Damped Oscillation

We have seen that the total energy of a harmonic oscillator remains constant. Once started, the oscillations continue forever with a constant amplitude (which is determined from the initial conditions) and a constant frequency (which is determined by the inertial and elastic properties of the system). Simple harmonic motions which persist indefinitely without loss of amplitude are called free or undamped.

However, observation of the free oscillations of a real physical system reveals that the energy of the oscillator gradually decreases with time and the oscillator eventually comes to rest. For example, the amplitude of a pendulum oscillating in the air decreases with time and it ultimately stops. The vibrations of a tuning fork die away with the passage of time. This happens because, in actual physical systems, friction (or damping) is always present. Friction resists motion.

The presence of resistance to motion implies that frictional or damping force acts on the system. The damping force acts in opposition to the motion, doing negative work on the system, leading to a dissipation of energy. When a body moves through a medium such as air, water, etc. its energy is dissipated due to friction and appears as heat either in the body itself or in the surrounding medium or both.

There is another mechanism by which an oscillator loses energy. The energy of an oscillator may decrease not only due to friction in the system but also due to radiation. The oscillating body imparts periodic motion to the particles of the medium in which it oscillates, thus producing waves. For example, a tuning fork produces sound waves in the medium which results in a decrease in its energy.

All sounding bodies are subject to dissipative forces, or otherwise, there would be no loss of energy by the body and consequently, no emission of sound energy could occur. Thus, sound waves are produced by radiation from mechanical oscillatory systems. We shall learn later that the electromagnetic waves are produced by radiations from oscillating electric and magnetic fields.

The effect of radiation by an oscillating system and of the friction present in the system is that the amplitude of oscillations gradually diminishes with time. The reduction in amplitude (or energy) of an oscillator is called damping and the oscillation are said to be damped.

Damping Forces

The damping of a real system is a complex phenomenon involving several kinds of damping forces. The damping force of a fluid (liquid or gas) to a moving object is some function of the velocity of the object. The damping force that depends on velocity is referred to as viscous damping force. The magnitude of this force is well described by the equation

F=p1v+p2v2F={{p}_{1}}v+{{p}_{2}}{{v}^{2}}

Where v is the magnitude of the velocity of the object. The direction of this resistive force is opposite to that of the velocity. If v is small compared to the ratio p1/ p2 the damping force will be proportional to the first power of v. Thus, for small velocities,

F = – pv

Where p, the viscous damping coefficient, represents the damping force per unit velocity. The negative sign indicates that the force opposes the motion, tending to reduce velocity. In other words, the viscous damping force is a retarding force. Since the velocity of most oscillating systems is usually small, the damping force exerted by the fluid in contact with the system is likely to be viscous. Viscous forces are generally much smaller than inertial and elastic forces in a system.

However, damping devices called dampers are sometimes deliberately introduced in a system for vibration control. The damping force exerted by such devices may be comparable in magnitude to the inertial and elastic forces.

In real systems, it is likely that the moving part is in contact with an unlubricated surface, as in the case of horizontal oscillations of a body attached to a spring (see figure). The oscillating body is always in contact with the horizontal surface. The resulting frictional force opposes the motion and can often be idealized as a force of constant magnitude. Such a force is usually referred to as a Coulomb friction force.

In a solid, some part of energy may be lost due to imperfect elasticity or internal friction of the material. It is very difficult to estimate this type of damping. Experiments suggest that a resistive force proportional to the amplitude and independent of the frequency may serve as a satisfactory approximation. This kind of damping in solids is referred to as structural damping.

Thus, the damping of a real system is a complex phenomenon involving several kinds of damping forces such as viscous damping, Coulomb friction and structural damping. Because it is generally very difficult to predict the magnitude of the damping forces, one usually has to rely on experience and experiment so as to make a reasonably good estimate. It is a common practice to approximate the damping of a system by equivalent viscous damping, for the simple reason that viscous damping is the most convenient to handle mathematically. Thus, according to this approximation, the magnitude of the viscous force to be used in a particular problem is chosen to be the one that would produce the same rate of energy dissipation as the actual damping forces. This usually provides a good estimate.

The inclusion of damping forces complicates the analysis considerably. Fortunately, in actual systems, the damping forces are usually small and can often be ignored. In situations, where they are not negligibly small, the viscous damping model is the most convenient mathematically. We shall use this model, under the simplifying assumption that the velocity of the moving part of the system is small so that the damping force is linear in velocity as in Eq. (3.1).

If the velocity is not small, the damping force exerted on the system may be represented more closely by a force proportional to the square of the velocity. We shall not deal with such forces. The effect of the linear viscous damping force on the free oscillations of simple systems, with one degree of freedom, is considered in the next section.

