Coupled Oscillations, Physics tutorial

Introduction:

Coupled Oscillations take place when two or more oscillating systems are joined in such a manner as to permit motion energy to be exchanged between them.  Coupled oscillators take place in nature (e.g., moon and earth orbiting each other) or can be found in man-made devices (like with pacemaker).

Oscillations of two coupled masses:

To analyze effect of coupling begin with model spring-mass system. Consider two such identical systems connected (coupled) by the spring, as shown in figure given below. In this system there are two equal masses joined to springs of stiffness constant k' and coupled to each other by the spring of stiffness constant k. In equilibrium position, springs don't apply any force on either mass. Motion of this system will depend on initial conditions. That is, motion may be transverse or longitudinal depending on how masses are disturbed. For simplicity, first consider longitudinal motion of the two coupled masses. We pull one of the masses longitudinally and then release H. Restoring force will tend to bring it back to its equilibrium position. As it overshoots equilibrium mark, the coupling spring will pull the other mass. Consequently both masses begin oscillating longitudinally. This signifies that motion imparted to one of the two coupled masses is not confined to it only; it is transmitted to other mass as well. We now establish equation of motion of the masses.

169_coupled oscillation.jpg

The Differential Equation:

We choose x -axis along the length of the spring with O as the origin. Let XA and XB be coordinates of centre of the masses A and B respectively. When mass B is displaced towards right and then released, mass A will also get pulled towards right because of coupling spring. Coupled system would then begin oscillating. Assume XA and XB are instantaneous positions of masses A and B respectively. Then their displacements from respective equilibrium positions are provided by

x2 = xB - XB and x1 = xA - XA

Now at any instant of time during oscillation, forces acting on mass A are

(i) Restoring force: -k'(xA - XA) = -k'x1 and

(ii) A coupling force: k(xB - xA) - (XB - XA) = k(x2 - x1)

We are here assuming that masses are moving on the frictionless surface. By Newton's second law, equation of motion of mass A is therefore given by:

md2xA/dt2 = -k'(xA - XA) + k[xB - XB - xA - XA]

Or m(d2(xA - XA)/dt2) = md2x1/dt2 = -k'x1 + k(x2 - x1)

Since dXA/dt = 0

Dividing through by m and rearranging terms, we obtain:

d2x1/dt2 + ω02x1 - ωs2 = k/m

Similarly, equation of motion of mass B is

md2x2/dt2 = -k'x2 - k(x2 - x1)

This can also be written as:

d2x2/dt2 + ω20x2 + ωs2(x2 - x1) = 0

We can't, in general, identify motion explained by the equations as simple harmonic due to presence of coupling term ωs2(x2 - x1).

If introduce two new variables stated as

ξ1 = x1 + x2 and ξ2 = x1 - x2

The motion of the coupled system can be explained in terms of two uncoupled and independent equations:

d2ξ1/dt2 + ω12ξ1 = 0 and d2ξ2/dt2 + ω22ξ2 = 0

Where we have put

ω12 = ω02 = k'/m and ω22 = ω02 + 2ωs2 = (k' + 2k)/m

We therefore find that new co-ordinates ξ1 and ξ2 have decoupled into two independent equations that explain simple harmonic motions of frequencies ω1 and ω2 and ω2 > ω1. New coordinates are referred to as normal coordinates and simple harmonic motion related with each coordinate is known as a normal mode. Each normal mode has its own characteristic frequency known as normal mode frequency.

Normal Coordinates and Normal Modes:

The normal coordinates ξ1 and ξ2 are not the measure of displacement like ordinary co-ordinates x1 and x2. Yet they specify configuration of the coupled system at any instant of time. Using analysis write general solution of

ξ1(t) = a1cos(ω1t + Φ1)

ξ2(t) = a2cos(ω2t + Φ2)

Where a1 and a1 are amplitudes of normal modes and, Φ1 and Φ2 are their initial phases. As x1(t) = (ξ1 + ξ2)/2, we can write displacement of mass A as

x1(t) = 1/2[a1cos(ω1t + Φ1) + a2cos(ω2t + Φ2)]

Similarly, we can write the displacement of the mass B as

x2(t) = 1/2[a1cos(ω1t + Φ1) - a2cos(ω2t + Φ2)]

Constants a1, a2, Φ1, Φ2 are fixed by initial conditions. Once we know these, we can entirely find out motion of coupled masses.

Modulation of Coupled Oscillations:

To understand what happen if only one of the coupled masses is pulled and then released have to solve Equationa. Suppose the initial condition is as follows:

Suppose the initial condition is as follows:

x1(0) = 2a, x2(0) = 0, dx1/dt|t=0 = 0 and dx2/dt|t=0 = 0

You will find that displacements of two coupled masses are provided by

x1(t) = a(cosω1t + cosω2t) and x2(t) = a(cosω1t - cosω2t)

Expressing sum (difference) of two cosine functions in their product, these equations can be rewritten in physically more familiar form:

x1(t) = 2acos((ω2 - ω1)/2)tcos((ω2 + ω1)/2)t and x2(t) = 2asin((ω2 - ω1)/2)tsin((ω2 + ω1)/2)t

As before, stated ωav = (ω1 + ω2)/2 as average angular frequency and ωmod = (ω2 - ω1)/2 as modulated angular frequency. Then Equation represent modulated oscillations respectively stated by

x1(t) = amod(t)cosωavt and x2(t) = bmod(t)sinωavt

Where amod = 2acosωmodt and bmod = 2asinωmodt are modulated amplitudes.

As sine and cosine functions vary by π/2, phase difference between displacements of coupled masses is π/2. Same is true of modulated amplitudes as well.

