classical mechanics fall 2011 chapter 11: coupled...
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Classical Mechanics Fall 2011
Chapter 11: Coupled Oscillators and Normal Modes
1. Two Masses Coupled By Three Springs
There are many interesting systems in which individual oscillators are coupled by some kind of interaction. Because of the coupling, the motion of each oscillator is influenced by others with which it interacts. An example of such a coupled system is a crystalline solid in which each of the many atoms that constitute the crystal interacts with other atoms in the crystal via an interatomic force. We begin our study of coupled oscillators with a much simpler system than a crystal but which nevertheless provides valuable information on the behavior of more complex systems: two objects coupled by three springs.
The springs are ideal springs, which obey Hooke’s law for all extensions (and compressions). The leftmost and rightmost springs are attached to rigid supports at one end. The objects, which we take to be carts, move on a frictionless horizontal surface. Since this is a relatively simple system, we will use Newton’s second law to write down the equations of motion. Note that we can find the force exerted on a cart by a spring by considering the displacement of the end of the spring attached to the cart relative to the other end of that spring. This is important in cases in which both ends of a spring are attached to moving carts so that the displacement of each end of the spring changes with time. The equations of motion are given below.
Cart 1: m1x1 = −k1x1 − k2 (x1 − x2 ).
Cart 2: m2x2 = −k3x2 − k2 (x2 − x1).
We can rewrite the equations of motion as follows:
m1x1 = −(k1 + k2 )x1 + k2x2 (11.1)
and
m2x2 = k2x1 − (k2 + k3)x2. (11.2)
Eqs. (11.1) and (11.2) can be collectively written using a matrix equation:
Mx = −Kx, (11.3)
where x is the column matrix
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x =x1x2
⎛
⎝⎜⎜
⎞
⎠⎟⎟, (11.4)
M is the mass matrix
m1 00 m2
⎛
⎝⎜⎜
⎞
⎠⎟⎟, (11.5)
and K is the spring-‐constant matrix
K =k1 + k2 −k2−k2 k2 + k3
⎛
⎝⎜⎜
⎞
⎠⎟⎟. (11.6)
Using matrices is a very powerful method of doing mechanics, as we shall presently see.
In-class Problem: Taylor, Problem 11.2
Since the system involves springs and masses and no friction, we look for oscillatory solutions to the equations of motion. Let us try
x1 = F1 cosωt +G1 sinωtx2 = F2 cosωt +G2 sinωt.
(11.7)
To facilitate finding the solutions, let us use complex numbers to represent the trial solutions. We have
z1 = C1e
i(ωt−δ1 ) = a1eiωt
z2 = C2ei(ωt−δ2 ) = a2e
iωt . (11.8)
In Eq. (11.8), a1 = C1e− iδ1 and a2 = C2e
− iδ2 , where a1, a2, C1, and C2 are constants. (a1 and a2 are complex, C1 and C2 are real.) The actual solutions (the displacements x1 and x2 ) are the real parts of z1 and z2 . In matrix form, we have
z =a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟eiωt = aeiωt , (11.9)
with x = Rez. Substituting the trial solution Eq. (11.9) in the matrix equation of motion (11.3) gives
−ω 2Maeiωt = −Kaeiωt ,
which can be rewritten as
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(K −ω 2M)a = 0. (11.10)
This is an eigenvalue-‐type equation for which solutions exist if
det(K −ω 2M) = 0. (11.11)
The eigenvalues are the values of ω 2 obtained from solving Eq. (11.11). After obtaining the eigenvalues, one can then find the column matrices a , i.e., the eigenvectors, by substituting each eigenvalue into Eq. (11.10). One then has the complete solution after taking the real part of z. Let us illustrate the technique by solving the problem of two identical masses coupled by three identical springs.
Two Identical Masses Coupled By Three Identical Springs
Let m1 = m2 = m and k1 = k2 = k3 = k.The mass and spring-‐constant matrices are
M = m 00 m
⎛⎝⎜
⎞⎠⎟
and K = 2k −k−k 2k
⎛⎝⎜
⎞⎠⎟
. (11.12)
Eq. (11.11) gives
2k −mω 2 −k−k 2k −mω 2
= 0,
which yields
(2k −mω 2 )2 − k2 = 0,
or,
(k −mω 2 )(3k −mω 2 ) = 0.
