Find the eigenfrequencies and describe the normal modes for a system of three equal masses m and four springs, all with spring constant k, with the system fixed at the ends as shown in the figure below. The motion can only take place in one dimension, along the axes of the springs.

Solution:

- Concepts:

Coupled oscillations, normal modes - Reasoning:

We are asked to find the normal modes of coupled harmonic oscillators. - Details of the calculation:

The kinetic energy is T = ½[m(dx_{1}/dt)^{2}+ m(dx_{2}/dt)^{2}+ m(dx_{3}/dt)^{2}],

and the potential energy is U = (k/2)[(x_{2}- x_{1})^{2}+ (x_{3}- x_{2})^{2}+ x_{1}^{2}+ x_{3}^{2}].

U = (k/2)[2x_{1}^{2}+ 2x_{2}^{2}+ 2x_{3}^{2}- x_{1}x_{2}- x_{2}x_{1}- x_{2}x_{3}- x_{3}x_{2}].

The x_{i}are the displacements from the equilibrium positions. We use the x_{i}as our generalized coordinates q_{i}.

The Lagrangian is L = T - U. This can be put into the form

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{11}= T_{22}= T_{33}= m, T_{ij}(i≠j) = 0,

k_{11}= k_{22}= k_{33}= 2k, k_{12}= k_{21}= k_{23}= k_{32}= -k, k_{13}= k_{31}= 0.

Solutions of the form x_{j}= Re(A_{j}e^{iωt}) can be found. For solutions of this form the equations of motion reduce to.

We can find the ω

^{2}from det(k_{ij}- ω^{2}T_{ij}) = 0. For a system with n degrees of freedom, n characteristic frequencies ω_{a}can be found. Our system has 3 degrees of freedom.**.**(2k - ω

^{2}m)^{3}– 2k^{2}(2k - ω^{2}m) = 0.

(i) (2k - ω^{2}m) = 0, ω_{1}^{2}= 2k/m, A_{2}= 0, A_{1}= -A_{3}.(ii) (2k - ω

^{2}m)^{2}– 2k^{2 }= 0. ω_{2}^{2}= (2 + √2)k/m, A_{3}= A_{1}, A_{2}= - √2A_{1}.

(iii) ω_{3}^{2}= (2 - √2)k/m, A_{3}= A_{1}, A_{2}= √2A_{1}.

Two particles of mass m and one particle of mass M are constrained to move on a line as shown. They are connected by massless springs with spring constant k.

Find the normal modes and eigenfrequencies of the system, keeping M/m finite.

Solution:

- Concepts:

Coupled oscillations, normal modes - Reasoning:

We are asked to find the normal modes of coupled harmonic oscillators. - Details of the calculations:

The kinetic energy is T = ½[ m(dx_{1}/dt)^{2}+ M(dx_{2}/dt)^{2}+ m(dx_{3}/dt)^{2}],

and the potential energy is U = (k/2)[(x_{2}- x_{1})^{2}+ (x_{3}- x_{2})^{2}].

U = (k/2)[x_{1}^{2}+ 2x_{2}^{2}+ x_{3}^{2}- x_{1}x_{2}- x_{2}x_{1}- x_{2}x_{3}- x_{3}x_{2}].

The x_{i}are the displacements from the equilibrium positions. We use the x_{i}as our generalized coordinates q_{i}.

The Lagrangian is L = T - U. This can be put into the form

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{11 }= T_{33}= m, T_{22}= M, T_{ij}(i≠j) = 0,

k_{11}= k_{33}= k, k_{22}= 2k, k_{12}= k_{21}= k_{23}= k_{32}= -k, k_{13}= k_{31}= 0.

Solutions of the form x_{j}= Re(A_{j}e^{iω t}) can be found. For solutions of this form the equations of motion reduce to:

(k - ω^{2}m)^{2}(2k - ω^{2}M) - 2k^{2}(k - ω^{ 2}m) = 0.

Solution 1: k - ω^{2}m = 0, ω = (k/m)^{½}.

If k - ω^{2}m ≠ 0, then (k - ω^{2}m)(2k - ω^{2}M) = 2k^{2}. ω^{2}(ω^{2}mM - kM - 2km) = 0.

Solution 2: ω = 0.

Solution 3: ω = (k/m + 2k/M)^{½}.

The displacements for each mode are determined from the equations of motion.

Solution 1: ω = (k/m)^{½}.

Equation 1 yields kA_{2}= 0, equation 2 then yields kA_{1}+ kA_{3}= 0.

A_{1}= - A_{3}, A_{2}= 0, the central mass is stationary, m_{1}and m_{3}move in opposite directions with equal amplitudes.

Solution 2: ω = 0.

A_{1}= A_{2}= A_{3}, translation of the CM, no relative motion.

