Find the normal, longitudinal modes of vibration for three
masses connected by identical springs of spring constant k. The masses are
collinear. The end masses have mass m, while the inner mass has mass 2m.

(a) Calculate the normal modes of the system.

(b) Describe the
relative motion of the particles for each normal mode.

Solution:

- Concepts:

Coupled oscillations, normal modes - Reasoning:

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

(a) The kinetic energy is T = ½[m(dx_{1}/dt)^{2}+ 2m(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}= 2m, 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-k 0 -k 2k-2ω ^{2}m-k 0 -k k-ω ^{2}mA _{1}A _{2}A _{3}= 0.

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.k-ω ^{2}m-k 0 -k 2k-2ω ^{2}m-k 0 -k k-ω ^{2}m= 0.

2(k-ω^{2}m)^{3}- 2k^{2}(k-ω^{2}m) = 0.

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

Solution 2: k-ω^{2}m ≠ 0, then (k-ω^{2}m)^{2}= k^{2}, (k-ω^{2}m) = ±k.

For (k-ω^{2}m) = +k, ω = 0.

Solution 3: (k-ω^{2}m) = -k, ω = (2k/m)^{½}.

(b) 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 = (2k/m)^{½}.

Equation 1 yields -kA_{1}- kA_{2}= 0, A_{1}= -A_{2}.

Equation 3 yields -kA_{2}- kA_{3}= 0, A_{3}= -A_{2}.

A_{1}= A_{3}= -A_{2}, the central mass move in a direction opposite to the direction of the outer masses. All masses oscillate with equal amplitudes.

A triple pendulum consists of masses αm, m and m attached to a single light
string at distances a, 2a, and 3a respectively from its points of suspension.
Consider only motion in a plane.

(a) Determine the value of α such that
one of the normal frequencies of the system will equal the frequency of a simple
pendulum of length a/2 and mass m. You may assume the displacements of the
masses from equilibrium are small.

(b) Find the mode corresponding to
this frequency and sketch it.

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:

We are asked to find the normal mode frequencies for small displacements. - Details of the calculation:

(a) Let Φ_{1}, Φ_{2}, and Φ_{3}be the generalized coordinates. Then

x_{1}= a sinΦ_{1}, x_{2}= a(sinΦ_{1}+ sinΦ_{2}), x_{3}= a(sinΦ_{1}+ sinΦ_{2}+ sinΦ_{3}),

y_{1}= a cosΦ_{1}, y_{2}= a(cosΦ_{1}+ cosΦ_{2}), x_{3}= a(cosΦ_{1}+ cosΦ_{2}+ cosΦ_{3}).

For small displacements we have sinΦ = Φ, cosΦ = 1 - Φ^{2}/2.

The kinetic energy of the system is

T = ½αm((dx_{1}/dt)^{2}+ (dy_{1}/dt)^{2}) + ½m((dx_{2}/dt)^{2}+ (dy_{2}/dt)^{2}) + ½m((dx_{3}/dt)^{2}+ (dy_{3}/dt)^{2}).

Let us only keep terms up to second order in Φ and dΦ/dt. Then

(dx_{1}/dt)^{2}+ (dy_{1}/dt )^{2}= a^{2}(dΦ_{1}/dt)^{2},

(dx_{2}/dt)^{2}+ (dy_{2}/dt )^{2}= a^{2}[(dΦ_{1}/dt)^{2}+ (dΦ_{2}/dt)^{2}+ 2(dΦ_{1}/dt)(dΦ_{2}/dt)],

(dx_{3}/dt)^{2}+ (dy_{3}/dt )^{2}= a^{2}[(dΦ_{1}/dt)^{2}+ (dΦ_{2}/dt)^{2}+ (dΦ_{3}/dt)^{2}

+ 2(dΦ_{1}/dt)(dΦ_{2}/dt) + 2(dΦ_{2}/dt)(dΦ_{3}/dt) + 2(dΦ_{1}/dt)(dΦ_{3}/dt)].

T = ½ma^{2}[(α + 2)(dΦ_{1}/dt)^{2}+ 2(dΦ_{2}/dt)^{2}+ (dΦ_{3}/dt)^{2 }+ 4(dΦ_{1}/dt)(dΦ_{2}/dt) + 2(dΦ_{2}/dt)(dΦ_{3}/dt) + 2(dΦ_{1}/dt)(dΦ_{3}/dt)].

The potential energy of the system is

U = -αmgy_{1}- mgy_{2 }- mgy_{3}+ arbitrary constant.^{ }U = ½mga[(α + 2)Φ_{1}^{2}+ 2Φ_{2}^{2}+ Φ_{3}^{2}].^{ . }We have

T = ½∑_{ij}[T_{ij}(dΦ_{i}/dt)(dΦ_{j}/dt), U = ½∑_{ij}k_{ij}Φ_{i}Φ_{j}.

T_{11}= ma^{2}(α + 2), T_{22}= 2ma^{2}, T_{33}= ma^{2}, T_{12}= T_{21}= 2ma^{2}, T_{13}= T_{31}= T_{23}= T_{32}= ma^{2}.

k_{11}= mga(α + 2), k_{22}= 2mga, k_{33}= mga, k_{ij}(i ≠ j) = 0.(b) Let λ = ω

^{2}a/g. Thenmga

(α + 2)(1 - λ) -2λ -λ -2λ 2(1 - λ) -λ -λ -λ (1 - λ) = 0.