Damped Oscillations Of A System Having One Degree Of Freedom

We shall investigate the effect of damping on the harmonic oscillations of a simple system having one degree of freedom. One such system is shown in the figure. When the system is displaced from its equilibrium state and released, it begins to move. The forces acting on the system are:

Damped Oscillations Of A System Having One Degree Of Freedom

(i) a restoring force –Kx, where K is the coefficient of the restoring force and x is the displacement, and

(ii) a damping force pdxdt,-p\frac{dx}{dt}, where p is the coefficient of the damping force and dxdt\frac{dx}{dt} is the velocity of the moving part of the system. From Newton’s law for a rigid body in translation, these forces must balance with Newton’s force md2xdt2,m\frac{{{d}^{2}}x}{d{{t}^{2}}}, where m is the mass of the oscillator and d2xd12,\frac{{{d}^{2}}x}{d_{1}^{2}}, its accelerations. Since, the restoring force and the damping force acts in a direction opposite to Newton’s force, we have

md2xdt2=Kxpdxdtm\frac{{{d}^{2}}x}{d{{t}^{2}}}=-Kx-p\frac{dx}{dt}

Remember, this equation holds only for small displacements and small velocities. This equation can be rewritten as:

d2xdt2+γdxdt+ω02x=0\frac{{{d}^{2}}x}{d{{t}^{2}}}+\gamma \frac{dx}{dt}+\omega _{0}^{2}x=0 … (3.2)

With γ=p/m\gamma =p/m … (3.3)

And ω02=K/m\omega _{0}^{2}=K/m … (3.4)

Notice that dimensionally γ=pm=forcevelocity  ⁣ ⁣× ⁣ ⁣ mass=MLT2LT1M=T1.\gamma =\frac{p}{m}=\frac{\text{force}}{\text{velocity }\!\!\times\!\!\text{ mass}}=\frac{\text{ML}{{\text{T}}^{-2}}}{\text{L}{{\text{T}}^{-1}}M}={{T}^{-1}}. the same as the dimension of frequency.

It is easy to see that in Eq. (3.2) the damping is characterized by the quantity γ, having the dimension of frequency, and the constant ω0 represents the angular frequency of the system in the absence of damping and is called the natural frequency of the oscillator. Equation (3.2) is the differential equation of the damped oscillator. To find out how the displacement varies with time, we need to solve Eq. (3.2) with constants γ and ω0 given respectively by Eqs. (3.3) and (3.4).

The General Solution

To solve Eq. (3.2) we make use of the exponential function again. Let us assume that the solution is

x=Aeαtx=A{{e}^{\alpha t}}

And solve for α. Constants A and α are arbitrary and as yet undetermined. Differentiating, we have

dxdt=αAeαt\frac{dx}{dt}=\alpha A\,\,{{e}^{\alpha t}} d2xdt2=α2Aeαt\frac{{{d}^{2}}x}{d{{t}^{2}}}={{\alpha }^{2}}A\,\,{{e}^{\alpha t}}

Substitution in Eq. (3.2) yields

(α2+γα+ω02)Aeαt=0\left( {{\alpha }^{2}}+\gamma \alpha +\omega _{0}^{2} \right)A\,{{e}^{\alpha t}}=0

For this equation to hold for all values of t, the term in the brackets must vanish, i.e.

α2+γα+ω02=0{{\alpha }^{2}}+\gamma \alpha +\omega _{0}^{2}=0

The two roots of this quadratic equation are

α1=γ2+12(γ24ω02)1/2{{\alpha }_{1}}=-\frac{\gamma }{2}+\frac{1}{2}{{\left( {{\gamma }^{2}}-4\omega _{0}^{2} \right)}^{1/2}}

And

α2=γ212(γ24ω02)1/2{{\alpha }_{2}}=-\frac{\gamma }{2}-\frac{1}{2}{{\left( {{\gamma }^{2}}-4\omega _{0}^{2} \right)}^{1/2}}

Thus the two possible solutions of Eq. (3.2) are

x1=A1eα1t=A1exp[γ2+12(γ24ω02)1/2]t{{x}_{1}}={{A}_{1}}\,\,{{e}^{\alpha }}{{1}^{t}}={{A}_{1}}\,\exp \left[ -\frac{\gamma }{2}+\frac{1}{2}{{\left( {{\gamma }^{2}}-4\omega _{0}^{2} \right)}^{1/2}} \right]t

And

x2=A2eα2t=A2exp[γ212(γ24ω02)1/2]t{{x}_{2}}={{A}_{2}}\,\,{{e}^{\alpha }}{{2}^{t}}={{A}_{2}}\,\exp \left[ -\frac{\gamma }{2}-\frac{1}{2}{{\left( {{\gamma }^{2}}-4\omega _{0}^{2} \right)}^{1/2}} \right]t

Since Eq. (3.2) is linear, the superposition principle is applicable. Hence, the general solution is given by the superposition of the two solutions, i.e.

x=x1+x2x={{x}_{1}}+{{x}_{2}}

Or

x=A1exp[γ2+(γ24ω02)1/2]tx={{A}_{1}}\,\exp \left[ -\frac{\gamma }{2}+{{\left( \frac{{{\gamma }^{2}}}{4}-\omega _{0}^{2} \right)}^{1/2}} \right]t +A2exp[γ2(γ24ω02)1/2]t+{{A}_{2}}\,\exp \left[ -\frac{\gamma }{2}-{{\left( \frac{{{\gamma }^{2}}}{4}-\omega _{0}^{2} \right)}^{1/2}} \right]t …..(3.5)

Here A1 and A2 are arbitrary constants to be determined from the initial conditions, namely, the initial displacement and the initial velocity.

