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Free oscillations

The restoring forces in any actual system are linear in the displace- ment only as an approximation. Nevertheless, in a vast majority of cases the

Oscillations 9

FIGURE 1.5: Beating of two oscillations with commensurable periods (6:7).

deformation produced results in restoring forces proportional to displacement and hence leads to simple harmonic motions. We consider several such cases by picking examples from various areas. But first we will carry out a deeper analysis of a basic mass-spring system, which serves as a prototype of many oscillatory systems. Specifically we consider a point mass attached to an ideal spring undergoing one-dimensional oscillatory motion. Two essential features necessary for oscillatory motion can immediately be identified. They are

• An inertial component capable of carrying kinetic energy and

• An elastic component capable of storing potential energy.

Assuming Hooke’s law to be valid, we can obtain the potential energy as proportional to the square of the displacement, while the kinetic energy is mv2/2. We can also write the equation of motion for the mass in either of two ways:

1. By Newton’s law (F=ma),

−kx=ma, or (1.27)

2. By conservation of total mechanical energy, 1

2mv2+1

2kx2=E. (1.28)

Note that Eq. (1.28) can be obtained from Eq. (1.27) by writing the first in the differential form, multiplying both sides by (dxdt) and integrating both sides.

1 1

2

2

FIGURE 1.6: Superposition of two perpendicular oscillations with the same frequency, but with phase differenceπ/4.

Indeed,

md2x dt2

dx

dt +kxdx

dt = 0, (1.29)

1 2m

dx dt

2

+1

2kx2=E. (1.30)

The solution to these equations can be written as

x=Acos(ωt+α), (1.31)

where ω2 =k/mand the unknown constantsA andα are to be determined from the initial conditionsx(t= 0) =x0 and (dxdt)t=0=v(t= 0) =v0. 1.3.1 General solution of the harmonic oscillator equation

Consider the equation for SHM given by d2x

dt22x= 0. (1.32)

We seek the solution in the form

x=Cept, (1.33)

which after substitution in Eq. (1.32) yields the characteristic equation forp as

p22= 0. (1.34)

Oscillations 11

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

−1 0 1

FIGURE 1.7: Superposition of two perpendicular oscillations with distinct frequencies with phase differenceδ = 0, π/4,π/2, 3π/4 and π(from left to right).

m k

FIGURE 1.8: Mass-spring system.

Eq. (1.34) immediately leads to a pair of purely imaginary solutions, namely, p=±iωand the general solution that can then be written as

x(t) =C1eiωt+C2eiωt. (1.35) Note that the general solution of a second-order ordinary differential equation has two constants figuring in it. The physical solution will correspond to the real part of Eq. (1.35), which can be written as

x= (C1 +C2) cos(ωt)−(C1′′−C2′′) sin(ωt), (1.36) where primes and double primes denote real and imaginary parts, respectively.

Introducing notations asC1+C2 =AcosαandC1′′−C2′′=Asinα, Eq. (1.36) can be cast in the form

x=Acos(ωt+α). (1.37)

The same result can be obtained using the rotating vector representation. The first (last) term in Eq. (1.35) corresponds to the vector C1 (C2) rotating in the counterclockwise (clockwise) direction. These combine to give a harmonic oscillation along thex-axis if the lengths are the same.C2 is rotated through

C

ωt

−ωt α

−α

x C

FIGURE 1.9: Superposition of complex solutions.

some angleαclockwise from−ωt, provided thatC1is rotated throughαwith respect toωt(see Fig. 1.9). This analysis clearly reveals that linear motion can be obtained as a superposition of circular motions, which is just the opposite of the case of Lissajous figures, where we superposed linear motions to get

‘circular’ motion. The last statement has deep meaning in the context of the interchangeability of linear and circular polarizations.

1.3.2 Elasticity, Hooke’s law and Young’s modulus

Stretching a rod or a wire provides the simplest example amenable to easy analysis. We assume the system to be in static equilibrium.

1. For a given material with a given cross-sectional areaA, the elongation

∆l under a given force is proportional to the original length l0. The dimensionless ratio ∆l/lois called the strain.

2. It is an experimental observation that for rods of a given material, but of differentA, the same strain is caused by forces proportional toA. The ratio ∆F/Ais called the stress and has the dimension of force per unit area, or pressure.

3. For small strains (≤ 0.1%), the relation between stress and strain is linear in accordance with Hooke’s law. The value of this constant for any given material is called Young’s modulus of elasticityY.