Solution 3: ω = (k/m + 2k/M)^{½}.

The displacements

Equation 1 yields –(2m/M)A_{1}= A_{2}.

Equation 3 yields –(2m/M)A_{3}= A_{2}.

A_{1}= A_{3}= -(M/(2m))A_{2}, the central mass move in a direction opposite to the direction of the outer masses.

Two masses a and b are on a horizontal surface. Mass b has a spring
connected to it and is at rest. Mass a has an initial velocity v_{0}
along the x-axis and strikes the spring of constant k, compressing it and thus
starting mass b in motion along the x-axis.

(a) Find the maximum force exerted.

(b) Find the resulting motion of mass a
while in contact with the spring.

Solution:

- Concepts:

Small oscillations, coupled oscillations, normal modes

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found. We can find the ω^{2}from det(k_{ij }- ω^{2}T_{ij}) = 0. - Reasoning:

During the time interval that both masses are in contact with the spring, they can be treated as coupled harmonic oscillators. - Details of the calculations:

(a) Assume that at t = 0 mass a makes contact with the spring. At t = 0 both masses are at their equilibrium position. Let x_{a}(t) and x_{b}(t) denote the their displacements from their equilibrium positions, respectively.

In the collision energy and momentum are conserved.

T_{i}= T_{f}+ U, P_{i}= P_{f}. The maximum force is exerted when the spring has maximum compression. At that instant both masses move with the same velocity.

½m_{a}v_{0}^{2}= ½(m_{a}+ m_{b})v^{2}+ ½k(x_{b}- x_{a})_{min}^{2}.

m_{a}v_{0}= (m_{a}+ m_{b})v.

Combining the two equations we have (x_{b}- x_{a})_{min}= (m_{a}m_{b}/k)^{½}v_{0}/(m_{a}+ m_{b})^{½}.

The maximum force is k(x_{b}- x_{a})_{min}= (km_{a}m_{b})^{½}v_{0}/(m_{a}+ m_{b})^{½}.

(b) T = ½m_{a}(dx_{a}/dt)^{2}+ ½m_{b}(dx_{b}/dt)^{2},

U = ½k(x_{b}- x_{a})^{2}= ½k(x_{b}^{2}+ x_{a}^{2}- x_{a}x_{b}- x_{b}x_{a}).

T_{11}= m_{a}, T_{22}= m_{b}, k_{11}= k_{22}= k, k_{12}= k_{21}= -k.

det(k_{ij }- ω^{2}T_{ij}) = 0 --> ω_{1}= 0, ω_{2}^{2}= k(m_{a}+ m_{b})/(m_{a}m_{b}).

For ω_{1}= 0 we have A_{1}= A_{2}. For ω_{2}^{2}= k(m_{a}+ m_{b})/(m_{a}m_{b}) we have A_{2}= -(m_{a}/m_{b})A_{1}.

The most general solutions are:

x_{a}= A_{1}t + B_{1}cos(ωt + φ), x_{b}= A_{1}t - (m_{a}/m_{b})B_{1}cos(ωt + φ),

dx_{a}/dt = A_{1}- ωB_{1}sin(ωt + φ), dx_{b}/dt = A_{1}+ ω(m_{a}/m_{a})B_{1}sin(ωt + φ).

The initial conditions at t = 0 are x_{a}= x_{b}= 0, dx_{a}/dt = v_{0}, dx_{b}/dt = 0.

This implies B_{1}cos(φ) = 0, φ = π/2.

v_{0}= A_{1}- ωB_{1}, A_{1}+ ω(m_{a}/m_{b})B_{1}= 0.

A_{1}= v_{0}m_{a}/(m_{a}+ m_{b}), B_{1}= -(1/ω)v_{0}m_{b}/(m_{a}+ m_{b}).

Therefore

x_{a}= (v_{0}/ω)m_{b}/(m_{a}+ m_{b})sinωt + m_{a}v_{0}/(m_{a}+ m_{b})t,

dx_{a}/dt = v_{0}m_{b}/(m_{a}+ m_{b})cosωt + m_{a}v_{0}/(m_{a}+ m_{b}),

as long as mass a is in contact with the spring.

[At t = 0 we have dx_{a}(lab) /dt = v_{0}m_{b}/(m_{a}+ m_{b}) + m_{a}v_{0}/(m_{a}+ m_{b}) = v_{0}.]

A different approach:

Solution:

- Concepts:

Simple harmonic motion, frame transformations, two interacting particles - Reasoning:

In the center of mass frame we have an elastic collision between two particles with an interaction energy U(x) = ½k(x-x_{0})^{2}, where x-x_{0}is the distance between the particles, as long as particle a is in contact with the spring . - Details of the calculations:

(a) In the center of mass frame we have an elastic collision. Assume mass a approaches the CM with with speed v_{a}from the left, and mass b approaches the CM with speed v_{b}from the right. The CM is at rest and mass a recedes from the CM with with speed v_{a}to the right, while mass b recedes from the CM with with speed v_{b}to the left after the collision.

