In metastable equilibrium, the system canโt return to the original equilibrium configuration, if
displaced sufficiently; for smaller displacement, however, it can return. Example- a balloon
that explodes above a certain gas pressure.
2. Equilibrium state and stability of a system-
Small Amplitude Oscillations: Minima of potential energy and points of stable equilibrium, expansion
of the potential energy around minimum, small amplitude oscillations about the minimum, normal
modes of oscillations example of N identical masses connected in a linear fashion to (N -1) - identical
springs.
2. In a conservative system the potential V is a function of position (qi) only and the system is said
to be equilibrium when the generalized forces Qi acting on the system vanish.
Therefore, ๐๐ = โ (
๐๐
๐๐๐
)
0
= 0 โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ (i)
i.e. the potential energy has an extremum (
maximum or minimum) at the equilibrium configuration of the system,
๐01, ๐02, ๐03, ๐04, โฆ โฆ โฆ โฆ โฆ . , ๐0๐ . When V is a minimum at equilibrium, any small deviation
from this position means an increase in V and consequently decrease in kinetic energy T i.e.,
velocity v, in a conservative system and the body ultimately comes to rest, indicating the small
bounded motion about the Vmin. Such equilibrium is known as the stable equilibrium. On the
other hand, a small departure from Vmax results in decrease in V and increase in kinetic energy
and velocity indefinitely, which corresponds to unstable motion. Thus the position of Vmax is the
position of unstable equilibrium. Therefore oscillations always occur about a position of stable
equilibrium i.e. about Vmin.
Let V(max) be the potential energy function for a particle and also let
the force F acting on the particle vanishes at x0 i.e.
๐น = โ (
๐๐
๐๐ฅ
)
๐ฅ0
= 0โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.. (ii)
In that case x0 is the point of equilibrium. To test the stability of the equilibrium we must examine
๐2๐
๐๐ฅ2 at x0. If the second derivative is positive, the equilibrium is stable. Thus in stable
equilibrium,
(
๐2
๐
๐๐ฅ2
)๐ฅ0
> 0
If the second derivative is negative, the equilibrium is unstable. Thus in unstable equilibrium,
(
๐2
๐
๐๐ฅ2
)๐ฅ0
< 0
And if the second derivative also vanishes, we must examine the higher derivatives at x0. If all the
derivatives vanish at x0, so that V(x) is constant in a region about x0, then the particle is
effectively free. In such a situation no force results from a displacement (from x0 ) and the system
to be in state of neutral stability. Thus in neutral equilibrium,
(
๐2
๐
๐๐ฅ2
)๐ฅ0
= 0
3. 3. The Stability of a Simple Pendulum (One Dimensional Oscillator)-
Let m be the mass of the bob of a pendulum and l be its length. The zero of the potential energy
scale is taken at the bottom of the swing.
If ำจ be its angular deflection, then the potential energy is given by,
๐(๐) = ๐๐๐(1 โ ๐ถ๐๐ ๐)โฆโฆโฆโฆโฆโฆโฆโฆโฆ.. (i)
where g is the acceleration due to gravity.
The equilibrium position of the pendulum is given by,
๐๐
๐๐
= ๐๐๐ ๐๐๐๐ = 0
๐. ๐. ำจ = ๐๐๐กโ๐๐ 0 ๐๐ ๐
Now, we have
๐2
๐
๐๐2
= ๐๐๐ ๐ถ๐๐ ๐
Now, (
๐2๐
๐๐2)๐=0 > 0, and it corresponds to the position of stable equilibrium.
(
๐2๐
๐๐2)๐=๐ < 0, and it corresponds to the position of unstable equilibrium.
Therefore ำจ=0 is the position of stable equilibrium at which the bob can hang
for an indefinite period, which is not at all possible at the position of unstable equilibrium at ๐ =
๐. The pendulum can therefore oscillate only about the position of stable equilibrium (ำจ=0).
4. General Problem of Small Oscillation-
a) Formulation of the Problem-
In tackling the general problem of small oscillations we perform an appropriate simplification of
Lagrangian method. We know that the oscillations whether large or small take place about a
position of equilibrium. Thus a pendulum oscillates about a verticval which is its stable
equilibrium position. When the pendulum is deflected from the stable position a restoring force
acts on the pendulum in a direction opposite to the direction of deflection. In the equilibrium
position this force is zero.
