FIGURE 1.14: Transients in forced harmonic oscillation (temporal evolu- tion of the amplitudes). (a) Corresponds to very large Q(∼20000). Almost nonexistent damping leads to beating of the natural frequency ω0 and the driving frequencyω. (b) and (c) are whenQ= 20 and show the off-resonant (ω= 0.85ω0) and the resonant (ω≈ω0) behavior, respectively.
Oscillations 21
A
A B
B
(a) (b)
FIGURE 1.15: Two coupled pendulums, (a) at rest and (b) when B is kept at equilibrium with A displaced and both then released.
be transferred to B and then this pattern of exchange of motion will continue.
Individual motion of A or B resembles that of beating with two frequencies.
Indeed, there are two characteristic frequencies of the coupled system and they are termed the normal modes. Any oscillation of the coupled system can always be written as a superposition of these two normal modes.
A change in the initial conditions makes it easy to recognize the normal modes (seeFig. 1.16). Suppose we draw both A and B to one side by equal amounts and release them. Let the distance between them be fixed and equal to the relaxed length of the spring. Both A and B will oscillate in phase. Since the spring is not stretched it does not affect oscillation frequency, which is just the individual oscillation frequency ω0 = p
(g/l) of the pendulums. In the absence of damping, these oscillations will continue forever. This is one of the two normal modes and ω0 is one normal mode frequency. The solutions in this case are given by
xA=Ccos(ω0t), xB=Ccos(ω0t). (1.74)
If A and B are drawn to opposite sides by equal amounts and then released, such oscillations can persist forever. This is the other normal mode with a higher frequency. We calculate this frequency as follows. If the pendulums were free, a displacement ofxwould correspond to a restoring force ofmω20x.
In the presence of coupling spring, either it is stretched or compressed by an amount 2x and hence the additional force is 2kx (k is the spring constant).
Thus the equation of motion for A is md2xA
dt2 +mω20xA+ 2kxA= 0, (1.75)
or d2xA
dt2 +ω02xA+ 2ωc2xA= 0, (1.76)
B
A A
B
(a) (b)
FIGURE 1.16: (a) Symmetric and (b) antisymmetric normal modes.
where ωc2 =k/m. We can easily recognize the resonance frequency ω′ from Eq. (1.76) as
ω′2=ω20+ 2ω2c = g
l +2k m
. (1.77)
For the said initial conditions, the solutions are
xA=Dcos(ω′t), xB=−Dcos(ω′t). (1.78) Note that at any given moment, the motion of B is the mirror image of motion of A. The motions of A and B areπout of phase.
It is important to observe that if any pendulum is clamped, the angular frequency of the other has two contributions, one from the gravity and the other from the spring. Thus the net frequency isp
(ω02+ω2c). If this frequency (characteristic of one oscillator) is taken as the reference, then the two normal mode frequencies lie on the two sides of it (i.e., one greater and the other smaller).
1.5.2 Superposition of normal modes
We mentioned earlier that once excited in any of the normal modes (of course for suitable initial conditions), the system continues in the same os- cillatory state. For any other initial conditions resulting in more complicated oscillatory pattern, the resultant oscillations can always be perceived as a su- perposition of the normal modes. In mathematical terms these normal modes form a suitable basis for representing any free motion of the coupled system.
Here we show how this is done.
Pick any arbitrary moment when A is displaced byxAand B is displaced by xB resulting in a stretching of the spring by an amount xA −xB (see Fig. 1.17). Thus the spring pulls on A and B with a force with magnitude
Oscillations 23
b
B A
a
FIGURE 1.17: Displacement of the two pendulums at any arbitrary mo- ment.
proportional tok(xA−xB). Since the direction of this force is opposite for A and B, the restoring forces on A and B, respectively, are given by
mω02xA+k(xA−xB) and mω02xB−k(xA−xB) (1.79) so that the equations of motion are written as
md2xA
dt2 +mω20xA+k(xA−xB) = 0, (1.80) md2xB
dt2 +mω02xB−k(xA−xB) = 0. (1.81) Using the notations introduced earlier, the equations above can be reduced to
d2xA
dt2 + (ω20+ω2c)xA−ω2cxB = 0, (1.82) d2xB
dt2 + (ω20+ω2c)xB−ωc2xA= 0. (1.83) Adding and subtracting the above two equations, we have the equations for the sum (xA+xB =q1) and difference (xA−xB =q2) displacements,
d2q1
dt2 +ω20q1= 0, (1.84) d2q2
dt2 + (ω02+ 2ωc2)q2= 0. (1.85) Using the notationω′=p
ω20+ 2ωc2, we can write one solution (not the general one) as
q1=Ccos(ω0t), q2=Dcos(ω′t), (1.86)
where C and D are to be evaluated from initial conditions. The frequencies ω0 andω′ are known as the normal frequencies.
It is important to note the differences between the pairs of equations (1.82), (1.83) and (1.84), (1.85), though they describe the same system. While the first two are coupled (in the sense that one cannot be solved without the other), in the second pair the equations are independent of each other, each one representing a normal mode of the system. This is why if any of the normal modes is excited, it persists forever, not being affected by the other.
