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.

The Differential Equation:

We choose x -axis along the length of the spring with O as the origin. Let X_A and X_B 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 X_A and X_B are instantaneous positions of masses A and B respectively. Then their displacements from respective equilibrium positions are provided by

x₂ = x_B - X_B and x₁ = x_A - X_A

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

(i) Restoring force: -k'(x_A - X_A) = -k'x₁ and

(ii) A coupling force: k(x_B - x_A) - (X_B - X_A) = 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:

md²x_A/dt² = -k'(x_A - X_A) + k[x_B - X_B - x_A - X_A]

Or m(d²(x_A - X_A)/dt²) = md²x1/dt² = -k'x1 + k(x₂ - x₁)

Since dX_A/dt = 0

Dividing through by m and rearranging terms, we obtain:

d²x₁/dt² + ω₀²x₁ - ω_s² = 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 ω_s²(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 + 2ω_s² = (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 ω_av = (ω₁ + ω₂)/2 as average angular frequency and ω_mod = (ω₂ - ω₁)/2 as modulated angular frequency. Then Equation represent modulated oscillations respectively stated by

x₁(t) = a_mod(t)cosω_avt and x₂(t) = b_mod(t)sinω_avt

Where a_mod = 2acosω_modt and b_mod = 2asinω_modt 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 ω_mod will be very small. Consequently a_mod and b_mod 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 ω_av. Now calculate energies of masses A and B using equations. We know that energy of the oscillator executing SHM is provided by

E₁ = 1/2mω²_ava²_mod(t) = 2ma²ω²_avcos²ω_modt

and E₂ = 1/2mω²_avb²_mod(t) = 2ma²ω²_avcos²ω_modt

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

E = E₁ + E₂ = 2ma²ω²_av 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 - ω_s²(x₁ - x₂) = 0 and d²x₂/dt² + -ω₀₂x₂ - ω_s²(x₁ -x₂) = 0

Where substituted ω₀² = g/l and ω_s² = 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.

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. ω_p = ω_s = ω₀, say, then

(ω₀² - 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 ψ_n-1, ψ_n and ψ_n-1 are respective displacements of (n - 1)^th, n^th and (n+1)^th masses from their mean positions, we can write equation of motion of n^th mass as

md²ψn/dt² = -k(ψn - ψ_n-1) - k(ψ_n - ψ_n-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[((ψ_n+1 - ψ_n)/d) - ((ψ_n - ψ_n-1)/d)]

If n^th 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.

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