FrontierScience / 328c6a47-15d7-470e-99f5-73e468634614 / Consider two simple non-interacting pendulums. Suppose each pendulum consists of…

Consider two simple non-interacting pendulums. Suppose each pendulum consists of a small mass `m` attached to a string of length `L` that hangs vertically, affected by gravitational acceleration `g`. The string has no mass. As the mass moves, it experiences a constant frictional force of magnitude `mLb` in the direction opposite to its motion. To compensate for the mechanical energy loss due to friction in each period, each pendulum receives a kick along its direction of motion once per period, and we assume the kick occurs at `θ=−b/ω²` when its angular velocity is positive. Assume that every time a kick is given, an amount of energy equal to `\( {\mathfrak{}}mL^{2}h^{2}/2 \)` is injected into the pendulum. Note that `ω≡√(g/L)`, and for convenience, any terms involving gravitational acceleration `g` will be expressed in terms of the angular velocity. At time `t`, let `θ₁(t)` and `θ₂(t) `denote the angular displacement of the two pendulums from the vertical direction.

The two pendulums have the same mass `m`, length `L`, friction parameter `b`, and kick magnitude `h`. `u` is the angular velocity of the pendulum at the instant immediately after the kick.

After steady state has been achieved, we allow the two pendulums to interact with each other. Suppose that when `θ₂=−b/ω²≈0` (you can assume \( b/\omega \lll u \)) and the angular velocity of pendulum 2 is positive, pendulum 2 imparts a small angular impulse of magnitude `mL²α` (\( L^2\alpha \lll u \)) to pendulum 1. Here, `α` is a small positive constant.

Similarly, when `θ₁=−b/ω²≈0` and the angular velocity of pendulum 1 is positive, pendulum 1 imparts a small angular impulse of magnitude `mL²α` to pendulum 2. Since these impulses are very small, only the lowest-order term in `α/u` needs to be considered in the calculations.

Suppose that at the beginning of the `n^(th)` cycle of pendulum 1, the phase of pendulum 2 lags behind relative to the phase of pendulum 1 by `ϕ_(n)(0<ϕ_(n)<π/2)`. This means that when pendulum 1 is at `θ₁=−b/ω²`, pendulum 2 has to go through `ϕ_(n)/ω` time before it reaches `θ₂=−b/ω²`. Find the recursive relation `ϕ_(n+1)` has with `ϕ_(n)` after using appropriate first-order approximations. The answer should be expressed in the form

`ϕ_(n+1)=f(ϕ_(n),α,u)`,

where `f` is a function of `ϕ_(n),α' and `u`.

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.

Consider two simple non-interacting pendulums. Suppose each pendulum consists of a small mass `\( m \)` attached to a string of length `\( L \)` that hangs vertically, affected by gravitational acceleration `\( g \)`. The string has no mass. As the mass moves, it experiences a constant frictional force of magnitude `\( mLb \)` in the direction opposite to its motion. To compensate for the mechanical energy loss due to friction in each period, each pendulum receives a kick along its direction of motion once per period, and we assume the kick occurs at `\( \theta=- b/\omega^{2} \)` when its angular velocity is positive. Assume that every time a kick is given, an amount of energy equal to `\( {\mathfrak{}}mL^{2}h^{2}/2 \)` is injected into the pendulum. Note that `\( \omega\equiv\sqrt{g/L} \)`, and for convenience, any terms involving gravitational acceleration `\( g \)` will be expressed in terms of the angular velocity. At time `\( t \)`, let `\( θ_1(t) \)` and `\(\theta_2(t)\) `denote the angular displacement of the two pendulums from the vertical direction.

The two pendulums have the same mass `\( m \)`, length `\( L \)`, friction parameter `\( b \)`, and kick magnitude `\( h \)`. `\( u \)` is the angular velocity of the pendulum at the instant immediately after the kick.

After steady state has been achieved, we allow the two pendulums to interact with each other. Suppose that when `\( \theta_{2}=- b/\omega^{2}\approx0 \)` (you can assume \( b/\omega \lll u \)) and the angular velocity of pendulum 2 is positive, pendulum 2 imparts a small angular impulse of magnitude `\( mL^{2}\alpha \)` (\( L^2\alpha \lll u \)) to pendulum 1. Here, `\( \alpha \)` is a small positive constant.

Similarly, when `\( \theta_{1}=- b/\omega^{2}\approx0 \)` and the angular velocity of pendulum 1 is positive, pendulum 1 imparts a small angular impulse of magnitude `\( mL^{2}\alpha \)` to pendulum 2. Since these impulses are very small, only the lowest-order term in `\( α/u \)` needs to be considered in the calculations.

Suppose that at the beginning of the `\( n^{th} \)` cycle of pendulum 1, the phase of pendulum 2 lags behind relative to the phase of pendulum 1 by `\( \phi_{n} (0<\phi_n<\pi/2)\)`. This means that when pendulum 1 is at `\( \theta_1 = -b/\omega^2 \)`, pendulum 2 has to go through `\( \phi_{n}/\omega \)` time before it reaches `\( \theta_2 = -b/\omega^2 \)`. Find the recursive relation `\( \phi_{n+1} \)` has with `\( \phi_n \)` after using appropriate first-order approximations. The answer should be expressed in the form

`\( \phi_{n+1} = f(\phi_n,\alpha, u) \)`,

where `\( f \)` is a function of `\( \phi_n,\alpha \)' and `\( u \)`.

Think step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.