The solutions are
ω1 =km
and ω 2 =3km
. (11.13)
These two frequencies are called the normal-mode frequencies or normal frequencies of the coupled oscillator system. Let us now substitute ω1 into Eq. (11.10). We have
k −k−k k
⎛⎝⎜
⎞⎠⎟
a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 0,
which gives two equations, both of which are equivalent to a1 − a2 = 0, i.e., a1 = a2. Eq. (11.8) tells us that
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z1 = z2 = aeiω1t = A1e
i(ω1t−δ1 ),
where we have taken a1 = a2 = a and a = A1eiδ1 . In matrix form, we have
z = aa
⎛⎝⎜
⎞⎠⎟eiω1t =
A1A1
⎛
⎝⎜⎜
⎞
⎠⎟⎟ei(ω1t−δ1 ).
Taking the real parts gives the solution for the first normal mode:
x1(t) = A1 cos(ω1t −δ1)x2 (t) = A1 cos(ω1t −δ1). [First normal mode]
(11.14)
Clearly, both masses oscillate precisely in phase, having the same displacement from equilibrium at all times. It follows that the middle spring is always relaxed and has no effect on the motion. That is why the first normal frequency is exactly that of a single mass attached to one spring.
We now substitute ω 2 into Eq. (11.10), yielding
−k −k−k −k
⎛⎝⎜
⎞⎠⎟
a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 0,
which gives a1 + a2 = 0, i.e., a1 = −a2. Thus, we obtain
z = a−a
⎛⎝⎜
⎞⎠⎟eiω2t =
A2−A2
⎛
⎝⎜⎜
⎞
⎠⎟⎟ei(ω2t−δ2 ),
where a = A2eiδ2 . Taking the real part gives the displacements as a function of time for the
second normal mode:
x1(t) = A2 cos(ω 2t −δ 2 )x2 (t) = −A2 cos(ω 2t −δ 2 ). [Second normal mode]
(11.15)
In this mode, the masses oscillate exactly out of phase. The displacements have the same magnitude but are opposite in direction. For any given displacement, the middle spring is stretched or compressed by an amount equal to twice the magnitude of that displacement. Along with the force exerted by the leftmost or rightmost spring, the magnitude of the net force on each mass is 3k x . This explains the value of the second normal frequency. The general solution is a superposition of the solutions for the first and second modes. (Remember that the equations of motion are linear so the principle of superposition is valid.) Thus, the general solution is
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x1(t) = A1 cos(ω1t −δ1)+ A2 cos(ω 2t −δ 2 )x2 (t) = A1 cos(ω1t −δ1)− A2 cos(ω 2t −δ 2 ). [General solution]
(11.16)
There must be four unknown constants because two second-‐order differential equations were solved. In matrix form, the general solution is
x(t) = A111
⎛⎝⎜
⎞⎠⎟cos(ω1t −δ1)+ A2
1−1
⎛⎝⎜
⎞⎠⎟cos(ω 2t −δ 2 ). (11.17)
We can rewrite Eq. (11.17) in the following form, which is sometimes more convenient:
x(t) = (B1 cosω1t +C1 sinω1t)11
⎛⎝⎜
⎞⎠⎟+ (B2 cosω 2t +C2 sinω 2t)
1−1
⎛⎝⎜
⎞⎠⎟. (11.18)
The general motion is therefore a combination of oscillations at both normal frequencies. Note that there are two normal frequencies because the system consists of two oscillators. Solving associated 2 × 2 determinant equation gives a quadratic equation in ω 2 , which has exactly two solutions.
In-class Problem: Taylor, Problem 11.7.
2. Normal Coordinates
Let us add Eqs. (11.1) and (11.2) with the masses equal (= m) and the spring constants all the same (= k). We obtain
m(x1 + x2 ) = −k(x1 + x2 ). (11.19)
Let us define the normal coordinate
ξ1 = x1 + x2. (11.20)
Substituting ξ1 in Eq. (11.19) gives
mξ1 = −kξ1. (11.21)
Eq. (11.21) is the equation of motion for a single harmonic oscillator with angular frequency ω1 = k /m . We see that rewriting the equations of motion in terms of the normal coordinate causes them to become uncoupled and the resulting motion is oscillation at a single frequency equal to the normal frequency ω1.