We have m_{a}v_{a}= m_{b}v_{b}in the CM frame before the collision. To find the relationship between the speeds in the CM and the lab frame we use

m_{a}(v_{0}- v_{b}) = m_{b}v_{b}. v_{b}= m_{a}v_{0}/(m_{a}+ m_{b})), v_{a}= m_{b}v_{0}/(m_{a}+ m_{b}).

The maximum force is exerted when all the kinetic energy is converted into potential energy.

F_{max}= k(x_{min}-x_{0}).

k(x_{min}-x_{0})^{2 }= m_{a}v_{a}^{2}+ m_{b}v_{b}^{2}= m_{a}(m_{b}v_{0}/(m_{a}+ m_{b}))^{2}+ m_{b}(m_{a}v_{0}/(m_{a}+ m_{b}))^{2}= m_{a}m_{b}v_{0}^{2}/(m_{a}+ m_{b}).

k(x_{min}-x_{0}) = (km_{a}m_{b})^{½}v_{0}/(m_{a}+ m_{b})^{½}.

(b) Assume that mass a makes contact with the spring at t = 0, at the origin of the CM and the lab frame. (We assume that at t = 0 the origin of the two frames coincide.) Assume that at t = 0 the distance between mass a and mass b is x_{0}.

In the CM frame when mass a moves a distance x_{a}to the right, mass b moves a distance x_{b}= x_{a}(m_{a}/m_{b}) to the left. The distance between the masses becomes x = x_{0}- (x_{a}+ x_{b}) = x_{0}- x_{a}(m_{a}+ m_{b})/m_{b}= x_{0}- x'.

For t > 0 we have U = ½kx^{'2}= ½[k(m_{a}+ m_{b})^{2}/m_{b}^{2}]x_{a}^{2}until the time when contact is lost again.

The kinetic energy of the particles is

T = ½m_{a}(dx_{a}/dt)^{2}+ ½m_{b}(dx_{b}/dt)^{2}= ½(m_{a}+ m_{a}^{2}/m_{b})(dx_{a}/dt)^{2}.

L = T - U = ½(m_{a}(m_{b}+ m_{a})/m_{b})(dx_{a}/dt)^{2}- ½)[k(m_{a}+ m_{b})^{2}/m_{b}^{2}]x_{a}^{2}.

This is a Lagrangian for simple harmonic motion.

d^{2}x_{a}/dt^{2}= -k(m_{b}+ m_{a})/(m_{a}m_{b})x_{a}= -(k/μ)x_{a}, with μ = m_{a}m_{b}/(m_{b}+ m_{a}).

We therefore have

x_{a}= Asinωt, ω = (k/m)^{½}, until t = π/ω (½ period of the motion).

dx_{a}/dt = ωAcosωt.

At t = 0 we have ωA = v_{a}= m_{b}v_{0}/(m_{a}+ m_{b}), A = m_{b}v_{0}/[ω(m_{a}+ m_{b})]

In the lab frame we have x_{a}(lab) = x_{a}+ v_{b}t.

The motion of mass a is described by

x_{a}(lab) = (v_{0}/ω)m_{b}/(m_{a}+ m_{b})sinωt + m_{a}v_{0}/(m_{a}+ m_{b})t,

dx_{a}(lab) /dt = v_{0}m_{b}/(m_{a}+ m_{b})cosωt + m_{a}v_{0}/(m_{a}+ m_{b}),

as long as mass a is in contact with the spring.

[At t = 0 we have dx_{a}(lab) /dt = v_{0}m_{b}/(m_{a}+ m_{b}) + m_{a}v_{0}/(m_{a}+ m_{b}) = v_{0}.]

Use Lagrange's equations to find the normal modes and normal frequencies for
linear vibrations of the CO_{2} molecule shown below.

Solution:

- Concepts:

Small oscillations, normal modes

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found. We can find the ω^{2}from det(k_{ij }- ω^{2}T_{ij}) = 0. - Reasoning:

We are asked to find the normal modes of the system. - Details of the calculations:

We are only considering linear vibrations.

Let q_{i}denote the displacement of the ith particle from its equilibrium position.

q_{i}= x_{i}- x_{i0}.

Then

T = ½[m_{o}(dq_{1}/dt)^{2}+ m_{c}(dq_{2}/dt)^{2}+ m_{o}(dq_{3}/dt)^{2}].

U = ½k_{1}[(q_{2}- q_{1})^{2}+ (q_{3}- q_{2})^{2}] + ½k_{2}[(q_{3}- q_{1})^{2}].

