IPhO 1987, Теория и експеримент — Задача 3. Infinite LC-grid

• a) $\varphi = 2 \cdot \arcsin\left(\frac{\omega \cdot \sqrt{L \cdot C}}{2}\right), \quad 0 \le \omega \le \frac{2}{\sqrt{L \cdot C}}$
• b) $v = \frac{\omega \cdot \ell}{2 \cdot \arcsin\left(\frac{\omega \cdot \sqrt{L \cdot C}}{2}\right)}$
• c) $\frac{\omega \cdot \sqrt{L \cdot C}}{2} \ll 1: \quad v_0 = \frac{\ell}{\sqrt{L \cdot C}}$

a)

Current law: $I_{L_{n-1}} + I_{C_n} - I_{L_n} = 0 \qquad (1)$

Voltage law: $V_{C_{n-1}} + V_{L_{n-1}} - V_{C_n} = 0 \qquad (2)$

Capacitive voltage drop: $V_{C_{n-1}} = \frac{1}{\omega \cdot C} \cdot \tilde{I}_{C_{n-1}} \qquad (3)$

Note: In eq. (3) $\tilde{I}_{C_{n-1}}$ is used instead of $I_{C_{n-1}}$ because the current leads the voltage by 90°.

Inductive voltage drop: $V_{L_{n-1}} = \omega \cdot L \cdot \tilde{I}_{L_{n-1}} \qquad (4)$

Note: In eq. (4) $\tilde{I}_{L_{n-1}}$ is used instead of $I_{L_{n-1}}$ because the current lags behind the voltage by 90°.

The voltage $V_{C_n}$ is given by: $V_{C_n} = V_0 \cdot \sin(\omega \cdot t + n \cdot \varphi) \qquad (5)$

Formula (5) follows from the problem.

From eq. (3) and eq. (5): $I_{C_n} = \omega \cdot C \cdot V_0 \cdot \cos(\omega \cdot t + n \cdot \varphi) \qquad (6)$

From eq. (4) and eq. (2) and with eq. (5)

Eqs. (6), (7) and (8) must satisfy the current law. This gives the dependence of $\varphi$ on $\omega$, L and C.

This condition must be true for any instant of time. Therefore it is possible to divide by $V_0 \cdot \cos(\omega \cdot t + n \cdot \varphi)$.

Hence $\omega^2 \cdot L \cdot C = 4 \cdot \sin^2\left(\frac{\varphi}{2}\right)$. The result is

b) The distance $\ell$ is covered in the time $\Delta t$ thus the propagation velocity is

c)

Slightly dependent means $\operatorname{arc\,sin}\left(\frac{\omega \cdot \sqrt{L \cdot C}}{2}\right) \sim \omega$, since v is constant in that case.

This is true only for small values of $\omega$. That means $\frac{\omega \cdot \sqrt{L \cdot C}}{2} \ll 1$ and therefore

d) The energy is conserved since only inductances and capacitances are involved. Using the terms of a) one obtains the capacitive energy

and the inductive energy

From this follows the standard form of the law of conservation of energy

The relation to mechanics is not recognizable in this way since two different physical quantities ($V_{C_n}$ and $I_{L_n}$) are involved and there is nothing that corresponds to the relation between the locus x and the velocity $v = \dot{x}$.

To produce an analogy to mechanics the energy has to be described in terms of the charge Q, the current $I = \dot{Q}$ and the constants L and C. For this purpose the voltage $V_{C_n}$ has to be expressed in terms of the charges $Q_{L_n}$ passing through the coil.

One obtains:

Mechanical analogue:

A (kinetic part): $\dot{Q}_{L_n} \longrightarrow v_n; \qquad L \longrightarrow m$

B (potential part): $Q_{L_n} \longrightarrow x_n$

$x_n$: displacement and $v_n$: velocity.

However, $Q_{L_n}$ could equally be another quantity (e.g. an angle). L could be e.g. a moment of inertia.

From the structure of the problems follows: Interaction only with the nearest neighbour (the force rises linearly with the distance). A possible model could be:

Another model is:

Grading Scheme