We now subtract add Eq. (11.2) from Eq. (11.1) giving
m(x1 − x2 ) = −3k(x1 − x2 ). (11.22)
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Defining a second normal coordinate
ξ2 = x1 − x2 (11.23)
and substituting it in Eq. (11.22) yields
mξ2 = −3kξ2. (11.24)
The second normal coordinate gives oscillation at the second normal frequency ω 2 = 3k /m only. We can initiate oscillations with a single frequency using initial conditions. If both masses are given the same initial displacement and released from rest, then the normal coordinate ξ1 will initially be non-‐zero whereas the normal coordinate ξ2 will initially be zero and will remain zero for all time. Hence, the system will oscillate with normal frequency ω1 .
In-class Question: How would you generate oscillations at a single frequency ω 2 ?
Note that if we write the equations of motion for the normal coordinates in matrix form, we obtain
Mξ = − ′K ξ, (11.25)
where
ξ =ξ1ξ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟,
M = m 00 m
⎛⎝⎜
⎞⎠⎟
and ′K = k 00 3k
⎛⎝⎜
⎞⎠⎟
.
Thus, the normal coordinates diagonalize the M and K matrices. (In this simple case, the M matrix was diagonal from the start.) This happens because the normal coordinates can be written in terms of the eigenvectors for the system. In fact, notice that we can write the normal coordinates of the oscillators in matrix form as
ξ = x1a1 + x2a2 = x111
⎛⎝⎜
⎞⎠⎟+ x2
1−1
⎛⎝⎜
⎞⎠⎟=
x1 + x2x1 − x2
⎛
⎝⎜⎜
⎞
⎠⎟⎟=
ξ1ξ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟. (11.26)
The normal coordinates can be expanded in a vector space in which the eigenvectors a1 and a2 that solve Eq. (11.10) play the role of unit vectors, i.e., they are the basis vectors that span the space. The displacements are the “components” or coefficients of the basis vectors. We can also write the displacements in terms the same basis vectors with the normal coordinates being the coefficients:
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x = 12 ξ1a1 + 1
2 ξ2a2 =12
ξ1 + ξ2ξ1 −ξ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟. (11.27)
Note that in this case, ξ1 is the coefficient of a1 and so is associated with the first normal mode only. (ai and ξi can be defined so that that the inelegant factor of 12 goes away.)
In-class problem: Taylor, Problem 11.11.
3. Two Weakly Coupled Oscillators
Let us assume that in our two-‐coupled-‐oscillator system, the masses are identical (m) and the rightmost and left most springs are identical (spring constant k). We further assume that the middle spring has a spring constant k2 that is different from that of the other springs. The spring constant matrix is
K =k + k2 −k2−k2 k + k2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
and
K −ω 2M =k + k2 −mω
2 −k2−k2 k + k2 −mω
2
⎛
⎝⎜⎜
⎞
⎠⎟⎟.
Setting the determinant equal to zero and solving yields
ω1 =km
and ω 2 =k + 2k2
m. (11.28)
We see that we obtain the same normal frequencies for the three identical springs case if k2 = k . As before, for the first normal mode, the masses oscillate in phase; for the second, they oscillate out of phase.