The nature of the motion depends on the character of the roots α1 and α2. The roots may be real or complex depending on whether γ > 2ω0 or γ < 2ω0 and γ < 2 ω0. Each condition describes a particular kind of behaviour of the system. We shall now treat each case separately.

Case I:

γ > 2 ω0 (Large Damping) In this case, the damping term γ/2 dominates the stiffness term ω0 and the term (γ2/4ω02)1/2{{\left( {{\gamma }^{2}}/4-\omega _{0}^{2} \right)}^{1/2}} in Eq. (3.5) is a real quantity with a positive value, say, q, i.e.

(γ24ω02)1/2=q{{\left( \frac{{{\gamma }^{2}}}{4}-\omega _{0}^{2} \right)}^{1/2}}=q

So that displacement ψ as a function of time is given by

x=A1exp(γ2+q)t+A2exp(γ2+q)tx={{A}_{1}}\,\exp \left( -\frac{\gamma }{2}+q \right)t+{{A}_{2}}\,\exp \left( -\frac{\gamma }{2}+q \right)t …..(3.6)

The velocity is given by

dxdt=(γ2+q)A1exp(γ2+q)t\frac{dx}{dt}=\left( -\frac{\gamma }{2}+q \right){{A}_{1}}\,\exp \left( -\frac{\gamma }{2}+q \right)t (γ2+q)A2exp(γ2+q)t-\left( \frac{\gamma }{2}+q \right){{A}_{2}}\,\exp \left( -\frac{\gamma }{2}+q \right)t ……(3.7)

These equations describe the behaviour of a heavily damped oscillator, as for example, a pendulum in a viscous medium such as a dense oil. As stated earlier, the constants A1 and A2 are determined from the initial conditions. Let us assume that the oscillator is at its equilibrium position) (ψ = 0) at time t = 0. At this instant it is given a kick so that it has a finite velocity, say, V0 at this time, i.e. at t = 0.

x = 0

Equations (3.6) and (3.7) then give (setting t = 0)

0=A1+A20={{A}_{1}}+{{A}_{2}} V0=(γ2+q)A1(γ2+q)A2{{V}_{0}}=\left( -\frac{\gamma }{2}+q \right){{A}_{1}}-\left( \frac{\gamma }{2}+q \right){{A}_{2}}

giving A1=A2=V02q{{A}_{1}}=-{{A}_{2}}=\frac{{{V}_{0}}}{2q}

Thus, under the above initial conditions, Eqs. (3.6) and (3.7) become

x=V02qeγt/2(eqteqt)x=\frac{{{V}_{0}}}{2q}{{e}^{-\gamma t/2}}\left( {{e}^{qt}}-{{e}^{-qt}} \right)

Or

x=V0qeγt/2sinh(qt)x=\frac{{{V}_{0}}}{q}{{e}^{-\gamma t/2}}\sinh \left( qt \right)

And

dxdt=V02eγt/2{(eqt+eqt)γ2q(eqteqt)}\frac{dx}{dt}=\frac{{{V}_{0}}}{2}{{e}^{-\gamma t/2}}\left\{ \left( {{e}^{qt}}+{{e}^{-qt}} \right)-\frac{\gamma }{2q}\left( {{e}^{qt}}-{{e}^{-qt}} \right) \right\}

Or

dxdt=V0eγt/2{cosh(qt)γ2qsinh(qt)}\frac{dx}{dt}={{V}_{0}}{{e}^{-\gamma t/2}}\left\{ \cosh \left( qt \right)-\frac{\gamma }{2q}\sinh \left( qt \right) \right\} ……(3.9)

Figure 1 illustrates the behaviour of a heavily damped system when it is disturbed from equilibrium by a sudden impulse at t = 0. It is the displacement – time graph of Eq. (3.8). For small values of time t, the term eγt/2{{e}^{-\gamma t/2}} is very nearly unity, the displacement increases with time since sinh (qt) increases as t increases. Very soon, however, the term eγt/2{{e}^{-\gamma t/2}} starts contributing and the displacement decays exponentially with time, eventually becoming zero. The turning point occurs at a time t=t0t={{t}_{0}} when dx/dt=0.dx/dt=0. Equation (3.9) tells us that this happens at a time t = t0 satisfying

tanh(qt0)=2qγ\tan \,h\left( q{{t}_{0}} \right)=\frac{2q}{\gamma }

Thus, the displacement increases until time t=t0,t={{t}_{0}}, after which it slowly returns to zero. Since, displacement ψ never becomes negative, there is no oscillation at all. Such a motion is called dead beat. We come across such a motion in the case of a dead beat galvanometer (see sec. 3.6).