U = ½(k_{1}+ k_{2})q_{1}^{2}+ ½(k_{1}+ k_{2})q_{3}^{2}+ k_{1}q_{2}^{2}- ½k_{1}(q_{2}q_{1}+ q_{1}q_{2}+ q_{3}q_{2}+ q_{2}q_{3})

- ½k_{2}(q_{3}q_{1}+ q_{1}q_{3}).

T_{11}= T_{33}= m_{o}, T_{22}= m_{c}, T_{ij}(i ≠ j) = 0.

k_{11}= k_{33}= k_{1}+ k_{2}, k_{22}= 2k_{1}, k_{12}= k_{21}= k_{32}= k_{23}= -k_{1}, k_{31}= k_{13}= -k_{2}.

det(k_{ij }- ω^{2}T_{ij}) = 0.

.

ω^{2}= 0, or ω^{2}= (k_{1}/m_{c}+ k_{1}/m_{o}+ k_{2}/m_{o}) ± (k_{1}/m_{c}- k_{2}/m_{o}).

To find the relative displacements we solve the system of equations

for each ω.

ω ^{2}= 0:uniform linear motion, A _{1}= A_{2}= A_{3}.ω ^{2}= 2k_{1}/m_{c}+ k_{1}/m_{o}:A _{3}= A_{1}, A_{2}= -(2m_{o}/m_{c})A_{1}.<--o o----> <--o ω ^{2}= k_{1}/m_{o}+ 2k_{2}/m_{o}:A _{3}= -A_{1}, A_{2}= 0.o---> o <---o

A particle of mass m is attached to a rigid support by a spring with a force constant k. At equilibrium, the spring hangs vertically downward. To this mass-spring combination is attached an identical oscillator, the spring of the latter being connected to the mass of the former.

(a) Show that by appropriate choice of coordinates and their zero-points the
equations of motion can be expressed as

md^{2}x_{1}/dt^{2}
+ 2kx_{1} – kx_{2} = 0,

md^{2}x_{2}/dt^{2}
+ kx_{2} – kx_{1} = 0.

(b) Calculate the characteristic
frequencies for one-dimensional vertical oscillations.

(c)
Qualitatively describe the normal modes of the system with a short discussion
plus drawings.

(d) Quantitatively compare the characteristic
frequencies with the frequencies when either of the particles is held fixed
while the other oscillates.

Solution:

- Concepts:

Lagrangian mechanics, coupled oscillations - Reasoning:

We are asked to find the normal modes of coupled harmonic oscillators. - Details of the calculation:

(a) T = ½m(dy_{1}/dt)^{2}+ ½m(dy_{2}/dt)^{2}, U = ½ky_{1}^{2}+ ½k(y_{2}– y_{1})^{2}– mgy_{1}– mgy_{2}.

Here y_{1}and y_{2}are the displacements of the springs from their respective equilibrium positions. The y-axis points down.

Let y_{1}= x_{1}+ a, y_{2}= x_{2}+ b.

T = ½m(dx_{1}/dt)^{2}+ ½m(dx_{2}/dt)^{2},

U = kx_{1}^{2}+ ka^{2}+ 2kx_{1}a + ½kx_{2}^{2}+ ½kb^{2}+ kx_{2}b – kx_{1}x_{2}– kab - kx_{1}b - kx_{2}a

– mgx_{1}– mgx_{2}- mga – mgb – constant.

The zero of the potential energy can be chosen arbitrarily.

To put the potential energy function into the form U = ½∑_{ij}k_{ij}x_{i}x_{j}, let ka^{2}+ ½kb^{2}– kab - mga - mgb = constant.

Choose 2ka - kb – mg = 0, kb - ka – mg = 0, which implies that ka = 2mg, kb = 3 mg.

Then x_{1}and x_{2}are the displacements from the equilibrium positions in the gravitational potential near the surface of Earth,

and T = ½m(dx_{1}/dt)^{2}+ ½m(dx_{2}/dt)^{2}, U = kx_{1}^{2}+ ½kx_{2}^{2}– kx_{1}x_{2}, L = T – U.

Lagrange's equations become

md^{2}x_{1}/dt^{2}+ 2kx_{1}– kx_{2}= 0,

md^{2}x_{2}/dt^{2}+ kx_{2}– kx_{1}= 0.(b) Assume x

_{1}= Re(A_{1}exp(iωt)), x_{2}= Re(A_{2}exp(iωt)). Then

.

ω^{2}= [3/2 ± √5/2](k/m), ω_{1}^{2}= [3/2 + √5/2](k/m), ω_{2}^{2}= [3/2 - √5/2](k/m).

ω_{1}= 1.62 (k/m)^{½}, ω_{2}= 0.62 (k/m)^{½}.(c) ω = ω

_{1}: A_{2}= -½A_{1}(√5 - 1) = -0.62 A_{1}.