Let us consider the case of a weak middle spring, i.e., k2 k . In this case, the second normal frequency is very close to the first. It is useful to observe that superposing two sinusoidal functions of the same amplitude but different frequencies gives
′A cosω1t + ′A cosω 2t = 2 ′A cos ω1 −ω 2
2⎛⎝⎜
⎞⎠⎟ t
⎡⎣⎢
⎤⎦⎥cos ω1 +ω 2
2⎛⎝⎜
⎞⎠⎟ t
⎡⎣⎢
⎤⎦⎥. (11.29)
Thus, if we use initial conditions to give A1 = A2 and δ1 = δ 2 (e.g., cart 2 is given an initial displacement from equilibrium and released from rest; cart 1 is at rest at equilibrium) in the general solution similar to Eq. (11.16), we obtain the solution
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x1(t) = Acosεt cosωtx2 (t) = Asinεt sinωt,
(11.30)
where
ε = ω1 −ω 2
2 and ω = ω1 +ω 2
2. (11.31)
For weak coupling, the average frequency ω is very close to both normal frequencies and the difference ε is very small. Eq. (11.30) underlies the phenomenon of beats, which represent a slow periodic variation of the amplitude when two waves with frequencies that are nearly the same are superposed. Here, the beats are due to the superposition of normal-‐mode oscillations (corresponding to the two normal coordinates) that have nearly equal frequencies. The period of the amplitude variation is 2π / ε . From Eq. (11.30), we see that when the time varying amplitude of mass 1 is maximum ( cosεt = 1), the amplitude of mass 2 is minimum and vice-‐versa. Thus, the energy of the system is continuously being transferred from one oscillator to the other.
In-class Problem: Taylor, Problem 11.13 (a)
4. The Double Pendulum: Lagrangian Approach
The double pendulum system (depicted below) is quite a bit more complicated than the cart-‐spring system that we have analyzed. Therefore, we will use the Lagrangian formulation to obtain the equations of motion.
Weakly coupled oscillators
x1
x2
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The two generalized coordinates are φ1 and φ2 . We take the zero of gravitational potential energy to be the fixed support to which the upper pendulum is attached. The potential energy of each mass is given by
U1 = −m1gL1 cosφ1 and U2 = −m2g(L1 cosφ1 + L2 cosφ2 ).
Hence, the total potential energy is
U = −(m1 +m2 )gL1 cosφ1 −m2gL2 cosφ2. (11.32)
The kinetic energy of mass 1 is given by
T1 =12 m1L1
2 φ12.
Note that the velocity of mass 2 relative to an inertial frame (e.g., one attached to the fixed support) is the vector sum of the velocity of mass 2 relative to mass 1 and the velocity of mass 1. Thus, the square of the velocity of mass 2 is calculated as
v22 = v2 ⋅
v2 = (v1 +v21) ⋅(
v1 +v21) = v1
2 + v212 + 2v1 ⋅
v21 = L12 φ1
2 + L22 φ2
2 + 2L1L2 φ1 φ2 cos(φ1 −φ2 ),
and T2 = 12 m2v2
2. Thus, the total kinetic energy is
T = 12 (m1 +m2 )L1
2 φ12 +m2L1L2 φ1 φ2 cos(φ1 −φ2 )+ 1
2 m2L22 φ2
2. (11.33)
To simply things, let us consider small oscillations. Thus, the generalized coordinates and their time derivatives will be small. We will retain terms up to second order in φi and φi . Thus, cosφi ≈1− 1
2φi2 . Further, the terms φi
2 and φiφ j in the kinetic energy are already second-‐order and so φ1 φ2 cos(φ1 −φ2 ) ≈ φ1 φ2. Then in the small-‐oscillation approximation, the potential energy is, after dropping constant terms,
U = 12 (m1 +m2 )gL1φ1
2 + 12 m2gL2φ2
2. (11.34)
The kinetic energy becomes
T = 12 (m1 +m2 )L1
2 φ12 +m2L1L2 φ1 φ2 + 1
2 m2L22 φ2
2. (11.35)
Note that in the small-‐oscillation approximation, the kinetic energy is a homogeneous quadratic function of the generalized velocities, i.e., every term is a product of two
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generalized velocities multiplied by a constant. Likewise, the potential energy is a homogeneous quadratic function of the generalized coordinates. Applying Lagrange’s equations gives
ddt
∂L∂ φ1
= ∂L∂φ1
, i.e.,
(m1 +m2 )L12φ1 +m2L1L2
φ2 = −(m1 +m2 )gL1φ1. (11.36)
Also,
ddt
∂L∂ φ1
= ∂L∂φ1
, i.e.,
m2L1L2φ1 +m2L2
2 φ2 = −m2gL2φ2. (11.37)
