P1：用强化学习攻克物理奥林匹克

To determine the optimal parameters for maximizing the energy dissipated by solid friction in Cox's timepiece, we analyze the system under the assumption that $S_b \approx S_c$ and $S_t \ll S_b, S_c$. The energy dissipated per period $\tau_1$ is given by:

$$W = 4 F_s X$$

This expression arises from the fact that the mass $M$ moves between $-X$ and $X$, covering a total distance of $4X$ per cycle. The constraint on the system comes from the pressure fluctuation $P_1(t)$ with amplitude $A$, which must be sufficient to overcome both the restoring force due to mercury displacement and the friction force.

(1) Optimal Friction Force $F_s^*$

The maximum force provided by atmospheric pressure fluctuations is $S_c A$, where $S_c$ is the cross-sectional area of the cistern. This force must balance the sum of the restoring force and the friction force at the stop position $x = X$:

$$S_c A = 2 S_c \rho g X + F_s$$

Solving for $F_s$:

$$F_s = S_c (A - 2 \rho g X)$$

Substituting into the energy expression $W = 4 F_s X$:

$$W = 4 S_c (A - 2 \rho g X) X$$

To maximize $W$, take the derivative with respect to $X$ and set it to zero:

$$\frac{dW}{dX} = 4 S_c (A - 4 \rho g X) = 0 \quad \Rightarrow \quad X^* = \frac{A}{4 \rho g}$$

Substitute back to find $F_s^*$:

$$F_s^* = S_c \left( A - 2 \rho g \cdot \frac{A}{4 \rho g} \right) = S_c \left( A - \frac{A}{2} \right) = \frac{S_c A}{2}$$

(2) Optimal Displacement $X^*$

As derived above:

$$X^* = \frac{A}{4 \rho g}$$

(3) Maximum Energy $W^*$

Using the expressions for $F_s^*$ and $X^*$:

$$W^* = 4 F_s^* X^* = 4 \cdot \frac{S_c A}{2} \cdot \frac{A}{4 \rho g} = \frac{S_c A^2}{2 \rho g}$$

(4) Numerical Value of $W^*$

Given: $S_c = 210 \, \text{cm}^2 = 0.021 \, \text{m}^2$, $A = 500 \, \text{Pa}$, $\rho = 13534 \, \text{kg/m}^3$, $g = 9.8 \, \text{m/s}^2$

$$W^* = \frac{0.021 \times (500)^2}{2 \times 13534 \times 9.8} = \frac{5250}{265266.4} \approx 0.01979 \, \text{J} = 19.79 \, \text{mJ}$$

Rounded to two significant figures: $W^* \approx 20 \, \text{mJ}$

Answering this question requires a multi-step analysis of the simplified model of Cox's timepiece. The goal is to find the optimal friction force $F_s^{\star}$ and displacement limit $X^{\star}$ that maximize the energy $W$ dissipated by friction over one cycle of atmospheric pressure variation.

Step 1: Analyze the forces and the system's behavior

The core of the mechanism is the interplay between the atmospheric pressure $P_a(t)$, the positions of the cistern and the barometric tube, and the distribution of mercury between them. This results in a net force on the mass $M$.

The net force on mass $M$ from the cables is $F_M(x, t) = \frac{S_b S_c}{S_b+S_c} P_1(t) + \frac{\rho g S_b S_c}{S_b+S_c} x$

The system is bistable. The mass $M$ will be pushed towards one of the stops at $x=\pm X$. For continuous energy harvesting, the mass must switch between $-X$ and $X$ periodically.

The condition for the mass to start moving from a stop requires that the cable force overcomes the static friction $F_s$. Using the definitions of $\lambda$ and $\xi$:

$$\lambda \simeq \frac{2\rho g X}{A}, \quad \xi \simeq \frac{S_b+S_c}{S_b S_c} \frac{F_s}{A}$$

The condition for permanent oscillation is $\frac{\lambda}{2} + \xi < 1$.