ω = ω_{2}: A_{2}= ½A_{1}(√5 + 1) = 1.62 A_{1}.

(d) When particle 1 is held fixed (x_{1}= 0), then particle 2 oscillates with angular frequency ω, where ω = (k/m)^{½}.

When particle 2 is held fixed (x_{2}= 0), then particle 1 oscillates with angular frequency ω, where ω = (2k/m)^{½}= 1.41 (k/m)^{½}.

The separation of the characteristic frequencies of the coupled system is greater than the separation of the single particle frequencies.

Four mass points of mass m move on a circle of radius R. Each mass point is
coupled to its two neighboring points by a spring constant k.

(a) Find the Lagrangian of the system, and derive the equations of motion of
the system.

(b) Calculate the eigen-frequencies of the system, and discuss the related
eigen-vibrations.

Solution:

- Concepts:

Coupled oscillations - Reasoning:

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{ii}= m, T_{ij}(i ≠ j) = 0, i, j = 1, 2, 3, 4, (cyclic).

k_{ii}= 2k, k_{ij}= -k if j = i ± 1, k_{ij}= 0 otherwise.

Equations of motion: d/dt(∂L/∂(dq_{i}/dt)) - ∂L/∂q_{i}= 0.

Solutions of the form q_{j}= Re(A_{j}e^{iω t}) can be found. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. Some frequencies may be degenerate. For a particular frequency ω_{α}we solve

∑_{j}[k_{ij}- ω_{α}T_{ij}]A_{jα}= 0

to find the A_{jα}. Since it is difficult to evaluate the determinant of a 4 x 4 matrix, we find the solutions of this system of coupled equations using physical insight.

(a) d/dt(∂L/∂(dq_{i}/dt)) - ∂L/∂q_{i}= 0.

(-kq_{i-1}+ 2kq_{i}- kq_{i+1}) + md^{2}q_{i}/dt^{2}= 0, i,j = 1, 2, 3, 4, (cyclic). - Details of the calculation:

(b) 1.) ω_{1}= 0, A_{1}= A_{2}= A_{3}= A_{4}, ∑_{j}k_{ij}= 2k – 2*k = 0.

All other modes can have no net angular momentum.

2.) Assume A_{1}and A_{3}are fixed. A_{1}= A_{3}= 0, A_{2}= -A_{4}.

The equation of motion for i = 2 is k_{21}A_{1}+ k_{22}A_{2}+ k_{23}A_{3}– ω_{2}^{2}T_{22 }A_{2}= 0,

2k - ω_{2}^{2}m = 0, ω_{2}^{2}= 2k/m.

3.) Assume A_{2}and A_{4}are fixed. A_{1}= -A_{3}= 0.

For this mode ω_{3}^{2}= 2k/m. Mode 2 and 3 are degenerate. Any linear combination of these modes is also a normal mode. For example, the mode with A_{2}= A_{3}= 0, A_{1}= A_{4}, , A_{1}= - A_{2}is a linear combination of modes 2 and 3 and is not linearly independent.

4.) Assume A_{1}= A_{3}, A_{2}= A_{4}, A_{1}= - A_{2}.

The equation of motion for i = 2 is k + 2k + k - ω_{2}^{2}m = 0, ω_{4}^{2}= 4k/m.For the eigen-vibrations we have:

1.) q_{i}= vt. (uniform rotation)

2.) q_{1}= q_{3}= 0, q_{2}= C_{2}cos(ω_{2}t + φ_{2}), q_{4}= -C_{2}cos(ω_{2}t + φ_{2})

3.) q_{2}= q_{4}= 0, q_{3}= C_{3}cos(ω_{3}t + φ_{3}), q_{1}= -C_{3}cos(ω_{3}t + φ_{3})

4.) q_{1}= q_{3}= C_{4}cos(ω_{4}t + φ_{4}), q_{2}= q = -C_{4}cos(ω_{4}t + φ_{4})

The C_{i}and φ_{i}depend on the initial conditions. The mode with the highest frequency is mode 4 (non-degenerate) and the mode with the lowest frequency is mode 1 (non-degenerate).

Consider six equal masses constrained to move on a circle of fixed radius and
connected by identical springs of spring constant k.

(a) Find the normal mode frequencies of the system for small displacement of
the masses.

(b) Find the (time dependent) displacement of the masses for each normal mode.
Give a physical description of the motion of the masses for normal modes with
the highest and lowest frequencies.

Solution:

- Concepts:

Coupled oscillations - Reasoning:

L = ½∑_{ij}[T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}] with T_{ij}= T_{ji}, k_{ij}= k_{ji}.

Here T_{ii}= m, T_{ij}(i ≠ j) = 0, i, j = 1, 2, 3, 4, 5, 6, (cyclic).

k_{ii}= 2k, k_{ij}= -k if j = i ± 1, k_{ij}= 0 otherwise.