Case II:

γ=2ω0\gamma =2\,{{\omega }_{0}} (Critical Damping). This is a special case of a heavily damped motion. Using the notation q=(γ2/4ω02)t/2q={{\left( {{\gamma }^{2}}/4-\omega _{0}^{2} \right)}^{t/2}} of case I, we see that, in this case, q = 0 and Eq. (3.6) becomes

x=(A1+A2)eγt/2x=\left( {{A}_{1}}+{{A}_{2}} \right){{e}^{-\gamma t/2}}

Or

x=Beγt/2x=B\,{{e}^{-\gamma t/2}} ……(3.10)

Where B = A1 + A2, is a constant. In other words, Eq. (3.10) is the solution of Eq (3.2) for γ = 2 ω0. In this case, the two roots α1 and α2 become identical. Notice that the solution (3.10) contains only one adjustable constant B. This solution is only a partial solution, since the solution of any second – order differential equation must contain two adjustable constants. This can be understood as follows. If Eq. (3.10) were a complete solution of Eq. (3.2), then the velocity of the oscillator would be given by

dxdt=Bγ2eγt/2\frac{dx}{dt}=-B\frac{\gamma }{2}{{e}^{-\gamma t/2}}

When the system is disturbed from equilibrium (x = 0) by giving an impulse (i.e. by imparting a velocity V0) at t = 0, we have, from the above two equations

B = 0

V0=γ2B{{V}_{0}}=-\frac{\gamma }{2}B

implying, thereby, that V0 is also zero, which is not our initial condition. Hence our trial solution yields only a partial solution in the case when q = 0.

We can verify that a second solution is represented by the trial solution

x=Cteγt/2x=Ct\,{{e}^{-\gamma t/2}}

Giving dxdt=Ceγt/2(1γt2)\frac{dx}{dt}=C\,{{e}^{-\gamma t/2}}\left( 1-\frac{\gamma t}{2} \right)

And

d2xdt2=Cγ2eγt/2(2+γt2)\frac{{{d}^{2}}x}{d{{t}^{2}}}=C\,\frac{\gamma }{2}{{e}^{-\gamma t/2}}\left( -2+\frac{\gamma t}{2} \right)

Substituting for x, dxdtanddx2dt2\frac{dx}{dt}\,and\,\,\frac{d{{x}^{2}}}{d{{t}^{2}}} in Eq. (3.2) with ωo2 replaced by γ24\frac{{{\gamma }^{2}}}{4} i.e.

d2xdt2+γdxdt+γ24x=0\frac{{{d}^{2}}x}{d{{t}^{2}}}+\gamma \frac{dx}{dt}+\frac{{{\gamma }^{2}}}{4}x=0

We have,

Cγ2eγt/2(2+γt2)+γCeγt/2(1γt2)+γ24Cteγt/2=0C\frac{\gamma }{2}{{e}^{-\gamma t/2}}\left( -2+\frac{\gamma t}{2} \right)+\gamma C\,{{e}^{-\gamma t/2}}\left( 1-\frac{\gamma t}{2} \right)+\frac{{{\gamma }^{2}}}{4}Ct\,{{e}^{-\gamma t/2}}=0

Or

γ2Ceγt/2(2+γt2+2γt+γt2)=0\frac{\gamma }{2}C\,{{e}^{-\gamma t/2}}\left( -2+\frac{\gamma t}{2}+2-\gamma t+\frac{\gamma t}{2} \right)=0

Or 0 = 0

Thus, Eqs. (3.10) and (3.11) are both possible solutions of Eq. (3.2) in the special case when γ=2ω0.\gamma =2\,{{\omega }_{0}}. From the superposition principle, the general solution is given by

x=Beγt/2+Cteγt/2=(B+Ct)eγt/2x=B\,{{e}^{-\gamma t/2}}+Ct\,{{e}^{-\gamma t/2}}=\left( B+Ct \right){{e}^{-\gamma t/2}} …..(3.12)

And

dxdt={Cγ2(B+Ct)}eγt/2\frac{dx}{dt}=\left\{ C-\frac{\gamma }{2}\left( B+Ct \right) \right\}{{e}^{-\gamma t/2}}

The constants B and C can be determined from the initial conditions. If at t = 0, x = 0 and dxdt=V0,\frac{dx}{dt}={{V}_{0}}, we have, from the above equations,

B = 0

C = V0

Thus, under these initial conditions, the displacement x in Eq, (3.12) is given by

x=V0teγt/2x={{V}_{0}}t\,{{e}^{-\gamma t/2}} …..(3.13)

And

dxdt=C(1γt2)eγt/2\frac{dx}{dt}=C\left( 1-\frac{\gamma t}{2} \right){{e}^{-\gamma t/2}} ….(3.14)

