#### Damped Harmonic Motion, Physics tutorial

Introduction:

In real systems, masses on springs do not continue to oscillate forever at same amplitude; ultimately oscillations die away and object stops. This is because of fact that springs are not very effective at storing and releasing energy. Much of the energy is dissipated as heat because of friction within spring.

Let us suppose that mass moves horizontally in the viscous medium, say inside lubricated cylinder, as shown in figure given below. As mass moves, it will experience drag, that we denote by Fd. For oscillations of adequately small amplitude, it is quite sensible to model damping force after Stokes' law. That is, we take Fd to be proportional to velocity and write

Fd = -γv.....................................Eq.1

The negative sign signifies that damping force opposes motion. Constant of proportionality γ is known as damping coefficient. Numerically, it is equal to force per unit velocity and is estimated in N/ms-1 = kgms-2/ms-1 = kgs-1

Let us take the x -axis to be along length of spring. We state origin of axis (x = 0) as equilibrium position of mass. Imagine that mass (in spring-mass system) is pulled longitudinally and then released. It gets displaced from equilibrium position. At any moment, forces applying on spring-mass system are:

(i) The restoring force: -kx where k is spring factor and

(ii) The damping force: -γv where v = dx/dt is instantaneous velocity of oscillator.

This signifies that for damped harmonic oscillator, equation of motion should comprise restoring force and the damping force. Therefore, in this case Eq. 2 is modified to

md2x/dt2 = -kx - γdx/dt.....................................Eq.2

After rearranging terms and dividing throughout by m, equation of motion of the damped oscillator takes form

d2x/dt2 + 2bdx/dt + ω02x = 0.....................................Eq.3

Where ω02 = k/m

Constant b has dimensions of force/(velocity x mass) = MLT-2/LT-1 = T-1

Therefore, its unit is s-1, which is same as that of ω0.

Solutions of differential equation:

To discover how damping influences amplitude of oscillation to solve Eq. when both the restoring force and the damping force are present. The general solution, as discussed above, should involve both exponential and harmonic terms. Let us therefore take a solution of the form

X(t) = a exp(αt) .....................................Eq.4

Where a and α

Differentiating Eq.4 twice with respect to time, we attain

dx/dt = aα exp(αt)

and d2x/dt2 = aα2exp(αt)

Substituting the expressions in Eq.3, we get

2 + 2bα + ω02)a exp(αt) = 0.....................................Eq.5

For this equation to hold at all times, we should either have

a = 0

Which is trivial, or

α2 + 2bα + ω02 = 0.....................................Eq.6

This equation is quadratic in α. Let us call the two roots α1 and α2

α1 = -b + (b2 - ω02)1/2.....................................Eq.7(a)

α2 = -b - (b2 - ω02)1/2.....................................Eq.7(b)

Since Eq.3 is linear, principle .of superposition is appropriate. Therefore, general solution is get by the superposition of x1 and x2:

x(t) = exp(-bt)[a1exp{(b2 - ω02)1/2}t + a2exp{exp{-(b2- ω02)1/2}].....................................Eq.8

Heavy Damping:

The system returns to equilibrium position very slowly, without any oscillation. Heavy damping takes place when resistive forces exceed those of critical damping.

When resistance to motion is very strong, system is said to be heavily damped. Springs joining wagons of the train constitute the most significant heavily damped system.

Mathematically, a system is said to be heavily damped if b > ω0. Then the quantity is positive definite. If we put

β = √b2 - ω02

The general solution for damped oscillator given by Eq.9 reduces to

x(t) = exp(-bt)[a1exp(βt)[a1 exp(βt) + a2exp(-βt)].....................................Eq.9

