Discussion
Let the Hamiltonian of a system be \(H = \dfrac{p^{2}}{2} + \dfrac{k^{2}q^{2}}{2}\). If \(C_{1}, C_{2}\) denote arbitrary constants the general solution of \(q\) is :
Answer: A
The general solution is \(q(t) = C_{1}\cos kt + C_{2}\sin kt\).
Hamilton's equations give
\[ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p} = p \\ \dot{p} &= -\frac{\partial H}{\partial q} = -k^{2}q \end{aligned} \]
so \(\ddot{q} = -k^{2}q\), simple harmonic motion with angular frequency \(k\). Its general solution is \(C_{1}\cos kt + C_{2}\sin kt\).
Exponentials \(e^{\pm kt}\) would solve \(\ddot{q} = +k^{2}q\), the sign of an inverted oscillator. Frequency \(k^{2}\) mistakes \(k^{2}\) in the Hamiltonian for the frequency, and \(kt + C_{1}\) is free motion.
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