Graph of x against t

Figure 2 is a graph of x against t in Eq. (3.13). It illustrates the displacement – time behaviour of a damped system with γ=2ω0,\gamma =2{{\omega }_{0}}, when it is disturbed from equilibrium by a sudden impulse For small values of t, the term eγt/2{{e}^{-\gamma t/2}} is very nearly unity and displacement [Eq. (3.13)] increases linearly with time t. After sometime eγt/2{{e}^{-\gamma t/2}} starts changing and the displacement decays exponentially with time, eventually become zero. The turning point occurs at a time t0, when dxdt=0.\frac{dx}{dt}=0. From Eq. (3.14) this happens at t = t0 given by

1γt02=01-\frac{\gamma {{t}_{0}}}{2}=0

Or

t0=2γ=1ω0{{t}_{0}}=\frac{2}{\gamma }=\frac{1}{{{\omega }_{0}}}

The displacement increases until time t = t0, after which it decays to zero. A comparison of Eqs. (3.8) and (3.13) reveals that the decay rate is much faster when γ=2ω0\gamma =2\,{{\omega }_{0}} then when γ>2ω0.\gamma >2\,{{\omega }_{0}}. In both cases, there is no oscillation at all, since ψ never becomes negative.

The motion described by Eq. (3.13) is called critically damped. The necessary condition for critical damping is γ=2ω0.\gamma =2\,{{\omega }_{0}}. Suppose we are faced with a problem in which we desire a high rate of decay without oscillation. Evidently, the optimum choice is critical damping. We come across such a problem in pointer-type galvanometers, where we would want the pointer to move immediately to the correct position and stay there without annoying oscillation (see sec. 3.6).

Case III γ<2ω0\gamma <2\,{{\omega }_{0}} (Small Damping).

When γ<2ω0,\gamma <2\,{{\omega }_{0}}, the damping is small and this gives the most important kind of behaviour, namely, oscillatory damped harmonic motion, for then, the expression (γ24ω02)1/2{{\left( \frac{{{\gamma }^{2}}}{4}-\omega _{0}^{2} \right)}^{1/2}} in the exponentials in Eq. (3.5) is an imaginary quantity. Writing this as

(γ24ω02)1/2=1(ω02γ24){{\left( \frac{{{\gamma }^{2}}}{4}-\omega _{^{0}}^{2} \right)}^{1/2}}=\sqrt{-1}\left( \omega _{0}^{2}-\frac{{{\gamma }^{2}}}{4} \right) =iω=i\omega *

Where, ω=(ω02γ24)1/2\omega *={{\left( \omega _{0}^{2}-\frac{{{\gamma }^{2}}}{4} \right)}^{1/2}} is a real positive quantity, the displacement Eq. (3.5) may be rewritten as

x=eγt/2{(A1exp(iωt)+A2exp(iωt))}x={{e}^{-\gamma t/2}}\left\{ \left( {{A}_{1}}\exp \left( i\omega *t \right)+{{A}_{2}}\exp \left( -i\omega *t \right) \right) \right\} ……(3.15)

To compare the behaviour of a damped oscillator with the ideal case in which damping is ignored, we will recast Eq. (3.15) into a more familiar form. We can do this by using the identities,

eiθ=cosθ+isinθ{{e}^{i\theta }}=\cos \theta +i\sin \theta eiθ=cosθisinθ{{e}^{-i\theta }}=\cos \theta -i\sin \theta

So that Eq. (3.15) can be written as

x=eγt/2{(A1+A2)cosωt+i(A1A2)sinωt}x={{e}^{-\gamma t/2}}\left\{ \left( {{A}_{1}}+{{A}_{2}} \right)\cos \omega *t+i\left( {{A}_{1}}-{{A}_{2}} \right)\sin \omega *t \right\}

If we choose

A1+A2=Acosδ{{A}_{1}}+{{A}_{2}}=A\cos \delta i(A1A2)=Asinδi\left( {{A}_{1}}-{{A}_{2}} \right)=A\sin \delta

Where A and δ are constants which depend upon the initial conditions, we find, after substitution,

x=Aeγt/2cos(ωtδ)x=A\,{{e}^{-\gamma t/2}}\cos \left( \omega *t-\delta \right) ……(3.16)

With

ω=ω0(1γ24ω02)1/2\omega *={{\omega }_{0}}{{\left( 1-\frac{{{\gamma }^{2}}}{4\omega _{0}^{2}} \right)}^{1/2}} …….(3.17)

Differentiating Eq. (3.16), we obtain an expression for the velocity of the oscillator, which reads

dxdt=Aeγt/2{ωsin(ωtδ)+γ2cos(ωtδ)}\frac{dx}{dt}=-A\,\,{{e}^{-\gamma t/2}}\left\{ \omega *\sin \left( \omega *t-\delta \right)+\frac{\gamma }{2}\cos \left( \omega *t-\delta \right) \right\} ……(3.18)

Equation (3.16) shows that the motion is oscillatory. The oscillation is not simple harmonic, since its, ‘amplitude’ Aeγt/2A\,{{e}^{-\gamma t/2}} is not constant but decreases with time. The motion is not periodic, since it never repeats itself, each swing being of smaller amplitude than the preceding one. However, if γ\gamma is very small compared to ω0,{{\omega }_{0}}, the amplitude will remain sensibly constant over a large number of cycles of the harmonic term cos(ωtδ)\cos \left( \omega *t-\delta \right) in which case, the motion is nearly periodic and simple harmonic.