For a simple pendulum with length a/2 we have ω = (2g/a)^{½}. We are looking for a mode with ω^{2}= 2g/a, or λ = 2.

α + 2 4 2 4 2 2 2 2 1 = 0 --> α = 2.

(b) The mode corresponding to this frequency is found from

α + 2 4 2 4 2 2 2 2 1 A _{1}A _{2}A _{3}= 0.

A

_{3}= -2A_{1}, A_{2}= 0.

Consider an ideal gas of N particles in a cylinder with a piston so that the
volume and pressure may change. Suppose that each particle has f = 5 quadratic degrees
of freedom in its energy. Consider the three paths in a PV diagram shown. Suppose ways 1
and 3 are straight lines on the PV diagram and way 2 is an adiabatic process.

(a) How much work W is done on the gas for ways 1, 2, and 3?

(b) How much heat Q is transferred to the gas for ways 1, 2, and 3?

(c) What is the change in internal energy ΔU for ways 1, 2, 3?

(d) What is the change in entropy ΔS for way 3?

Solution:

- Concepts:

The ideal gas law PV = NkT, work done by the system W = ∫PdV, the first law of thermodynamics, entropy - Reasoning:

We are asked to compute ΔU, ΔQ, W, and ΔS for along different path on a PV diagram. - Details of the calculation:

(a) W = -∫PdV = work done on the gas = -area under the curve on a PV diagram.

First law of thermodynamics: ΔU = ΔQ + W.

way 1: W = -½(P_{1}+ P_{2})(V_{2}- V_{1}).

way 2: W = ΔU = (5/2)Nk(T_{2}- T_{1}) = (5/2)(P_{2}V_{2}- P_{1}V_{1}).

way 3: W = -P_{1}(V_{2}- V_{1}).

(b) ΔQ = ΔU - W.

way 1: ΔQ = (5/2)(P_{2}V_{2}- P_{1}V_{1}) + ½(P_{1}+ P_{2})(V_{2}- V_{1})

= 3(P_{2}V_{2}- P_{1}V_{1}) - ½(P_{2}V_{1}- P_{1}V_{2}).

way 2: ΔQ = 0.

way 3: ΔQ = (5/2)(P_{1}V_{2}- P_{1}V_{1}) + P_{1}(V_{2}- V_{1})

= (7/2)P_{1}(V_{2}+ V_{1}).

(c) ΔU is a physical property of the system. It depend only on the state of the system, not on the way the system was put into this state.

ways 1 and 2: ΔU = (5/2)(P_{2}V_{2}- P_{1}V_{1})

way 3: ΔU = (5/2)(P_{1}V_{2}- P_{1}V_{1})

(d) Change in entropy: ΔS = ∫_{i}^{f}dS = ∫_{i}^{f}dQ_{r}/T.

dQ = dU - dW = (5/2)NkdT + NkdT = (7/2)NkdT.

ΔS = (7/2)Nkln(T_{2}/T_{1}).

way 3: ΔS = (7/2)ln(V_{2}/V_{1}).

Consider a ring of radius, r, with 4 identical point particles of mass m
interconnected by identical springs with spring constant k to their nearest
neighbors. The particles move without friction. The interconnected springs can
be viewed as causing harmonic oscillations.

(a) Determine the number of normal modes of oscillations, and establish the
Lagrangian.

(b) Find the frequencies for small oscillations and describe the corresponding
eigen-vibrations.

Solution:

- Concepts:

Coupled oscillations - Reasoning:

This is a one-dimensional problem. We have 4 masses, so there will be 4 normal modes. - Details of the calculation:

(a) L = ½m∑_{i=1}^{4}(dq_{i}/dt)^{2}- ½k∑_{i=1}^{4}(q_{i+1}- q_{i})^{2}.

Here the q_{i}are the clockwise displacements from equilibrium,

and the index I is cyclic. We can write L asL = 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}^{2}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.

(b) Equation of motion for mass i:

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

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).

A metal block of mass m and specific heat c with temperature T_{b} is
placed into the ocean which has temperature T. Assume T_{b} > T. What
is the total change in the entropy, ΔS_{total}, for the system? Express
your answer in terms of m, c, T_{b}, and T.

Solution:

- Concepts:

Change in entropy: dS = dQ/T - Reasoning:

We are asked calculate the change in entropy along a reversible path. - Details of the calculation:

For the metal block ΔS_{block}is negative. ΔS_{block = }∫_{Tb}^{T }dQ/T =_{ }∫_{Tb}^{T }mcdT/T = mc ln(T/T_{b}).

For the ocean ΔS_{ocean}is positive. ΔS_{ocean}= Q/T = mc(T_{b}- T)/T = mc(T_{b}/T - 1).

ΔS_{total}= ΔS_{block}+ ΔS_{ocean}= mc(T_{b}/T - ln(T_{b}/T) - 1), ΔS_{total}is positive.