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IPhO 1986, Теория — Задача 3. Q3

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IPhO 1986 — Theory

Условие

Three particles, each of mass $m$ , are in equilibrium and joined by unstretched massless springs, each with Hooke’s Law spring constant $k$ . They are constrained to move in a circular path as indicated in Figure 3.1.

Three masses m on a circle joined by three springs k; the displacements u₁, u₂ and u₃ of the masses along the circle are marked with small arrows. — Figure 3.1

(i) If each mass is displaced from equilibrium by small displacements $u_1$ , $u_2$ and $u_3$ respectively, write down the equation of motion for each mass.

(ii) Verify that the system has simple harmonic solutions of the form

u_n = a_n\cos\omega t\ ,

with accelerations, $(-\omega^2 u_n)$ where $a_n$ $(n = 1,2,3)$ are constant amplitudes, and $\omega$ , the angular frequency, can have 3 possible values,

\omega_o\sqrt{3}, \omega_o\sqrt{3}\ \text{and}\ 0. \ \text{ where }\ \omega_o^2 = \frac{k}{m}.

(iii) The system of alternate springs and masses is extended to $N$ particles, each mass $m$ is joined by springs to its neighbouring masses. Initially the springs are unstretched and in equilibrium. Write down the equation of motion of the $n$ th mass ( $n$ = 1,2... $N$ ) in terms of its displacement and those of the adjacent masses when the particles are displaced from equilibrium.

u_n(t) = a_s\sin\left(\frac{2ns\pi}{N} + \phi\right)\cos\omega_s t,

are oscillatory solutions where $s$ = 1, 2,... $N$ , $n$ = 1, 2, ... $N$ and where $\phi$ is an arbitrary phase, providing the angular frequencies are given by

\omega_s = 2\omega_o\sin\left(\frac{s\pi}{N}\right),

where $a_s$ $(s = 1,......N)$ are constant amplitudes independent of $n$ .

State the range of possible frequencies for a chain containing an infinite number of masses.

(iv) Determine the ratio

u_n / u_{n+1}

for large $N$ , in the two cases:

(a) low frequency solutions

(b) $\omega = \omega_{\max}$ , where $\omega_{\max}$ is the maximum frequency solution.

Sketch typical graphs indicating the displacements of the particles against particle number along the chain at time $t$ for cases (a) and (b).

(v) If one of the masses is replaced by a mass $m' \ll m$ estimate any major change one would expect to occur to the angular frequency distribution.

Describe qualitatively the form of the frequency spectrum one would predict for a diatomic chain with alternate masses $m$ and $m'$ on the basis of the previous result.

Reminder

\sin(A + B) = \sin A\cos B + \cos A\sin B

\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)

2\sin^2 A = 1 - \cos 2A

Решение

Answer Q3

Equations of motion:

m\frac{d^2u_1}{dt^2} = k(u_2 - u_1) + k(u_3 - u_1)

m\frac{d^2u_2}{dt^2} = k(u_3 - u_2) + k(u_1 - u_2)

m\frac{d^2u_3}{dt^2} = k(u_1 - u_3) + k(u_2 - u_3)

Substituting $u_n(t) = u_n(0)\cos\omega t$ and $\omega_o^{\ 2} = \dfrac{k}{m}$ :

(2\omega_o^{\ 2} - \omega^2)u_1(0) - \omega_o^{\ 2}u_2(0) - \omega_o^{\ 2}u_3(0) = 0 \qquad \text{(a)}

-\omega_o^{\ 2}u_1(0) + (2\omega_o^{\ 2} - \omega^2)u_2(0) - \omega_o^{\ 2}u_3(0) = 0 \qquad \text{(b)}

-\omega_o^{\ 2}u_1(0) - \omega_o^{\ 2}u_2(0) + (2\omega_o^{\ 2} - \omega^2)u_3(0) = 0 \qquad \text{(c)}

Solving for $u_1(0)$ and $u_2(0)$ in terms of $u_3(0)$ using (a) and (b) and substituting into (c) gives the equation equivalent to