4. We now suppose that the force is acting on a system which is wholly
conservative and hence it is derivable from potential. In the equilibrium condition of the system
the generalized force Qi will be equal to zero.
The potential energy is a function of position only i.e., the generalized co-ordinates
q1,q2,q3,q4,โฆโฆโฆโฆ.,qn. Hence,
(
๐๐
๐๐๐
)
0
= 0 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ . (๐)
Let us denote the deviation of the generalized co-ordinates from equilibrium by ฮทi.
Therefore, ๐๐ = ๐0๐ + ๐๐ ๐๐, ๐๐ = ๐๐ โ ๐0๐โฆโฆโฆโฆโฆโฆโฆโฆ.......(ii)
q0i denotes the equilibrium position of the co-ordinate system.
Expanding the potential V (q1,q2,q3,q4,โฆโฆโฆโฆ.,qn ) in Taylorโs series about q0i we may write,
๐ (๐1, ๐2, โฆ โฆ โฆ . , ๐๐)
= ๐(๐01, ๐02, โฆ โฆ โฆ . , ๐0๐) + โ (
๐๐
๐๐๐
)
0
๐
๐
๐๐ +
1
2
โ โ (
๐2
๐
๐๐๐๐๐๐
)
0
๐๐
๐,๐
๐,๐
๐๐ + โฏ
โฆโฆโฆโฆโฆโฆโฆ.(iii)
The first term on the right hand side of the equation is constant and hence may be taken to be
equal to zero and the second term vanishes due to equation (i). So the first approximation we may
write,
๐ (๐1, ๐2, โฆ โฆ โฆ . , ๐๐) =
1
2
โ โ (
๐2
๐
๐๐๐๐๐๐
)
0
๐๐
๐,๐
๐,๐
๐๐ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ . (๐๐ฃ)
Writing Vij for (
๐2๐
๐๐๐๐๐๐
)
0
we get,
๐ (๐1, ๐2, โฆ โฆ โฆ . , ๐๐) =
1
2
โ โ ๐๐๐
๐,๐
๐๐๐๐ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฃ)
We shall now examine the expression for kinetic energy of the system of particles which is given
by,
๐ = โ โ
1
2
๐๐๐๐ฬ๐๐ฬ๐
๐,๐
Here, ๐๐๐ = ๐๐๐, ๐. ๐. ๐๐๐ form a symmetric matrix. From Taylorโs series we may write,
๐๐๐(๐1, ๐2, โฆ โฆ โฆ . , ๐๐) = ๐๐๐(๐01, ๐02, โฆ โฆ โฆ . , ๐0๐) + โ(
๐๐๐๐
๐๐๐
)0
๐
๐๐ + โฏ
โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ(vi)
Hence the kinetic energy becomes
๐ =
1
2
{๐๐๐(๐01, ๐02, โฆ โฆ โฆ . , ๐0๐) + โ (
๐๐๐๐
๐๐๐
)
0
๐
๐๐ + โฏ }๐ฬ๐๐ฬ๐
This equation is quadratic in ๐ฬ๐โs and the lowest non-vanishing approximation to T is obtained by
dropping all the terms on the right hand side of the above equation except the first term.
Therefore,
5. ๐ =
1
2
โ โ ๐๐๐(๐01, ๐02, โฆ โฆ โฆ . , ๐0๐)๐ฬ๐๐ฬ๐ =
1
2
โ โ ๐๐๐ ๐ฬ๐๐ฬ๐
โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ..(vii)
Here the constant value of mij at equilibrium is denoted by Tij.
Hence the Lagrangian function is,
๐ฟ = ๐ โ ๐ =
1
2
โ โ(๐๐๐
๐,๐
๐ฬ๐๐ฬ๐ โ ๐๐๐๐๐๐๐)
Thus the n-equations of motion derived from the function L is
๐
๐๐ก
(
๐๐ฟ
๐๐ฬ๐
) โ (
๐๐ฟ
๐๐๐
) = 0 (๐ = 1,2,3, โฆ . . , ๐)
๐๐, โ ๐๐๐ ๐ฬ๐ + โ ๐๐๐๐๐ = 0โฆโฆโฆโฆโฆโฆโฆ..(viii)
In the above equation both Tij and Vij have symmetric properties.