This represents non-oscillatory behavior. Such a motion is called dead-beat. The actual displacement will, however, be determined by the initial conditions. Let us suppose that to begin with, the oscillator is at its equilibrium position, i.e x = 0 at t = 0. Then we give it a sudden kick so that it acquires a velocity v0, i.e. v = v0 at t = 0. Then from Eq.9 we have

a1 + a2 = 0 and -b(a1 + a2) +β(a1 - a2) = v0

These equation may be solved to give

a1 = -a2 = v0/2β

On substituting these results in Eq.9, we can write solution in compact form:

x(t) = v0/2βexp(-bt)[exp(βt)-exp(-βt)]

= v0/βexp(-bt)sinhβt...............Eq.10

Where sinhβt = (1/2)[exp(βt) - exp(-βt)] is hyperbolic sine function. From Eq. 10 it is clear that x(t) will be determined by interplay of the increasing hyperbolic function and decaying exponential. These are plotted separately. Eq.10 shows for heavily damped system when it is rapidly disturbed from equilibrium position. You will note that at first displacement increases with time. But soon exponential term becomes significant and displacement starts to decrease slowly.

Critical Damping:

You may have observed that on hitting an isolated road bump, a car bounces up and down and the occupants feel uncomfortable. To minimise this discomfort, the bouncing caused by the road bumps must be damped very rapidly and the automobile restored to equilibrium quickly. For this we use critically damped shock absorbers. Critical damping is also useful in recording instruments such as a galvanometer (pointer type as well as suspended coil type) which experience sudden impulses. We require the pointer to move to the correct position in minimum time and stay there without executing oscillations. Similarly, a ballistic galvanometer coil is required to return to zero displacement immediately.

Mathematically, we say that a system is critically damped if b is equal to the natural frequency, ω0, of the system. This means that b2 - ω02 = 0, so that eq.8 reduces to x(t) = (a1 + a2)exp(-βt)

= aexp(-bt)

Where a = a1 + a2

So, two terms in Eq.8 give same time dependence and reduce to one term. It can be easily verified that in this case general solution of Eq.3

x(t) = (P + qt)exp(-bt)

Where p and q are constants, p has dimensions of length and q those of velocity. These can be find out easily from initial conditions.

Let us suppose that system is disturbed from mean equilibrium position by the sudden impulse. (Coil of a suspended type galvanometer receives some electric charge at t = 0.) That is, at t = 0, x(0) and dx/dt|t=0 = v0. This provides p = 0 and q = v0, so that complete solution is

x(t) = v0texp(-bt)

Average energy of a weakly damped oscillator:

The presence of damping the amplitude of oscillation decreases with the passage of time. This means that energy is dissipated in overcoming resistance to motion. From Unit I we recall that at any time, the total energy of a harmonic oscillator is made up of kinetic and potential components. We can still use the same definition and write

E(t) = K.E.(t) + U(t)

= 1/2m(dx/dt)2 + 1/2kx2

Where dx/dt denotes instantaneous velocity

For the weakly damped harmonic oscillator, instantaneous displacement is given by Equation:

x(t) = a0exp(-bt)cos(wdt + Φ). By differentiating it with respect to time, instantaneous velocity is:

dx(t)/dt = v = -a0exp(-bt)[bcos(ωdt + Φ) + ωdsin(ωdt + Φ)]2

Therefore, kinetic energy of oscillator is

K.E. = 1/2ma20exp(-2bt)[b2cos2dt + Φ) + ωd2sin2dt + Φ) + bωdsin2(ωdt + Φ)]

Potential energy of the oscillator is:

U = 1/2kx2 = 1/2mω02x2

Therefore, total energy of oscillator at any time t is provided by

E(t) =  1/2ma20exp(-2bt)[(b2 + ω02)cos2dt + Φ) + ωd2sin2(ωdt + Φ) + bωdsin2(ωdt + Φ)]

Methods of characterizing damped systems:

In the viscous damping model, damped oscillator is characterized by γ and ω0. We also know that model applies to greatly different physical systems. In certain cases it is more suitable to employ other parameters to characterize damped motion.