Energy of Two Coupled Masses:

If coupling between two masses is weak, ω2 will be only slightly different from ω1, so that ωmod will be very small. Consequently amod and bmod will take quite some time to demonstrate observable change. That is, mod a and mod b will be practically constant over cycle of angular frequency ωav. Now calculate energies of masses A and B using equations. We know that energy of the oscillator executing SHM is provided by

E1 = 1/2mω2ava2mod(t) = 2ma2ω2avcos2ωmodt

and E2 = 1/2mω2avb2mod(t) = 2ma2ω2avcos2ωmodt

Total energy of two masses coupled through the spring that stores almost no energy is provided by

E = E1 + E2 = 2ma2ω2av that remains constant with time.

These equations show that at t = 0, E1 = E and E2 = 0. That is, to begin with, mass at A possesses all energy. As time passes, energy of mass at A starts decreasing. But mass at B begins to gain energy such that total energy of system remains constant.

When (ω2 - ω1)t = π/2, two masses share energy equally. When (ω2 - ω1)t = π, E1 = 0 and E2 = E, i.e. mass B possesses all energy. As time passes, energy exchange procedure continues. That is, total energy flows backward and forward twice between two masses in time T, provided by

T = 2π/(ω2 - ω1)

Normal mode analysis of other coupled systems:

Two Coupled Pendulums:

Consider two identical simple pendulums A and B having bobs of equal mass, m and suspended by strings of equal length l. Bobs of two pendulums are joined by weightless, spring of force constant k. In equilibrium position, distance between bobs is equal to length of unstretched spring.

Assume that both bobs are displaced to right from their respective equilibrium positions. Let x1(t) and x2(t) be displacements of the bobs at time t. Tension in coupling spring will be k(x1 - x2). It opposes acceleration of A but will support acceleration of B. For small amplitude approximation, equation of motion a simple pendulum is

md2x/dt2 = -mg/lx

In present case, equations of motion of bobs A and B are

md2x1/dt2 = -(mg/l)x1 - k(x1 - x2) and md2x2/dt2 = -(mg/l)x2 + k(x1 - x2)

Term ±k(x1 - x2) arises because of presence of coupling. Dividing throughout by m and rearranging terms, we get

d2x1/dt2 + -ω02x1 - ωs2(x1 - x2) = 0 and d2x2/dt2 + -ω02x2 - ωs2(x1 -x2) = 0

Where substituted ω02 = g/l and ωs2 = k/m

Inductively Coupled LC circuits:

Two electrical circuits are said to be inductively coupled when a change in the magnetic flux linked with one circuit induces emf (and hence gives rise to a current) in the other circuit. The coupling coefficient is given by

μ = M/√L1L2, where M is mutual inductance and L1 and L2 are self inductances of two coupled circuits.

1165_Inductively coupled identical LC circuits.jpg

Let I1 and I2 be instantaneous values of currents in two circuits. Equation of motion of charge in circuit 1 is attained by

q1/C1 = -L1dI1/dt + MdI2/dt

Where MdI2/dt is emf produced in circuit 1 because of current I2 in second circuit. Obviously, it tends to increase I1.

Expression can be written as

(ωp2 - ω2)(ωs2 - ω2) = M2/L1L2ω4 = μ2ω2

Where μ is coupling coefficient.

This equation is quadratic in ω2; its roots provide normal mode frequencies. For simplicity, we suppose that circuits are identical so that their natural frequencies are equal, i.e. ωp = ωs = ω0, say, then

02 - ω2)2 = μ2ω4 or ω02 - ω2 = ±μω2

Acceptable normal mode frequencies are those values of ω which correspond to positive roots and are provided by:

ω = ω0√1-μ and ω = -ω0√1-μ

When coupling is weak (μ<<1), ω1≈ω2≈ω0 and two circuits behave as basically independent. But when coupling is strong, ω1 and ω2 will be much different.

Longitudinal oscillations of n coupled masses: the wave equation:

We know that every fluid or solid has more than two coupled atoms held by intermolecular forces. To know normal modes of such a system extend preceding analysis to three or in general to N coupled oscillators that may not all be of same mass. For simplicity, first consider the system of N identical masses held together by (N + 1) identical springs, each of force constant k. The free ends of the system are rigidly fixed at x = 0 and x = l . In the equilibrium state, the masses are situated at x = a, 2a, ..., Na so that l = (N + 1) d. If ψn-1, ψn and ψn-1 are respective displacements of (n - 1)th, nth and (n+1)th masses from their mean positions, we can write equation of motion of nth mass as

md2ψn/dt2 = -k(ψn - ψn-1) - k(ψn - ψn-1)

Spring constant k is stated as restoring force per unit extension. So we can write k = F/d, where d is extension in spring. Using this relation we obtain:

d2ψn/dt2 = F/m[((ψn+1 - ψn)/d) - ((ψn - ψn-1)/d)]

If nth mass is located at the distance x from origin, then in limit Δx→0, we have

d2ψn/dt2 = F/m[(dψ/dx)n+1 - (dψ/dx)n]

This means that ψ is now function of t as well as x. Any continuous function f(x + Δx) can be defined in terms of function stated at x and its derivatives using given expansion:

f(x + Δx,t) = f(x,t) + ∂f(x,t)/∂xΔx + 1/2!(∂2f(x,t)/∂x2)(Δx)2+.....

Taking dψ/dx as f and retaining terms only up to first order in Ax, we can write

dψ/dx|x+Δx = ∂ψ/∂x|x + ∂2ψ/∂x2Δx +.....

On re-arranging terms, we get

d2ψ/dt2 = Fρld2ψ/dx2

Where ρl = m/Δx. above equation is the partial differential equation. Furthermore, quantity F/ρl has dimensions of square of velocity. For this reason, this equation is referred to as wave equation.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)

Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology.  Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online physics tutoring. Chat with us or submit request at info@tutorsglobe.com