We can write this equation in the following matrix form,
๐๐ฬ + ๐๐ = 0โฆโฆโฆโฆโฆ...............โฆโฆ..(ix)
where ๐ = (
๐11 ๐12 โฆ ๐1๐
๐21 ๐22 โฆ ๐2๐
โฆ โฆ โฆ โฆ
๐๐1 ๐๐2 โฆ ๐๐๐
) ; ๐ = (
๐11 ๐12 โฆ ๐1๐
๐21 ๐22 โฆ ๐2๐
โฆ โฆ โฆ โฆ
๐๐1 ๐๐2 โฆ ๐
๐๐
) ; ๐ = (
๐1
๐2
โฎ
๐๐
)
Here T denotes the inertial coefficient matrix and V the stiffness coefficient matrix. V plays the
role of restoring force which tends to bring the system to the position of equilibrium which may
be called restoring tensor.
b) Eigen Value Problem and Normalization-
For similarity to the equation of simple harmonic oscillator, we try an oscillator solution of
equation (viii) in the form,
๐๐ = ๐๐๐๐๐๐ก
โฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆโฆ.(x)
The idea behind the above trial solution is that ฯ is independent of j, i.e. all the co-ordinates are
assumed to execute SHM with same period but different amplitude (and either in-phase or out-of-
phase only, depending on the sign of aj ). aj are in general complex and only the real parts of them
correspond to the actual motion. Putting equation (x) in equation (viii) we get,
โ(๐๐๐ โ ๐2
๐๐๐)๐๐ = 0
๐
โฆโฆโฆโฆโฆโฆโฆ..(xi)
(๐ = 1,2,3, โฆ โฆ โฆ โฆ โฆ , ๐)
which is a set of n linear homogeneous equations for
the ajโs and consequently for a non-trivial solution (i.e., not all aj =0) the determinant of the
coefficients must vanish i.e. โน๐๐๐ โ ๐2
๐๐๐โน = 0
๐๐, |
๐11 โ ๐2
๐11 โฆ โฆ ๐1๐ โ ๐2
๐1๐
โฎ โฎ
โฎ โฎ
๐๐1 โ ๐2
๐๐1 โฆ โฆ ๐
๐๐ โ ๐2
๐๐๐
| = 0
....................................... (xii)
This is called the secular determinant. This determinant condition is an algebraic equation of nth
degree for ๐2
and hence provides n- values of ๐2
for which equation (x) represents the correct
6. solutions to the equations of motion. For each of these values of ๐2
the equation (xi) may be
solved for the amplitudes or more precisely for (n-1) of the amplitudes in terms of the remaining
one. This is due to the fact that the coefficient aj are determined only up to a common multiplier.
This is apparent from equation (xi). Thus the ratios of ajโs are definite (for a particular frequency)
but not their absolute values. This arbitrariness in the values of aj will be utilized in the
normalization procedure.
Since there exist n values of ๐๐, we also get n- set of values of
amplitudes. Each of these set (i.e., n- values of ajโs corresponding to a particular frequency(๐๐)
can be considered to define the components of a n- dimensional vector ๐
โ๐. This ๐
โ๐ is called the
eigen vector of the system associated with the eigen frequency๐๐. The symbol ajk may now be
used to represent the j-th component of the k-th eigen vector. Using the principle of
superposition, the general solution may now be written as,
๐๐(๐ก) = โ ๐๐๐๐๐๐๐๐ก
๐
....................................... (xiii)
where motion of each ฮทj is compounded of motions with each of the n values of frequencies ๐๐.
The actual motion is the real part of the complex equation (xiii) which can be expressed as,
๐๐(๐ก) = โ ๐๐๐ ๐ถ๐๐ (
๐
๐๐๐ก + ๐๐)
The motion is oscillatory about the stable equilibrium position only for ๐๐
2
> 0. If two or more
values of ๐๐ happen to be the same, then the phenomenon is known as degeneracy.
c) Orthogonality of Eigenvectors-
Equation (xi) represents a special type of eigen value equation as,
๐ฝ๐ = ๐๐ป๐ ..........................................(xiv) with ๐2
= ๐
where the matrix V,T and a are
๐ฝ = (
๐11 ๐12 โฆ ๐1๐
๐21 ๐22 โฆ ๐2๐
โฆ โฆ โฆ โฆ
๐๐1 ๐๐2 โฆ ๐
๐๐
), ๐ป = (
๐11 ๐12 โฆ ๐1๐
๐21 ๐22 โฆ ๐2๐
โฆ โฆ โฆ โฆ
๐๐1 ๐๐2 โฆ ๐๐๐
), ๐ = (
๐1
๐2
โฎ
๐๐
)
In ordinary eigenvalue problem, ๐ฝ๐ = ๐๐ i.e. the effect of V on eigenvector a is just to generate a
vector which is ๐ times a, ๐ being a scalar. Here, however, V acting on a, generates a multiple (๐)
of the result of T acting on a.