Note that because T and U are homogeneous quadratic functions, the equations of motion are linear and so can be solved using matrix methods and other techniques. As usual, we can rewrite the equations of motion in matrix form:
Mφ = −Kφ, (11.38)
where
M =(m1 +m2 )L1
2 m2L1L2m2L1L2 m2L2
2
⎛
⎝⎜⎜
⎞
⎠⎟⎟, (11.39)
K =(m1 +m2 )gL1 0
0 m2gL2
⎛
⎝⎜⎜
⎞
⎠⎟⎟, (11.40)
and
φ =φ1φ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟. (11.41)
Let us take m1 = m2 = m and L1 = L2 = L. This gives
M = mL2 2 11 1
⎛⎝⎜
⎞⎠⎟
and K =2mgL 0
0 mgL
⎛
⎝⎜
⎞
⎠⎟ = mL2 2g / L 0
0 g / L
⎛
⎝⎜
⎞
⎠⎟ . (11.42)
For non-‐trivial solutions to the equations of motion to exist, we must have det(K −ω 2M) = 0 . This gives
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mL22(ω 0
2 −ω 2 ) −ω 2
−ω 2 ω 02 −ω 2
= 0,
where ω 0 = g / L is the frequency for a simple pendulum. Solving the resulting equation gives the normal frequencies
ω12 = (2 − 2)ω 0
2 and ω 22 = (2 + 2)ω 0
2, (11.43)
In-class Problem: Fill in the details leading to Eq. (11.43).
To find the eigenvector matrix a corresponding to ω1 , we set (K −ω12M)a = 0 , which gives
mL2ω 02 2( 2 −1) 2 − 2
2 − 2 2 −1
⎛
⎝⎜⎜
⎞
⎠⎟⎟
a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= mL2ω 0
22 2 −1( ) 2 1− 2( )2 1− 2( ) 2 −1
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
= mL2ω 02 2 −1( ) 2 − 2
− 2 1
⎛
⎝⎜⎜
⎞
⎠⎟⎟
a1a2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= 0.
This gives the equations
2a1 − 2a2 = 0 and − 2a1 + a2 = 0,
which are the same equation apart from a constant factor of − 2 . Thus, a2 = 2a1. The solution for the first normal mode is therefore φ = aeiω1t , or,
φ1
φ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= A1
12
⎛
⎝⎜
⎞
⎠⎟ cos ω1t −δ1( ). [First mode] (11.44)
The two masses vibrate in phase but the amplitude of the lower pendulum is 2 times that of the upper one. For the second normal mode, we set (K −ω 2
2M)a = 0 . One finds, using the same procedure as above, that
φ1
φ2
⎛
⎝⎜⎜
⎞
⎠⎟⎟= A2
1− 2
⎛
⎝⎜
⎞
⎠⎟ cos ω 2t −δ 2( ). [Second mode] (11.45)
In this case, the pendulums oscillate out of phase, with the lower pendulum again having an amplitude that is 2 times the upper one.
In-class Problem: Taylor, Problem 11.16.
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5. Lagrangian Approach: General Case
We consider a holonomic system with n degrees of freedom, so that there are n generalized coordinates q1,q2,....,qn . If there are N particles, the Cartesian positions
rα are functions of
the generalized coordinates:
rα = rα (q1,q2,....,qn ). α=1, 2, ...., N (11.46)
To find the velocities, we take the time derivative of the positions and use the chain rule. For example, the time derivative of the x -‐coordinate of particle α is given by
xα = ∂xα
∂qiqi
i=1
n
∑ ,
and
xα2 = ∂xα
∂qi
∂xα∂qjqi qj
j=1
n
∑i=1
n
∑ = Aijα qi qj
ij∑ .
Similar equations can be written down for yα2 and zα
2 . We have assumed in the above that the generalized coordinates are not explicit functions of time, i.e., ∂qi / ∂t = 0.The total kinetic energy is given by
T = 1
2α∑ mαrα2 . Using the expression above, we can therefore
write the kinetic energy of the system as
T = 1
2 Aij qi qjij∑ , (11.47)
where Aij is a function of the generalized coordinates. To obtain an expression for the potential energy, we will perform a Taylor expansion about the equilibrium values of the generalized coordinates, which we assume are zero:
qi = 0 (equilibrium) i = 1,2,...,n
Hence, we have
U(q1,q2,..,qn ) =U(0)+∂U∂qi
⎛⎝⎜
⎞⎠⎟i
∑0
qi +∂2U
∂qi ∂qj
⎛
⎝⎜⎞
⎠⎟ 0qiqj
j∑
i∑ + ... (11.48)
U(0) is a constant, which we will set to zero. Since the derivatives are evaluated at the equilibrium position, the first derivatives (∂U / ∂qi )0= 0. Further, we will assume that the oscillations are small and drop terms that are higher than those quadratic in the generalized coordinates. Hence, the potential energy becomes
U = 12 Kijqiqj
ij∑ , (11.49)
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where
Kij =∂2U
∂qi ∂qj
⎛
⎝⎜⎞
⎠⎟ 0.