The angular frequency of the oscillation is ω* given Eq. (3.17) which is less than the natural angular frequency of free undamped oscillations. Strictly speaking, we are really not justified in using the terms ‘amplitude’ and ‘frequency’ for a motion which is not periodic. But, when damping is small, the motion is nearly periodic, we may use these terms with some reservations.

To illustrate the behaviour of a weakly damped oscillator, let us choose the initial conditions, namely, that at t = 0, x = 0 and dxdt=V0.\frac{dx}{dt}={{V}_{0}}. Using these conditions in Eqs. (3.16) and (3.18) we get

0=Acosδ0=A\cos \delta

And

V0=A(γ2cosδωsinδ){{V}_{0}}=-A\left( \frac{\gamma }{2}\cos \delta -\omega *\sin \delta \right)

yielding δ=π2\delta =\frac{\pi }{2} (A = 0; being a trivial case)

And

A=V0ωA=\frac{{{V}_{0}}}{\omega *}

Using these values of A and δ in Eqs (3.16) and (3.18) we find that, under the above initial conditions, the displacement and velocity of the oscillator are, respectively, given by

x=V0ωeγt/2sinωt=A(t)sinωtx=\frac{{{V}_{0}}}{\omega *}{{e}^{-\gamma t/2}}\sin \omega *t=A\left( t \right)\sin \omega *t

With A(t)=V0ωeγt/2=A0eγt/2A\left( t \right)=\frac{{{V}_{0}}}{\omega *}{{e}^{-\gamma t/2}}={{A}_{0}}{{e}^{-\gamma t/2}}

A0 being the value of A(t) when γ = 0

And

dxdt=V0eγt/2(cosωtγ2ωsinωt)\frac{dx}{dt}={{V}_{0}}\,{{e}^{-\gamma t/2}}\left( \cos \omega *t-\frac{\gamma }{2\omega *}\sin \omega *t \right) ….(3.20)

Figure 3.4 depicts the behaviour of a weakly damped oscillator. It is a graph of ψ against t of the motion described by Eq. (3.19). The constant A0 is the value of A(t)=V0ωeγt/2A\left( t \right)=\frac{{{V}_{0}}}{\omega *}{{e}^{-\gamma t/2}} in the absence of damping (γ = 0), i.e. A0=V0/ω.{{A}_{0}}={{V}_{0}}/\omega *. Since the maximum values of sin(ωt)\sin \left( \omega *t \right) are + 1 and -1 alternately, the displacement-time graph of oscillation is bounded by the dotted curves A0eγt/2{{A}_{0}}{{e}^{-\gamma t/2}} and A0eγt/2-{{A}_{0}}{{e}^{-\gamma t/2}}.

Thus, although, the amplitude decreases exponentially with time, the weekly damped oscillator executes some sort of oscillatory motion. The motion does not repeat itself and is, therefore, not periodic in the usual sense of the term. However, it still has a time period T=2π/ω,T*=2\pi /\omega *, which is the time interval between two alternate zeros of displacement. The time period between two successive zeros of displacement is T*/2. This is also the time interval between a maximum and minimum value of the displacement, but the maxima and minima are not exactly halfway between the zeros. This is obvious from Eq. (3.20), because at a maximum or a minimum of displacement, the velocity is zero, giving

cosωtγ2ωsinωt=0\cos \omega *t-\frac{\gamma }{2\omega *}\sin \omega *t=0

Or

tanωt=2ωγ\tan \omega *t=\frac{2\omega *}{\gamma }

Displacement-time behaviour of a weakly damped oscillator

Displacement-time behaviour of a weakly damped oscillator

The values of t satisfying this equation are the instants at which ψ is either a positive maximum or a negative maximum. In the case when γ<2ω0,2ω0γ,\gamma <2{{\omega }_{0}},\frac{2{{\omega }_{0}}}{\gamma }\to \infty , so that

ωtπ2,3π2,5π2,,\omega *t\to \frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},…,

The first maximum of ψ occurs at a time t=t1t={{t}_{1}} given by

ωt1=π2\omega *{{t}_{1}}=\frac{\pi }{2}

Or

t1=π2ω=T4{{t}_{1}}=\frac{\pi }{2\omega *}=\frac{T*}{4}

i.e. the maximum is exactly midway between the two zeros of ψ. Thus, only in the case of negligibly small damping, are the maxima and minima halfway between the zeros of displacement as in the case of simple harmonic motion.