Solutions of the form q_{j}= Re(A_{j}e^{iω t}) can be found. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. Some frequencies may be degenerate. For a particular frequency ω_{α}we solve

∑_{j}[k_{ij}- ω_{α}T_{ij}]A_{jα}= 0

to find the A_{jα}. Since it is difficult to evaluate the determinant of a 6 x 6 matrix, we find the solutions of this system of coupled equations using physical insight. - Details of the calculation:

(a)

1.) ω_{1}= 0, A_{1}= A_{2}= A_{3}= A_{4}= A_{5}= A_{6}, ∑_{j}k_{ij}= 2k – 2k = 0.

All other modes must have zero total or net angular momentum

2.) Assume A_{1}and A_{4}are fixed. A_{1}= A_{4}= 0, A_{2}= A_{3}, A_{5}= A_{6}, A_{2}= - A_{5}.

The equation of motion for i = 2 is k_{21}A_{1}+ k_{22}A_{2}+ k_{23}A_{3}– ω_{2}^{2}T_{22 }A_{2}= 0,

2k - k - ω_{2}^{2}m = 0, ω_{2}^{2}= k/m.

3.) Assume A_{2}and A_{5}are fixed. A_{2}= A_{5}= 0, A_{3}= A_{4}, A_{1}= A_{6}, A_{3}= - A_{6}.

For this mode ω_{3}^{2}= k/m. Mode 2 and 3 are degenerate. Any linear combination of these modes is also a normal mode. The mode with A_{3}= A_{6}= 0, A_{1}= A_{2}, A_{4}= A_{5}, A_{1}= - A_{4}is a linear combination of modes 2 and 3 and is not linearly independent.

4.) Assume A_{1}and A_{4}are fixed. A_{1}= A_{4}= 0, A_{2}= A_{5}, A_{3}= A_{6}, A_{2}= - A_{3}.

The equation of motion for i = 2 is 2k + k - ω_{2}^{2}m = 0, ω_{4}^{2}= 3k/m.

5.) Assume A_{2}and A_{5}are fixed. A_{2}= A_{5}= 0, A_{3}= A_{6}, A_{1}= A_{4}, A_{3}= - A_{4}.

For this mode ω_{5}^{2}= 3k/m. Mode 4 and 5 are degenerate. Any linear combination of these modes is also a normal mode.

6.) Assume A_{1}= A_{3}= A_{5}, A_{2}= A_{4}= A_{6}, A_{1}= - A_{2}.

The equation of motion for i = 2 is k + 2k + k - ω_{2}^{2}m = 0, ω_{6}^{2}= 4k/m.

(b)

1.) q_{i}= vt.

2.) q_{1}= q_{4}= 0, q_{2}= q_{3}= C_{2}cos(ω_{2}t + φ_{2}), q_{5}= q_{6}= -C_{2}cos(ω_{2}t + φ_{2})

3.) q_{2}= q_{5}= 0, q_{3}= q_{4}= C_{3}cos(ω_{3}t + φ_{3}), q_{6}= q_{1}= -C_{3}cos(ω_{3}t + φ_{3})

4.) q_{1}= q_{4}= 0, q_{2}= q_{5}= C_{4}cos(ω_{4}t + φ_{4}), q_{3}= q_{6}= -C_{4}cos(ω_{4}t + φ_{4})

5.) q_{2}= q_{5}= 0, q_{3}= q_{6}= C_{5}cos(ω_{5}t + φ_{5}), q_{1}= q_{4}= -C_{5}cos(ω_{5}t + φ_{5})

6.) q_{1}= q_{3}= q_{5}= C_{6}cos(ω_{6}t + φ_{6}), q_{2}= q_{4}= q_{6}= -C_{6}cos(ω_{6}t + φ_{6})

The C_{i}and φ_{i}depend on the initial conditions. The mode with the highest frequency is mode 6 (non-degenerate) and the mode with the lowest frequency is mode 1 (non-degenerate).

Three point masses of mass m move on a circle of radius
R. The equilibrium positions are shown in the figure. Each point
mass is coupled to its two neighboring points

by a spring with spring constant k.

(a) Write down the Lagrangian of the system.

(b) Find the normal modes of the system.

(c) If at t = 0 mass 1 is displaced from its equilibrium position

clockwise by 9^{o}, what is the subsequent motion of all the
masses?