In other words if through some algebraic manipulations we have reduced a coupled system into its independent components, then we have reduced the system to its normal modes. The dependent variablesq1andq2are sometimes referred to as normal coordinates, and this procedure is termed normal mode decomposition.
Going back to the original displacementsxAandxB, the solutions can be written as
xA=1
2(q1+q2) = 1
2(Ccos(ω0t) +Dcos(ω′t)), (1.87) xB =1
2(q1−q2) = 1
2(Ccos(ω0t)−Dcos(ω′t)). (1.88) It is clear that ifC= 0, both pendulums oscillate at one normal frequencyω′, whileD= 0 implies oscillation at the other frequencyω0. Thus one important characteristic of the normal frequency is that both the bobs can oscillate at that frequency.
For initial conditions given by xA=A0, dxA
dt = 0, xB = 0, dxB
dt = 0, (1.89)
which correspond to B at the equilibrium position with null velocity while A is moved toA0and released (initial null velocity), we can solve for the unknown constants to obtainC=A0, D=A0. Hence
xA= 1
2A0(cos(ω0t) + cos(ω′t)), (1.90) xB= 1
2A0(cos(ω0t)−cos(ω′t)). (1.91) Eqs. (1.90) and (1.91) can be reduced to the following form:
xA=A0cos
ω′−ω0
2 t
cos
ω′+ω0
2 t
, (1.92)
xB=A0sin
ω′−ω0
2 t
sin
ω′+ω0
2 t
. (1.93)
Both of these represent oscillation at the average frequency (ω′+ω0)/2 with a low frequency modulation. The amplitude of one goes to the peak value while that of the other goes to zero.
Oscillations 25 1.5.3 Coupled oscillations as an eigenproblem: Exact
analysis
Let us now address the coupled mode problem from a different angle, namely, as an eigenvalue problem. We first reduce the two second-order equa- tions, Eqs. (1.82) and (1.83), to a set of four coupled first-order equations by writing the dependent variables as
x1=xA, x2=dxA
dt , x3=xB, x4= dxB
dt . (1.94)
With these definitions Eqs. (1.82) and (1.83) take the following compact ma- trix form
d dt
x1
x2
x3
x4
=
0 1 0 0
−ωoc2 0 ωc2 0
0 0 0 1
ωc2 0 −ω2oc 0
x1
x2
x3
x4
. (1.95)
Here we have definedω02+ω2c =ωoc2. The solution of Eq. (1.95) is determined by the eigenvalues and the corresponding eigenvectors of the 4×4 matrix on the right-hand side. The eigenvalues of the above 4×4 matrix are given by the roots of the characteristic equation
λ4+ 2λ2ωoc2 +ω4oc−ωc4= 0. (1.96) The roots of this equation are given by
λ2=ωoc2 ±ω2c. (1.97)
One pair of roots is given as λ=±iω0, while the other pair is ±iω′, where ω′=p
ω02+ 2ω2c as before. We can thus recover the oscillation frequencies of the normal modes expressed in Eqs. (1.84) and (1.85). Further, calculation of the eigenvectors will lead to the normal modesq1 andq2described earlier.
1.5.4 Coupled oscillations as an eigenproblem: Approximate analysis
In the literature, an approximation, referred to as the slowly varying en- velope approximation (SVEA), is often used. The purpose SVEA serves is to get rid of the higher-order derivatives on physical grounds and thus reduce the complexity of the problem. Here we demostrate how SVEA works in the context of normal modes discussed earlier. We will make use of the complex notations in order to solve Eqs. (1.82) and (1.83). We write the solutions as
xA=xa(t)e−iω0t, (1.98) xB =xb(t)e−iω0t, (1.99)
and treat xa(t) and xb(t) as slowly varying so that the change in x(t) (for both the subscripts) over one high-frequency period (= 2π/ω0) is negligible compared to the function itself,
d2x(t) dt2
≪ω0
dx(t) dt
≪ω20|x(t)|. (1.100) Making use of the approximation (1.100) and Eqs. (1.98) and (1.99), the set of coupled Eqs. (1.82) and (1.83) can be written in matrix form as
x˙a
˙ xb
= i 2ω0
−ωc2 ωc2 ωc2 −ωc2
xa
xb
=A xa
xb
. (1.101)
The eigenvalues of the matrix A determine the frequencies for the normal modes. The eigenvaluesλ1 andλ2 are found to be
λ1= 0, (1.102)
λ2=−iωc2 ω0
. (1.103)
Thus, one of the normal modes oscillates atω0, and the other atω0+ (ωc2/ω0).
Note that the latter can easily be recognized as ω′ = p
ω02+ 2ω2c ≈ ω0+ (ωc2/ω0) when the coupling is weak (ωc ≪ ω0). In fact, SVEA is applicable in this weak coupling situation. Further, we can find out the corresponding eigenvectors that will correspond to the normal modes mentioned earlier. We can refer to Goldstein [2] for a detailed analysis of similar cases.