Effect of Damping:

The effect of damping is two-fold: (a) The amplitude of oscillation decreases exponentially with time as

A(t)=A0eγt/2A\left( t \right)={{A}_{0}}{{e}^{-\gamma t/2}}

Where A0 is the amplitude in the absence of damping and (b) The angular frequency ω* of the damped oscillator is less than ω0, the frequency of the undamped oscillation. The relation between them is

ω=ω0(1γ24ω02)1/2\omega *={{\omega }_{0}}{{\left( 1-\frac{{{\gamma }^{2}}}{4\omega _{0}^{2}} \right)}^{1/2}}

Energy Of A Weakly Damped Oscillator

We shall now develop an expression for the average energy of a weakly damped oscillator at any instant of time. We have seen that, in the case of weak damping (γ<2ω0),\left( \gamma <2{{\omega }_{0}} \right), the displacement and velocity of the oscillator are respectively given by Eqs. (3.16) and (3.18). If m is the mass of the oscillator, its instantaneous kinetic energy is

12m(dxdt)2\frac{1}{2}m{{\left( \frac{dx}{dt} \right)}^{2}}

Which with the help of Eq. (3.18) becomes

KE =12mA2eγt{ωsin(ωtδ)+γ2cos(ωtδ)}2=\frac{1}{2}m{{A}^{2}}{{e}^{-\gamma t}}{{\left\{ \omega *\sin \left( \omega *t-\delta \right)+\frac{\gamma }{2}\cos \left( \omega *t-\delta \right) \right\}}^{2}} =12mA2eγt{ω2sin2(ωtδ)+ωγsin(ωtδ)cos(ωtδ)+γ24cos2(ωtδ)}=\frac{1}{2}m{{A}^{2}}{{e}^{-\gamma t}}\left\{ \omega {{*}^{2}}{{\sin }^{2}}\left( \omega *t-\delta \right)+\omega *\gamma \sin \left( \omega *t-\delta \right)\cos \left( \omega *t-\delta \right)+\frac{{{\gamma }^{2}}}{4}{{\cos }^{2}}\left( \omega *t-\delta \right) \right\}

The instantaneous potential energy of the oscillator is given by

PE=0xKxdx=12Kx2PE=\int\limits_{0}^{x}{Kx}dx=\frac{1}{2}K{{x}^{2}}

Using Eq. (3.16) we have, since K=mω02,K=m\omega _{0}^{2}, PE=12mω02A2eγtcos2(ωtδ)PE=\frac{1}{2}m\omega _{0}^{2}{{A}^{2}}{{e}^{-\gamma t}}{{\cos }^{2}}\left( \omega *t-\delta \right)

The total energy of the oscillator at any instant of time is then given by

E(t)=KE+PEE\left( t \right)=KE+PE =12mA2eγt{ω2sin2(ωtδ)+ωγ2sin2(ωtδ)+(γ24+ω02)cos2(ωtδ)}=\frac{1}{2}m\,{{A}^{2}}{{e}^{-\gamma t}}\left\{ \omega {{*}^{2}}{{\sin }^{2}}\left( \omega *t-\delta \right)+\frac{\omega *\gamma }{2}\sin 2\left( \omega *t-\delta \right)+\left( \frac{{{\gamma }^{2}}}{4}+\omega _{0}^{2} \right){{\cos }^{2}}\left( \omega *t-\delta \right) \right\} ….(3.21)

If damping is very small (γ<2ω0),\left( \gamma <2{{\omega }_{0}} \right), as is usually the case, the term eγt{{e}^{-\gamma t}} in Eq. (3.21) does not change appreciably during one time period T=2π/ωT*=2\pi /\omega * of the oscillation. Thus, assuming that eγt{{e}^{-\gamma t}} is sensibly constant during period T* of the oscillations, the time-averaged energy of the oscillator is given by

<E(t)>=12mA2eγt{ω2<sin2(ωtδ)>+ωγ2<sin2(ωtδ)>+(γ24+ω02)<cos2(ωtδ)>}<E\left( t \right)>=\frac{1}{2}m{{A}^{2}}\,{{e}^{-\gamma t}}\left\{ \omega {{*}^{2}}<{{\sin }^{2}}\left( \omega *t-\delta \right)>+\frac{\omega *\gamma }{2}<\sin 2\left( \omega *t-\delta \right)>+\left( \frac{{{\gamma }^{2}}}{4}+\omega _{0}^{2} \right)<{{\cos }^{2}}\left( \omega *t-\delta \right)> \right\} ……(3.22)

Where notation < > implies averaging over one time period T*. A function f(t) averaged over T, is by definition, given by

<f(t)>=0Tf(t)dt0Tdt=1T0Tf(t)dt<f\left( t \right)>=\frac{\int\limits_{0}^{T}{f\left( t \right)dt}}{\int\limits_{0}^{T}{dt}}=\frac{1}{T}\int\limits_{0}^{T}{f\left( t \right)dt}

Thus,

<sin2(ωtδ)>=1T0Tsin2(2πtTδ)2dt<{{\sin }^{2}}\left( \omega *t-\delta \right)>=\frac{1}{T*}\int\limits_{0}^{T*}{{{\sin }^{2}}}{{\left( \frac{2\pi t}{T*}-\delta \right)}^{2}}dt