Assuming that the equilibrium positions correspond to a minimum in U (we are looking for oscillations after all!) , then the Kij are positive numbers. Also, Kij = Kji since the order of differentiation is immaterial.
Going back to the kinetic energy, recall that the Aij parameters are functions of the generalized coordinates. We can perform a Taylor expansion about the equilibrium positions for these functions as we did for the potential energy. The first term in the expansion is a constant [ Aij (0) ] and the other terms are first and higher orders in the generalized coordinates. Since we are assuming that the qi values are small, the associated generalized velocities qi will also be small. Thus, the product qi qj in the kinetic energy is already second-‐order in the small generalized velocities. Since we are keeping terms in the kinetic energy only up to second order, we must keep only the first (constant) term Aij (0)in the Taylor expansion of Aij. Thus, the kinetic energy becomes
T = 1
2 Mij qi qjij∑ , (11.50)
where Mij = Aij (0) is a positive constant. Note that in the small-‐oscillation approximation, the kinetic energy is a homogenous quadratic function of the generalized velocities and the potential energy is a homogeneous quadratic function of the generalized coordinate, which guarantees that the equations of motion will be linear and solvable.
Equations of Motion
Lagrange’s equations for the general case are
ddt
∂L∂ qi
= ∂L∂qi
. i = 1,2,...,n (11.51)
Before proceeding, we note that
T = 12 Mij qi qj
ij∑ = 1
2 Mii qi2
i∑ + 1
2 Mij qi qji< j∑ + M ji qj qi
i< j∑⎡
⎣⎢
⎤
⎦⎥
= 12 Mii qi
2
i∑ + Mij qi qj
i< j∑ ,
(11.52)
where the last equality follows because Mij = M ji . The expression for U in Eq. (11.49) can be rewritten in a similar way. Using Eq. (11.52), we obtain
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∂L∂ qi
= ∂T∂ qi
= Miiqi + Miji< j∑ qj = Mij
j∑ qj .
Thus,
ddt
∂L∂ qi
= Mijj∑ qj (11.53)
Using similar manipulations, we find that
∂L∂qi
= − ∂U∂qi
= − Kijj∑ qj . (11.54)
Thus, Lagrange’s equations give
Mijj∑ qj = − Kij
j∑ qj , i = 1,2,....,n (11.55)
or, in matrix form,
Mq = −Kq, (11.56)
where q is a 1× n column matrix containing the generalized coordinates. If we take the trial solution to be the complex column matrix z = aeiωt with q = Rez , the matrix form of the equation of motion becomes the eigenvalue equation
(K −ω 2M)a = 0, (11.57)
for which solutions exist if
det(K −ω 2M) = 0. (11.58)
The n normal-‐mode frequencies satisfy Eq. (11.58), which is called the secular equation (or characteristic equation). Substituting each normal frequency into Eq. (11.57) gives the column matrix a (the eigenvector) which represents the solution for that normal mode. The general solution to the equation of motion is a linear superposition of all the normal-‐mode solutions. If we take the eigenvectors to be column matrices consisting of the numerical coefficients from solving Eq. (11.57), then the ith normal coordinate can be obtained from
ξi = q ⋅ai , (11.59)
where ai is the eigenvector corresponding to the ith normal mode. In Eq. (11.59), q is a 1× n row matrix and ai is a n ×1 column matrix. Eq. (11.59) tells us that the ith normal coordinate is the ith projection (“component”) of the expansion of the solutions q in vector space consisting of the orthogonal eigenvectors a.