Logarithmic Decrement:

The most suitable method to find out amount of damping present in the system is to estimate rate at which amplitude of oscillation dies away. Let us assume damped vibration shown graphically in Figure given below. Let a0 and a1 be first two successive amplitudes of oscillation separated by one period.

You will note that the amplitudes lie in same direction/quadrant. If T is period of oscillation, then using Equation for weakly damped oscillator, we can write

a1 = a0exp(-bt)

So that a0/a1 = exp(bt) = exp(γT/2m)

The ratio a0/a1, larger amplitude is in numerator. That is why this ratio is known as decrement. It is denoted by the symbol d.

Relaxation Time:

In physics we estimate decay of the quantity in terms of fraction e-1 of initial value. This provides us another way of defining damping effect by means of time taken by amplitude to decay to e-1 = 0.368 of original value. This time is known as relaxation time. To understand this, we recall that amplitude of damped oscillation is given by

a(t) = a0exp(-bt)

If we signify amplitude of oscillation after the interval of time r by a (t + T), we can write

a(t + τ) = a0exp(-b(t + T))

By taking the ratio a(t + T)/a(t), we get

1/e for bT = 1

This illustrates that for b = T-1amplitude drops to 1/ e = 0.368 of initial value. Using result in Eq., we obtain:

<P> = 2<E>/T

Relaxation time, T, is thus a measure of rapidity with which motion is damped.

Quality Factor:

Yet another way of stating damping effect is by means of rate of decay of energy. Average energy of weakly damped oscillator decays to E0e-1 in time t = 1/2b = m/γ seconds. If ωd is angular frequency, then in this time oscillator will vibrate through ωdm/γ radians. Number of radians by which weakly damped system oscillates as its average energy decays to E0e-1 is the measure of quality factor, Q:

Q = ωdm/γ = ωd/2b = ωdT/2

For weakly damped mechanical oscillator, quality factor can be stated in terms of spring factor and damping constant. For weak damping,

ωd ≈ ω0 = √k/m

Therefore Q = √km/γ

That is, quality factor of the weakly damped oscillator is directly proportional to square root of k and inversely proportional to γ

Examples of damped systems:

All harmonic oscillators in nature have some damping, which in general, is quite small. To understand effect of damping, consider two specific cases: (a) Oscillations of charge in the LCR circuit, and (b) motion of coil in the suspension type galvanometer. These are of particular interest to us as former has wide applications in radio engineering and latter is utilized in physics laboratory.

An LCR Circuit:

In the ideal LC circuit, charge excites SHM. If a current I flows through circuit because of discharging/charging of capacitor, voltage drop across resistor is RI. Therefore Equation modifies to -

q/c = -LdI/dt - dq/dt

Equation can be written as:

Ld2q/dt2 + Rdq/dt + q/c = 0

L, R and 1/C are respectively analogous to m, γ and k.

This signifies that resistor in electric circuit has exactly analogous effect as that of viscous force in the mechanical system. To continue further, divide Equation given above throughout by L attaining

d2q/dt2 + (R/L)dq/dt + 1q/LC = 0

ω02 = 1/LC and b = R/2L

We know that b has dimensions of time inverse. This means that R/L has unit of s-1, same as that of ω0. That is why ω0L is measured in ohm.

For the weakly damped circuit, charge on capacitor plates at time t is

q(t) = q0exp(-Rt/2L)cos(ωdt + Φ) with angular frequency

ωd = √(1/LC - R2/4L2)

Equation given above illustrates that charge amplitude q0exp(-R/2Lt) will decay at rate that depends on resistance. Therefore in LCR circuit, resistance is only dissipative element; an increase in R increases rate of decay of charge and decreases frequency of oscillations.

Since ω0L is measured in ohms, 1/ω0C is also measured in ohms. These are respectively referred to as inductive reactance and capacitive reactance.

Q value of weakly damped LCR circuit is

Q = ωd/2b ≈ ω0L/R = 1/R√L/C

This equation shows that for a purely inductive circuit (R = 0), quality factor will be infinite.

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