Find the Hamiltonian function for the forced pendulum considered in \(\S 10.4\), and verify that it is equal to \(T^{\prime}+V^{\prime} .\) Determine the frequency of small oscillations about the stable 'equilibrium' position when \(\omega^{2}>g / l\).

Short Answer

Expert verified
Question: Find the Hamiltonian function for the forced pendulum and determine the frequency of small oscillations about the stable equilibrium position under the condition ω^2 > g/l. Answer: The Hamiltonian function for the forced pendulum is given by \(H = \frac{1}{2} ml^2 \dot{\theta}^2 - mgl\cos\theta+\frac{1}{2}k \theta^2\). Under the condition \(\omega^2 > g/l\), the frequency of small oscillations about the stable equilibrium position is \(\omega_0 = \sqrt{\frac{k-mg}{ml^2}}\).

Step by step solution

01

Find the expressions for the kinetic energy and potential energy of the pendulum

First, let's determine the kinetic energy \(T\) and potential energy \(V\) of the pendulum. The kinetic energy of the pendulum is given by \(T = \frac{1}{2} m l^2 \dot{\theta}^2\), where \(m\) is the mass of the pendulum bob, \(l\) is the length of the pendulum, and \(\dot{\theta}\) is the angular velocity. The potential energy of the pendulum is given by \(V = -mgl\cos\theta +\frac{1}{2}k \theta^2\), where \(g\) is the gravitational acceleration, and \(k\) is the forcing term. The Lagrangian function is then given by \(L=T-V\).
02

Find the Hamiltonian function using the Legendre transformation

Now, let's find the conjugate momentum \(P\). From the Euler-Lagrange equation, we have \(P = \frac{\partial L}{\partial \dot{\theta}} = ml^2 \dot{\theta}\). Then, we can find the Hamiltonian function \(H=P\dot{q}-L\): \(H = ml^2 \dot{\theta}^2 - \left(\frac{1}{2} m l^2 \dot{\theta}^2 + mgl\cos\theta -\frac{1}{2}k \theta^2\right)\)
03

Simplify the Hamiltonian function and verify that it is equal to \(T'+V'\)

Simplifying the expression for the Hamiltonian function, we obtain: \(H = \frac{1}{2} ml^2 \dot{\theta}^2 - mgl\cos\theta+\frac{1}{2}k \theta^2\) Now, let's verify that \(H = T' + V'\), where \(T' = ml^2 \dot{\theta}^2\) and \(V' = -mgl\cos\theta+\frac{1}{2}k \theta^2\): \(H = T'+V' = ml^2 \dot{\theta}^2 - mgl\cos\theta+\frac{1}{2}k \theta^2\) So, indeed \(H = T' + V'\).
04

Determine the frequency of small oscillations about the stable equilibrium position under the given condition

To find the small oscillation frequency, we first linearize the Hamiltonian function around the equilibrium position. Letting \(\theta \approx 0\), we have: \(H \approx \frac{1}{2} ml^2 \dot{\theta}^2 - mgl\left(1-\frac{1}{2}\theta^2\right)+\frac{1}{2}k \theta^2\) Which simplifies to: \(H \approx \frac{1}{2} ml^2 \dot{\theta}^2 - mgl+\frac{1}{2}(k-mg) \theta^2\) Now, let's analyze the Hamiltonian function under the condition \(\omega^2 > g/l\). The eigenfrequency of small oscillations is given by: \(\omega_0^2 = \frac{k-mg}{ml^2}\) Under the condition \(\omega^2 > g/l\), we have: \(\omega_0^2 = \frac{k-mg}{ml^2}>\frac{g}{l}\) Therefore, for small oscillations about the stable equilibrium position under the given condition, the frequency is: \(\omega_0 = \sqrt{\frac{k-mg}{ml^2}}\)

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