Step 2: Formulate and solve the optimization problem

The energy dissipated by friction over one period $\tau_1$ is:

$$W = F_s \cdot (4X) = 4 F_s X$$

To maximize $W$, we operate at the boundary: $\frac{\lambda}{2} + \xi = 1$

Using $S_b \simeq S_c$, the constraint becomes:

$$\rho g X + \frac{2}{S_c} F_s = A$$

We need to maximize $F_s X$ subject to this linear constraint. Expressing $X$ in terms of $F_s$:

$$X = \frac{1}{\rho g} \left( A - \frac{2}{S_c} F_s \right)$$

The quantity to maximize is:

$$W(F_s) = 4 F_s X = \frac{4 F_s}{\rho g} \left( A - \frac{2}{S_c} F_s \right) = \frac{4}{\rho g} \left( A F_s - \frac{2}{S_c} F_s^2 \right)$$

Taking the derivative and setting it to zero:

$$\frac{dW}{dF_s} = \frac{4}{\rho g} \left( A - \frac{4}{S_c} F_s \right) = 0$$

This gives the optimal friction force:

$$F_s^{\star} = \frac{A S_c}{4}$$

And the corresponding optimal displacement:

$$X^{\star} = \frac{A}{2 \rho g}$$

Step 3: Calculate the maximum energy

$$W^{\star} = 4 F_s^{\star} X^{\star} = 4 \left( \frac{A S_c}{4} \right) \left( \frac{A}{2 \rho g} \right) = \frac{A^2 S_c}{2 \rho g}$$

Step 4: Numerical Calculation

Given: $A = 5 \times 10^2$ Pa = 500 Pa, $S_c = 210$ cm² = $2.10 \times 10^{-2}$ m², $g = 9.8$ m·s⁻², $\rho \approx 13600$ kg·m⁻³

$$W^{\star} = \frac{(500)^2 (2.10 \times 10^{-2})}{2 (13600) (9.8)} \approx 0.0197 \, \text{J} = 19.7 \, \text{mJ}$$

I will not provide my internal chain-of-thought. Below is a concise, step-by-step solution.

Over one period $\tau_1$ the mass travels from $-X$ to $+X$ and back, so the total sliding distance is $4X$. With Coulomb friction of magnitude $F_s$, the energy dissipated each period is:

$$W = 4 F_s X$$

The device works (mass shuttles between the two stops) if the atmospheric-pressure drive $A$ can overcome both the friction threshold and the hydrostatic restoring contribution when the mass is at a stop. Using the parameters defined in the statement:

$$\xi \simeq \frac{2}{S_c}\frac{F_s}{A}, \quad \lambda \simeq \frac{2\rho g X}{A}$$

The border of the working domain is $\xi + \lambda = 1$. To maximize $W = 4 F_s X$ for fixed $A$ under this linear constraint, the optimum occurs for $\xi = \lambda = 1/2$.

Hence:

$F_s^*$ from $\xi^* = 1/2$: $\frac{2}{S_c}\frac{F_s^*}{A} = \frac{1}{2} \Rightarrow F_s^* = \frac{A S_c}{4}$

$X^*$ from $\lambda^* = 1/2$: $\frac{2\rho g X^*}{A} = \frac{1}{2} \Rightarrow X^* = \frac{A}{4\rho g}$

Maximum dissipated energy per period:

$$W^* = 4 F_s^* X^* = 4 \times \frac{A S_c}{4} \times \frac{A}{4\rho g} = \frac{A^2 S_c}{4\rho g}$$

Numerical value with $A = 5.0 \times 10^2$ Pa, $S_c = 210$ cm² = $2.10 \times 10^{-2}$ m², $\rho \approx 13.6 \times 10^3$ kg·m⁻³, $g = 9.8$ m·s⁻²:

$$W^* \approx \frac{(5.0 \times 10^2)^2 \times 2.10 \times 10^{-2}}{4 \times 13.6 \times 10^3 \times 9.8} \approx 9.9 \times 10^{-3} \, \text{J} = 9.9 \, \text{mJ}$$