To integrate, let us use the transformation

2πtTδ=α\frac{2\pi t}{T*}-\delta =\alpha

So that dt=T2πddt=\frac{T*}{2\pi }d\infty

Then,

<sin2(ωtδ)>=12πδ2πδsin2αdx=14π02π(1cos2α)dx=12<{{\sin }^{2}}\left( \omega *t-\delta \right)>=\frac{1}{2\pi }\int\limits_{-\delta }^{2\pi -\delta }{{{\sin }^{2}}\alpha dx}=\frac{1}{4\pi }\int\limits_{0}^{2\pi }{\left( 1-\cos 2\alpha \right)dx=\frac{1}{2}}

Similarly,

<cos2(ωtδ)>=12<{{\cos }^{2}}\left( \omega *t-\delta \right)>=\frac{1}{2}

And

<sin2(ωtδ)>=0<\sin 2\left( \omega *t-\delta \right)>=0

Substituting for these time-averaged values in Eq. (3.22), we get

<E(t)>=14mA2eγt(ω2+γ24+ω02)<E\left( t \right)>=\frac{1}{4}m{{A}^{2}}{{e}^{-\gamma t}}\left( \omega {{*}^{2}}+\frac{{{\gamma }^{2}}}{4}+\omega _{0}^{2} \right)

Now, since ω2=ω02γ24,\omega {{*}^{2}}=\omega _{0}^{2}-\frac{{{\gamma }^{2}}}{4}, we have

<E(t)>=12mA2ω02eγt<E\left( t \right)>=\frac{1}{2}m{{A}^{2}}\omega _{0}^{2}{{e}^{-\gamma t}}

Or

<E(t)>=E0eγt<E\left( t \right)>={{E}_{0}}{{e}^{-\gamma t}}

Where E0=12mA2ω02,{{E}_{0}}=\frac{1}{2}m{{A}^{2}}\omega _{0}^{2}, is the total energy of an undamped oscillator. Hence, the energy of a weakly damped oscillator diminishes exponentially with time. The decay of the total energy is illustrated in Figure.

Exponential decay of total energy during damping of harmonic oscillations

Figure: Exponential decay of total energy during damping of harmonic oscillations

The average power dissipation during one time period is given by

< P (t) > = rate of loss of energy

=ddt<E(t)>=\frac{d}{dt}<E\left( t \right)> =γE0eγt=\gamma {{E}_{0}}\,{{e}^{-\gamma t}} =γ<E(t)>=\gamma <E\left( t \right)>

This expression may also be obtained as follows: Since the loss of energy is due to the work done by the oscillator to overcome the force of friction F=pdxdt,F=-p\frac{dx}{dt}, the instantaneous power P(t) is given by

P(t)=worktime=F.δxδt=FdxdtP\left( t \right)=\frac{\text{work}}{\text{time}}=\frac{F.\delta x}{\delta t}=F\frac{dx}{dt}

Where δψ is the change in displacement in time δt. Thus

P(t)=p(dxdt)2=mγ(dxdt)2P\left( t \right)=p{{\left( \frac{dx}{dt} \right)}^{2}}=m\gamma {{\left( \frac{dx}{dt} \right)}^{2}} ….(3.24)

Now using Eq. (3.18) we have,

P(t)=mγA2eγt{ω2sin2(ωtδ)+ωγ2sin2(ωtδ)+γ24cos2(ωtδ)}P\left( t \right)=m\gamma {{A}^{2}}{{e}^{-\gamma t}}\left\{ \omega {{*}^{2}}{{\sin }^{2}}\left( \omega *t-\delta \right)+\frac{\omega *\gamma }{2}\sin 2\left( \omega *t-\delta \right)+\frac{{{\gamma }^{2}}}{4}{{\cos }^{2}}\left( \omega *t-\delta \right) \right\}

Hence, the power dissipation during one time period of oscillation is given by

<P(t)>=mγA2eγt{ω2<sin2(ωtδ)>+ωγ2<sin2(ωtδ)>+γ24<cos2(ωtδ)>}<P\left( t \right)>=m\gamma {{A}^{2}}{{e}^{-\gamma t}}\left\{ \omega {{*}^{2}}<{{\sin }^{2}}\left( \omega *t-\delta \right)>+\frac{\omega *\gamma }{2}<\sin 2\left( \omega *t-\delta \right)>+\frac{{{\gamma }^{2}}}{4}<{{\cos }^{2}}\left( \omega *t-\delta \right)> \right\} =12γmA2eγt(ω2+γ24)=\frac{1}{2}\gamma \,m\,{{A}^{2}}\,{{e}^{-\gamma t}}\left( \omega {{*}^{2}}+\frac{{{\gamma }^{2}}}{4} \right) =12γmA2ω02eγt=\frac{1}{2}\,\gamma m{{A}^{2}}\omega _{0}^{2}{{e}^{-\gamma t}} =γ<E(t)>=\gamma <E\left( t \right)> …..(3.25)

As mentioned earlier, this loss of energy is due to the friction in the system (leading to heating) and the emission of radiation from the system (resulting in waves).