Solution:

- Concepts:

Coupled oscillations - Reasoning:

If L = ½∑_{ij}(T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}), the solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found. We can find the ω^{2}from det(k_{ij}- ω^{2}T_{ij}) = 0. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. While the secular equation det(k_{ij}- ω^{2}T_{ij}) = 0 can in principle always be solved, it is often simpler to find the normal modes by using physical insight and noting the symmetries of the system. - Details of the calculation:

(a) T = ½m((dq_{1}/dt)^{2}+ (dq_{2}/dt)^{2}+ (dq_{3}/dt)^{2}), U = ½k((q_{2}– q_{1})^{2}+ (q_{3}– q_{2})^{2}+ (q_{1}– q_{3})^{2}),

L = T – U = ½∑_{ij}(T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}).

The q_{i}are the displacements of the masses from their equilibrium positions, q_{i}= rθ_{i}.

Here T_{ii}= m, T_{ij}(i≠j) = 0, k_{ii}= 2k, k_{ij}(i≠j) = -k for i, j = 1, 2, 3.

(b) Equations of motion:

Solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found.

We can find the ω^{2}from det(k_{ij }- ω^{2}T_{ij}) = 0. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. Some frequencies may be degenerate.

For a particular frequency ω_{α}we solve ∑_{j}(k_{ij }- ω^{2}T_{ij}) A_{jα}= 0 to find the A_{jα}.

ω^{2}(ω^{4 }- 6(k/m) ω^{2}+ 9(k/m)^{2}= 0, ω_{1}^{2}= 0, ω_{2}^{2}= ω_{3}^{2}= ω^{2}= 3k/m.

ω_{1}^{2}= 0: A_{1}= A_{2}= A_{3}, constant displacement or uniform circular motion.

For the degenerate frequency we pick two linearly independent modes.

ω_{2}^{2}= 3k/m. A_{2}= 0, A_{1}= -A_{3}.

ω_{3}^{2}= 3k/m. A_{3}= 0, A_{1}= -A_{2}.

(c) The most general solution for each coordinate q_{j}is a sum of simple harmonic oscillations in all of the frequencies ω_{α}.

q_{1}(t) = Re(A + (B + C)exp(iω t)), q_{2}(t) = Re(A - Cexp(iω t)), q_{3}(t) = Re(A - Bexp(iω t)),

dq_{1}(t)/dt = Re(iω(B+C)exp(iω t)), dq_{2}(t)/dt = Re(-iωCexp(iω t)), dq_{3}(t)/dt = Re(-iωBexp(iω t)).

Given: q_{1}(0) = Re(A + B + C) = D, D = R*9^{o}= 9*0.05π,

q_{2}(0) = Re(A - C) = 0, q_{3}(0) = Re(A - B) = 0.

Therefore Re(A) = Re(B) = Re(C) = D/3.

dq_{1}/dt = dq_{2}/dt = dq_{3}/dt = 0 at t = 0. Therefore Im(A) = Im(B) = Im(C) = 0.

q_{1}(t)/ = (D/3)(1 + 2cos(ωt)), θ_{1}(t) = 3^{o}(1 + 2cos((3k/m)^{½}t)),

θ_{2}(t) = θ_{3}(t) = 3^{o}(1 - cos((3k/m)^{½}t)).

Three point masses, one of mass 2m and two of mass m are constrained to move on a circle of radius R. Each mass point is coupled to its two neighboring points by a spring. The springs coupling mass 1 and 3 and mass 1 and 2 have spring constant k, and the spring coupling mass 2 and mass 3 has spring constant 2k. Find the normal modes of the system.

Solution:

- Concepts:

Coupled oscillations - Reasoning:

If L = ½∑_{ij}(T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}), the solutions of the form q_{j}= Re(A_{j}e^{iωt}) can be found. We can find the ω^{2}from det(k_{ij}- ω^{2}T_{ij}) = 0. For a system with n degrees of freedom, n characteristic frequencies ω_{α}can be found. While the secular equation det(k_{ij}- ω^{2}T_{ij}) = 0 can in principle always be solved, it is often simpler to find the normal modes by using physical insight and noting the symmetries of the system. - Details of the calculation:

Formal approach:

T = m(dq_{1}/dt)^{2}+ ½m(dq_{2}/dt)^{2}+ ½m(dq_{3}/dt)^{2},

U = ½k(q_{2}– q_{1})^{2}+ k(q_{3}– q_{2})^{2}+ ½k(q_{1}– q_{3})^{2}).

L = T – U = ½∑_{ij}(T_{ij}(dq_{i}/dt)(dq_{j}/dt) - k_{ij}q_{i}q_{j}).

The q_{i}are the displacements of the masses from their equilibrium positions, q_{i}= rθ_{i}.

Here T_{11}= 2m, T_{22}= T_{33}= m, T_{ij}(i≠j) = 0,

k_{11}= 2k, k_{22}= k_{33}= 3k, k_{12}= k_{21}= k_{13}= k_{31}= -k, k_{23}= k_{32}= -2k.