A typical equation for the eigenvalue ๐๐ can be written from equation (xiv) as,
โ ๐๐๐๐๐๐ = ๐๐ โ ๐๐๐๐๐๐
๐
๐
......................................(xv)
The complex conjugate of similar equation for ฮปl gives,
โ ๐๐๐๐๐๐
โ
= ๐๐
โ
โ ๐๐๐
๐
๐
๐๐๐
โ
.........................................(xvi)
(since Vij and Tij are real and also symmetric)
Multiplying equation (xv) by ๐๐๐
โ
and summing over i and multiplying equation (xvi) by ajk and
summing over j, and next subtracting the latter from the former, we get
7. (๐๐ โ ๐๐
โ
) โ ๐๐๐๐๐๐
๐๐
๐๐๐
โ
= 0 โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฅ๐ฃ๐๐)
๐๐, (๐๐ โ ๐๐
โ
) โ ๐๐๐๐๐๐
๐๐
๐๐๐
โ
= 0
............................................(xviii) for l=k
Now, ๐๐๐ = ๐ผ๐๐ + ๐๐ฝ๐๐ as ajk is complex.
and
โ ๐๐๐๐๐๐๐๐๐
โ
= โ ๐๐๐๐ผ๐๐
๐๐
๐ผ๐๐ + โ ๐๐๐๐ฝ๐๐
๐๐
๐ฝ๐๐ + ๐ โ ๐๐๐(๐ฝ๐๐๐ผ๐๐ โ ๐ผ๐๐๐ฝ๐๐) โฆ โฆ โฆ โฆ . . (๐ฅ๐๐ฅ)
Tij being symmetric, the interchange of dummy indices i and j changes sign
of the third summation on the right. Therefore, the imaginary term vanishes in equation (xix) and
the two real sums are twice the kinetic energy when the velocities ๐ฬ๐ have values ๐ผ๐๐ and ๐ฝ๐๐
respectively. The kinetic energy can never be zero for real velocities. Thus, the eigenvalues ๐๐
must be real.
Eigenvalues and eigenvectors being real, the equation (xvii) may be written as
(๐๐ โ ๐๐) โ ๐๐๐๐๐๐
๐๐
๐๐๐ = 0 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ . (๐ฅ๐ฅ)
When the roots of the characteristic equation (xii) are all distinct, ๐๐ โ ๐๐, we shall get,
โ ๐๐๐๐๐๐ ๐๐๐ = 0 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฅ๐ฅ๐)(๐๐๐ ๐ โ ๐)
This shows that the eigenvectors are orthogonal.
The eigenvalue equation (xi) canโt completely fix the values of ajk, as has already been stated. The
uncertainty can however be removed be removed by setting, purely arbitrarily,
โ ๐๐๐๐๐๐
๐๐
๐๐๐ = 1 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฅ๐ฅ๐๐)(๐๐๐ ๐ = ๐)
Each of n such equations uniquely fix the arbitrary component in each eigenvector
(corresponding to a given eigenfrequency). Combining the equations (xxi) and (xxii), we get
โ ๐๐๐๐๐๐
๐๐
๐๐๐ = ๐ฟ๐๐
๐. ๐. ๐
ฬ๐ป๐ = ๐ฐ................................(xxiii)
where ๐
ฬ being the transpose of a.
Introducing a diagonal matrix ฮป with elements ๐๐๐ = ๐๐๐ฟ๐๐, the eigenvalue equation (xv) becomes
โ ๐๐๐๐๐๐ = ๐๐ โ ๐๐๐๐๐๐
๐
๐
= โ ๐๐๐
๐๐
๐๐๐๐๐๐
๐. ๐. ๐ฝ๐ = ๐ป๐๐ โฆ โฆ โฆ โฆ โฆ โฆ โฆ . . (๐ฅ๐ฅ๐๐ฃ)
Therefore, ๐
ฬ๐ฝ๐ = ๐
ฬ๐ป๐๐ = ๐..........................(xxv)
This shows that the matrix a diagonalises both T and V simultaneously.