We thus have

dF/A dl/l0

=−Y. (1.38)

Oscillations 13 If we choose a different notation (x for displacement, and F for force) Eq. (1.38) can be recast in the standard form of SHM,

F =− AY

l0

x. (1.39)

Here the spring constant can be identified ask=AY /l0. 1.3.3 Pendulums

A conventional simple pendulum is shown in Fig. 1.10 and we must note that the motion of the bob is essentially two-dimensional in contrast to the problems discussed earlier. Indeed, though the motion is predominantly hori- zontal (alongx), there is a vertical displacement (alongy) associated with a change in the gravitational potential energy. This situation is well suited for a discussion based on the formulas for the conservation of energy

1

2mv2+mgy=E, where v2= dx

dt 2

+ dy

dt 2

. (1.40)

It is clear from Fig. 1.10 thatl2= (l−y)2+x2orx2= 2ly−y2and for small θ,y≪xand we havex2= 2ly so that

y≈x2

2l. (1.41)

Using this approximate relation and the other consequence dxdt

dy dt

, we can recast the energy conservation relation Eq. (1.40) in the form

1 2m

dx dt

2

+1 2

mg

l x2=E. (1.42)

θ

x P y O

l

FIGURE 1.10: Schematic view of a simple pendulum.

Eq. (1.42) can easily be recognized as one describing SHM with frequency

ω=p

g/l.

Damping of free oscillations

All physical systems are subjected to dissipative processes. For example, the motion of the bob of the pendulum is subjected to air resistance (frictional force). In a general case, in the lowest approximation we can write the damping force as

Fdamping=−bv. (1.43)

Note that this force is proportional to the magnitude of velocity and acts in the opposite direction of velocity. Accounting for this force, Newton’s equation can be reduced to

d2x

dt2 + 2γdx

dt +ω20x= 0, (1.44)

where 2γ = b/m, ω20 = k/m (in case of a mass on a spring). Assuming a solution of the form exp(iβt), the characteristic equation can be written as

ω02−β2+ 2iγβ= 0, (1.45)

which can be easily solved forβ, yielding the pair of roots given by

β1,2=iγ±ω, ω202−γ2. (1.46) It is clear that damping modifies the oscillation frequency. In terms of system parameters,ω can be expressed as

ω= s

k

m −

b 2m

2

. (1.47)

Using Eq. (1.46) one can write the general solution as

x=C1e1t+C2e2t, (1.48)

=eγt C1eiωt+C2eiωt

. (1.49)

Finally, taking the real part, Eq. (1.49) can be reduced to the following:

x=Aeγtcos(ωt+α). (1.50)

Thus, the solution represents ‘damped harmonic’ oscillations with frequency ωdistinct from the natural frequencyω0(modification due to damping) with an amplitude that decays in time in an exponential fashion. As expected this damping is determined byγ characterized by the damping force constant b.

This is shown in Fig. 1.11. From this graph as well as Eq. (1.50), we can recognizeγas the reciprocal of the time required for the amplitude to decay to 1/eof its initial value.

Oscillations 15

0 0.01 0.02 0.03

−1

−0.5 0 0.5 1

x

t (s) e-γt

-e-γt

FIGURE 1.11: Damped harmonic oscillation withf = 600 Hz,γ= 2πf /100 andα= 0.

Depending on how ω0 and γ compare with each other, we can distin- guish three different regimes, namely, (i) under-damped, (ii) over-damped and (iii) critically damped. In fact, Eqs. (1.46)–(1.50) describe theunder-damped regime whenγ < ω0leads to real values forω. In many casesγis much smaller than the natural frequencyω0so that we almost have ‘harmonic’ oscillation, albeit with exponential damping as shown in Fig. 1.11. Keeping in mind that the total energy of harmonic oscillation is given by kA2/2, we can find the temporal evolution of total energy as

E(t) = 1

2kA20e2γt=E0e2γt, (1.51) where A0 andE0 are the initial amplitude and energy, respectively, at time t = 0. In the context of damped oscillatory systems, we often talk about a universal figure of merit (dimensionless), otherwise known as the quality factor or simply the Q-factor. It is defined as

Q= ω0

2γ. (1.52)

It is clear that low values ofγ imply large Q-factors, meaning thereby that the system is likely to sustain oscillations longer. The modified oscillation frequency can be expressed in terms of the Q-factor as

ω202

1− 1 4Q2

. (1.53)

Thus ifQ≫1 it follows thatω≈ω0 and Eq. (1.50) can be written as x=Aeω0t/(2Q)cos(ω0t+α). (1.54) In theover-damped case,γ > ω0, and instead of Eq. (1.46) we have

β1,2=i(γ±ξ), ξ22−ω20, (1.55)

and the general solution can be written as

x=c1e(γ+ξ)t+c2eξ)t. (1.56) The critically damped case corresponds to the equality γ = ω0 and in that case the solution is written as

x= (A+Bt)eγt. (1.57)