(3\omega_o^{\ 2} - \omega^2)^2\omega^{\ 2} = 0

\omega^{\ 2} = 3\omega_o^{\ 2},\ \ 3\omega_o^{\ 2}\ \text{ and }\ 0

\omega\ = \sqrt{3}\omega_o,\ \sqrt{3}\omega_o\ \text{ and } 0

(ii) Equation of motion of the n’th particle:

m\frac{d^2u_n}{dt^2} = k(u_{1+n} - u_n) + k(u_{n-1} - u_n) \qquad n = 1,2\ldots\ldots N

\frac{d^2u_n}{dt^2} = k(u_{1+n} - u_n) + \omega_o^{\ 2}(u_{n-1} - u_n)

Substituting $u_n(t) = u_n(0)\sin\left(2ns\dfrac{\pi}{N}\right)\cos\omega_s t$

-\omega_s^{\ 2}\left(\sin\left(2ns\frac{\pi}{N}\right)\right) = \omega_o^{\ 2}\left[\sin\left(2(n+1)s\frac{\pi}{N}\right) - 2\sin\left(2ns\frac{\pi}{N}\right) + \sin\left(2(n-1)s\frac{\pi}{N}\right)\right]

-\omega_s^{\ 2}\left(\sin\left(2ns\frac{\pi}{N}\right)\right) = 2\omega_o^{\ 2}\left[\frac{1}{2}\sin\left(2(n+1)s\frac{\pi}{N}\right) + \sin\left(2ns\frac{\pi}{N}\right) - \frac{1}{2}\sin\left(2(n-1)s\frac{\pi}{N}\right)\right]

-\omega_s^{\ 2}\left(\sin\left(2ns\frac{\pi}{N}\right)\right) = 2\omega_o^{\ 2}\left[\sin\left(2ns\frac{\pi}{N}\right)\cos\left(2s\frac{\pi}{N}\right) - \sin\left(2ns\frac{\pi}{N}\right)\right]

\therefore \omega_s^{\ 2} = 2\omega_o^{\ 2}\left[1 - \cos\left(2s\frac{\pi}{N}\right)\right]: \qquad (s = 1,2,.....N)

As $2\sin^2\theta = 1 - \cos 2\theta$

This gives

\omega_s = 2\omega_o\sin\left(\frac{s\pi}{N}\right) \qquad (s = 1,2,...N)

$\omega_s$ can have values from 0 to $2\omega_o = 2\sqrt{\dfrac{k}{m}}$ when $N \to \infty$ ; corresponding to range $s = 1$ to $\dfrac{N}{2}$ .

(iv) For s’th mode

\frac{u_n}{u_{n+1}} = \frac{\sin\left(2ns\frac{\pi}{N}\right)}{\sin\left(2(n+1)s\frac{\pi}{N}\right)}

\frac{u_n}{u_{n+1}} = \frac{\sin\left(2ns\frac{\pi}{N}\right)}{\sin\left(2ns\frac{\pi}{N}\right)\cos\left(2s\frac{\pi}{N}\right) + \cos\left(2ns\frac{\pi}{N}\right)\sin\left(2s\frac{\pi}{N}\right)}

(a) For small $\omega$ , $\left(\dfrac{s}{N}\right) \approx 0$ , thus $\cos\left(2ns\dfrac{\pi}{N}\right) \cong 1$ and $= \sin\left(2ns\dfrac{\pi}{N}\right) \approx 0$ , and so $\dfrac{u_n}{u_{n+1}} \cong 1$ .

(b) The highest mode, $\omega_{\max} = 2\omega_o$ , corresponds to $s = N/2$

\therefore \frac{u_n}{u_{n+1}} = -1 \quad \text{as} \quad \frac{\sin(2n\pi)}{\sin(2(n+1)\pi)} = -1

Case (a)

Case (b)
N odd

N even

(vi) If $m' \ll m$ , one can consider the frequency associated with $m'$ as due to vibration of $m'$ between two adjacent, much heavier, masses which can be considered stationary relative to $m'$ .