d) Normal co-ordinates-
The normalization of ajk leaves no ambiguity in the solution for ๐๐. To remove the loss of
generality due to the introduction of arbitrary normalization, a scalar factor ฮฒk may be used when
the equation (xiii) becomes
8. ๐๐(๐ก) = โ ๐๐๐๐ฝ๐๐๐๐๐๐ก
= โ ๐๐๐๐๐ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฅ๐ฅ๐ฃ๐)
๐
๐
where ๐๐ = ๐ฝ๐๐๐๐๐๐ก
............................(xxvii)
Therefore, ๐ผ
ฬ = ๐
ฬ๐ธ
ฬ..........................(xxviii)
Equation (xxvii) shows that Qk oscillates with one frequency only, called normal frequency,
namely ๐๐. These Qkโs are called the normal co-ordinates of the system. The normal co-ordinates
are thus defined as the generalized co-ordinates where each one of them executes oscillations
with a single frequency. If expressed in terms of normal co-ordinates, potential energy and kinetic
energy take rather simple forms as given below,
๐. ๐ธ. = ๐ =
1
2
โ ๐๐๐๐๐๐๐ =
1
2
๐๐
๐ผ
ฬ๐ฝ๐ผ =
1
2
๐
ฬ๐
ฬ๐ฝ๐๐ =
1
2
๐
ฬ๐๐
=
1
2
โ ๐๐๐๐๐๐๐
๐๐
=
1
2
โ ๐๐๐๐๐ฟ๐๐๐๐ =
๐๐
1
2
โ ๐๐
2
๐๐
2
๐
โฆ โฆ โฆ โฆ . . (๐ฅ๐ฅ๐๐ฅ)
(using equations (xxviii) and (xxv))
๐พ. ๐ธ = ๐ =
1
2
โ ๐ฬ๐๐๐๐
๐๐
๐ฬ๐ =
1
2
๐ฬ
ฬ๐๐ฬ =
1
2
๐ฬ
ฬ๐
ฬ๐ป๐๐ฬ =
1
2
๐ฬ
ฬ๐ฬ =
1
2
โ ๐ฬ๐
2
โฆ โฆ โฆ โฆ โฆ โฆ . (๐ฅ๐ฅ๐ฅ)
๐
So, the new Lagrangian is,
๐ฟ = ๐ โ ๐ =
1
2
โ(๐ฬ๐
2
โ
๐
๐๐
2
๐๐
2
) โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ . . (๐ฅ๐ฅ๐ฅ๐)
Therefore, the Lagrangeโs equation of motion are given by,
๐
๐๐ก
(
๐๐ฟ
๐๐ฬ๐
) โ
๐๐ฟ
๐๐
= 0
๐๐, ๐ฬ๐ + ๐๐
2
๐๐ = 0 โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ โฆ (๐ฅ๐ฅ๐ฅ๐๐) (k=1,2,3......)
This is the set of n-independent equations of SHM with single normal frequency ๐๐.
The solution of the equation (xxxii) is given by,
๐๐ = ๐๐๐ถ๐๐ (๐๐๐ก) + ๐๐๐๐๐(๐๐๐ก), ๐คโ๐๐ ๐๐
2
> 0
๐๐ = ๐๐๐ก + ๐๐๐ก, ๐คโ๐๐ ๐๐
2
= 0
๐๐ = ๐๐๐๐๐๐ก
+ ๐๐๐โ๐๐๐ก
, ๐คโ๐๐ ๐๐
2
< 0
For ๐๐
2
> 0, the co-ordinates always remain finite. Such solutions refer to stable
equilibrium. For ๐๐
2
= 0 and ๐๐
2
< 0, the co-ordinates become infinite as time advances and
such solutions refer to unstable equilibrium.