ω^{2}(ω^{4 }- 7(k/m)ω^{2}+ 10(k/m)^{2}) = 0, ω_{1}^{2}= 0, ω_{2}^{2}= 2k/m, ω_{3}^{2}= 5k/m.

ω_{1}^{2}= 0 --> A_{1}= A_{2}= A_{3}, constant displacement or uniform circular motion.

ω_{2}^{2}= 2k/m --> A_{2}= A_{3}= -A_{1}. ω_{3}^{2}= 5k/m --> A_{1}= 0, A_{2}= -A_{3}.

Using physical insight:

Equations of motion:

(2k - 2mω^{2})A_{1}– kA_{2}– kA_{3}= 0, (3k - mω^{2})A_{2}– 2kA_{3}– kA_{1}= 0,

(3k - mω^{2})A_{3}– kA_{1}– 2kA_{2}= 0.

(i) An angular displacement of the system as a whole does not stretch any spring, A_{1}= A_{2}= A_{3}. Then ω^{2}= 0.

(ii) Mass 1 can stay fixed (A_{1}= 0) if A_{3 }= -A_{2}. Then (3k - mω^{2})A_{2}+2kA_{2}= 0, ω^{2}= 5k/m.

(iii) The system cannot have an angular displacement as a whole. Assume A_{2}= A_{3}= -A_{1}.

Then (2k - 2mω^{2})A_{1}+ kA_{1}+ kA_{1}= 0, ω^{2}= 2k/m.

A large number N (N = even) of point masses m are connected by identical
springs of equilibrium length a and spring constant k. Let q_{i}
(i = 0 to N - 1) denote the displacement of the ith mass from its equilibrium
position. Assume periodic boundary conditions, q_{i} = q_{i+N}.
(You can, for example imagine the masses arranged on a large circle of
circumference Na.)

(a) Write down the Lagrangian for the system of N
point masses.

(b) Find the equation of motion for the jth point mass.

(c) Assume solutions of the form q_{j}(t) = |A|exp(iφ_{j})
exp(-iωt) = |A|exp(i(φ_{j} - ωt)) exist, where ω is a normal mode
frequency.

Assume the phase of the amplitude depends on the position
of the mass and write φ_{j }= p*ja.

What are the restrictions on p
due to the boundary conditions?

(d) Find the N normal mode frequencies
ω_{n}. Make a sketch of ω_{n} as a function of mode number
n.

Solution:

- Concepts:

Lagrangian Mechanics, coupled oscillations, normal modes - Reasoning:

This is a one-dimensional problem. We are asked to find the normal modes of coupled harmonic oscillators. - Details of the calculation:

(a) L = T - U = ∑_{i = 0}^{N-1}[½m(dq_{i}/dt)^{2}- ½k(q_{i+1}- q_{i})^{2}]

= ∑_{i = 0}^{N-1}[½m(dq_{i}/dt)^{2}- ½k(q_{i+1}^{2}+ q_{i}^{2}- 2q_{i}q_{i+1}].

(b) d/dt(∂L/∂(dq_{j}/dt)) - ∂L/∂q_{j}= 0.

The terms in the Lagrangian that depend on q_{j }an dq_{j}/dt are

½m(dq_{j}/dt)^{2}- ½k(2q_{j}^{2}- 2q_{j-1}q_{j}- 2q_{j}q_{j+1}).

(They come from the terms in the sum i = j and i + 1 = j.)

The equation of motion for the jth point mass therefore is

md^{2}q_{j}/dt^{2}- k(q_{j+1}- 2q_{j}+ q_{j-1}) = 0.

(c) Assume q_{j}(t) = |A|exp(i(p*ja - ωt). The periodic boundary conditions imply that

p*ja = p*(j + N)a ± n*2π, p_{n}Na = n*2π, p_{n}= n2π/(Na), n = integer.

(d) Inserting solutions of the form q_{j}= |A|exp(i(p_{n}*ja - ωt)) into the equation of motion we obtain

-mω_{n}^{2}exp(ip_{n}*ja) - k[exp(ip_{n}*(j+1)a) - 2exp(ip_{n}*ja) + exp(ip_{n}*(j-1)a)] = 0,

or

ω_{n}^{2}= -(k/m)[exp(ip_{n}*a) - 2 + exp(-ip_{n}*a)] = -¼ω_{0}^{2}[2cos(p_{n}*a) - 2]

= ½ω_{0}^{2}[1 - cos(p_{n}*a)] = ω_{0}^{2}sin^{2}(p_{n}*a/2) = ω_{0}^{2}sin^{2}(nπ/N).

Here ω_{0}^{2}= 2k/m.

ω_{n}= ω_{0}|sin(nπ/N)| are the normal mode frequencies.

There are N frequencies which we can label with a mode number from n = -N/2 to (N/2 -1). All but two of the frequencies are two-fold degenerate.