e) Normal co-ordinates Qk in terms of co-ordinates ฮทj
The complex quantity ฮฒk may be written as,
9. ๐ฝ๐ = ๐๐ + ๐๐๐.................................................(xxxiii)
Therefore, equation (x) becomes,
๐๐(๐ก) = โ ๐๐๐(
๐
๐๐ + ๐๐๐)๐๐๐๐๐ก
............................(xxxiv)
On differentiating with respect to time we get,
๐
ฬ ๐(๐ก) = โ ๐๐๐๐๐๐(
๐
๐๐ + ๐๐๐)๐๐๐๐๐ก
................................................(xxxv)
The real part of equation (xxxiv) gives at t=0,
๐๐(0) = โ ๐๐๐
๐
๐๐
............................................(xxxvi)
Multiplying throughout by Tjiair and summing over ij, and using equation (xxxvi), we get,
โ ๐๐(0)๐๐๐๐๐๐ = โ ๐๐
๐
โ ๐๐๐
๐๐
๐๐
๐๐๐๐๐๐ = โ ๐๐๐ฟ๐๐ = ๐๐
๐
..........................(xxxvii)
Similarly, the real part of equation (xxxv) at t=0 gives,
๐ฬ๐(0) = โ โ ๐๐
๐
๐๐๐๐๐
...................................(xxxviii)
Multiplying both sides by Tjiair and summing over ij, and using equation(xxxvi), we get,
โ ๐ฬ๐(0)๐๐๐๐๐๐ = โ โ ๐๐๐๐
๐
โ ๐๐๐
๐๐
๐๐
๐๐๐๐๐๐ = โ โ ๐๐๐๐๐ฟ๐๐ = โ๐๐
๐
๐๐
.......................(xxxix)
Thus, the normal co-ordinates Qr may be expressed as the real part of the expression in,
๐๐(๐ก) = ๐ฝ๐๐๐๐๐๐ก
= (๐๐ + ๐๐๐)๐๐๐๐๐ก
= โ ๐๐๐๐๐๐[
๐๐
๐๐(0) โ
๐
๐๐
๐ฬ๐(0)]๐๐๐๐๐ก
Therefore, for any arbitrary ๐๐(0) and ๐ฬ๐(0), a set of normal co-ordinates Qr may be found, each
of which varies harmonically with single frequency ๐๐.
f) Energy in normal co-ordinates
Here,
๐ =
1
2
โ ๐๐
2
๐๐
2
๐ and ๐ =
1
2
โ ๐ฬ๐
2
๐
So, total mechanical energy E, in normal co-ordinates, is given by,
๐ธ = ๐ + ๐ =
1
2
โ(๐ฬ๐
2
+ ๐๐
2
๐๐
2
)
๐
This is true for small oscillations with any number n of degrees of freedom.
Examples-
10. 1. The potential energy V(x) of a particle is given by,
๐(๐ฅ) = 3๐ฅ4
โ 8๐ฅ3
โ 6๐ฅ2
+ 24๐ฅ
Determine the points of stable and unstable equilibrium.
Solution-
For the points of equilibrium(๐๐/๐๐ฅ) = 0. Since, we have
๐(๐ฅ) = 3๐ฅ4
โ 8๐ฅ3
โ 6๐ฅ2
+ 24๐ฅ
So,
๐๐
๐๐ฅ
= 12๐ฅ3
โ 24๐ฅ2
โ 12๐ฅ + 24
At the points of equilibrium, therefore, 12๐ฅ3
โ 24๐ฅ2
โ 12๐ฅ + 24 = 0
๐๐, ๐ฅ3
โ 2๐ฅ2
โ ๐ฅ + 2 = 0
๐๐, (๐ฅ + 1)(๐ฅ โ 1)(๐ฅ โ 2) = 0
Therefore, the points of equilibrium are
x= -1,1 and 2.
To test the stability of equilibrium at those points, we find
๐2๐
๐๐ฅ2 at the points and if it is positive, the
equilibrium is stable; if negative, it is unstable.
Now,
๐2๐
๐๐ฅ2 = 36๐ฅ2
โ 48๐ฅ โ 12
So, (
๐2๐
๐๐ฅ2)
๐ฅ=โ1
= 72 (๐๐๐ ๐๐ก๐๐ฃ๐)
(
๐2
๐
๐๐ฅ2)
๐ฅ=1
= โ24 (๐๐๐๐๐ก๐๐ฃ๐)
(
๐2
๐
๐๐ฅ2)
๐ฅ=2
= 36 (๐๐๐ ๐๐ก๐๐ฃ๐)
Therefore, equilibrium are stable at x = -1 and x = 2; unstable at x = 1.
2. Consider two identical simple harmonic oscillators coupled together. Their K.E. and P.E. are
given by ๐ =
1
2
๐(๐ฬ1
2
+ ๐ฬ2
2
) and ๐ =
1
2
๐(๐1
2
+ ๐2
2) โ โ๐1๐2
(i) Find the Hamiltonian of the system.
(ii) Fine the frequencies for normal modes and the ratio of the amplitudes of each of the
coordinates q1 and q2 in those modes.
(iii) Introduce the normal coordinates and thereby set up the corresponding Lagrangeโs
equation.
Solution-
(i) Here, V is independent of the velocity, so it represents a conservative system.
Thus the Hamiltonian is given by,
๐ป = ๐ + ๐ =
1
2
๐(๐ฬ1
2
+ ๐ฬ2
2) +
1
2
๐(๐1
2
+ ๐2
2) โ โ๐1๐2
(ii) Here, Lagrangian is
๐ฟ = ๐ โ ๐ =
1
2
๐(๐ฬ1
2
+ ๐ฬ2
2) โ
1
2
๐(๐1
2
+ ๐2
2) + โ๐1๐2....................(i)
Now, we get,
๐๐ฟ
๐๐ฬ1
= ๐๐ฬ1
โ
๐๐ฟ
๐๐1
= ๐๐1 โ โ๐2
11. ๐๐ฟ
๐๐ฬ2
= ๐๐ฬ2
โ
๐๐ฟ
๐๐2
= ๐๐2 โ โ๐1
Putting these values in equation (i) we get,
๐๐ฬ1 + ๐๐1 โ โ๐2 = 0...................(ii)
๐๐ฬ2 + ๐๐1 โ โ๐2 = 0...................(iii)
Let ๐1 = ๐ด1๐๐๐๐ก
. Putting this value in equation (ii) we get,
๐ด1(๐ โ ๐2
๐) โ ๐ด2โ = 0
Again, let ๐2 = ๐ด2๐๐๐๐ก
. Putting this value in equation (iii) we get,
โ๐ด1โ + ๐ด2(๐ โ ๐2
๐) = 0
For non-zero solution, we must have the determinant,
|๐ โ ๐2
๐ โโ
โโ ๐ โ ๐2
๐
| = 0
๐๐, (๐ โ ๐2
๐)2
โ โ2
= 0
๐๐, ๐ โ ๐2
๐ = ยฑโ
Therefore, ๐2
= (๐ โ โ)
So, ๐1 = โ
(๐โโ)
๐
, ๐2 = โ
(๐+โ)
๐
Now,
๐ด2
๐ด1
=
(๐โ๐1
2
๐)
โ
๐๐,
(๐โ๐2
2
๐)
โ
(iii) Let the normal coordinates are ๐๐ = ๐ด๐๐๐๐๐๐ก
Differentiating with respect to t we get,
๐ฬ๐ + ๐๐
2
๐๐ = 0, ๐ = 1,2.
Thus Lagrangeโs equation is,
๐ฟ๐ =
1
2
๐ฬ๐
2
โ
1
2
๐๐๐๐
2
Exercise-
1. The potential energy of a particle is given by
๐(๐ฅ) = ๐ฅ4
โ 4๐ฅ3
โ 8๐ฅ2
+ 48๐ฅ
Find the points of stable and unstable equilibrium. (KNU-2019)
2. Discuss the general theory of small oscillation of a system to obtain the equation of motion near
its equilibrium position. How the eigen frequencies of the system can be obtained for such a
system. (KNU-2019)
3. What is normal mode and normal frequency? (KNU-2019)
4. Discuss the stability of a simple pendulum and show that it can oscillate only about the position
of its stable equilibrium.
5. Establish the relation
๐๐ฬ + ๐๐ = 0,for small oscillations of a system, the symbols
having their usual meanings.
6. A particle moves in potential energy given by
12. ๐(๐ฅ) = ๐๐ฅ2
+
๐
๐ฅ2, a,b>0. Show that its frequency of oscillation is โ
8๐
๐
.
7. What is meant by equilibrium, stable and unstable equilibrium? Show that oscillations can occur
only about a stable equilibrium.
8. Write down the differences between stable and unstable equilibrium.
References-
1) S.N.Maithi & D.P.Raychaudhuri, Classical Mechanics and General Properties of Matter, New Age International Publishers,
2015
2) A.B.Gupta, Fundamentals of Classical Mechanics, 2nd
Edition, Books And Allied (P) Ltd. (2019)
3) A. Beiser, Perspective of Modern Physics, 3rd
Edition, New York, McGraw Hill (1969).
AJAY KUMAR SHARMA
Assistant Professor (Physics)
B.C.College, Asansol
Email-id- ajay888sharma@yahoo.co.in