The relativistic equivalent of the simple harmonic oscillator equation for a spring with constant \(k\) and a rest mass \(m_{0}\) attached is $$ \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m_{0} y}{\sqrt{1-y^{2} / c^{2}}}\right)+k x=0 \quad \text { with } \quad \dot{x}=y $$ where \(c\) is the speed of light. Show that the phase trajectories are given by $$ m_{0} c^{2} / \sqrt{1-y^{2} / c^{2}}+\frac{1}{2} k x^{2}=\text { constant } $$ and sketch the phase portrait for this system.

Q: Show that the phase trajectories for the given relativistic spring equation are given by the expression \(\frac{m_{0} c^{2}}{\sqrt{1 - y^{2} / c^{2}}} + \frac{1}{2} k x^{2} = const\). A: To show that the phase trajectories for the given relativistic spring equation match the required expression, we followed these steps: 1. Analyze the given equation: The equation is a relativistic spring system with force proportional to displacement, \(x\), and rest mass, \(m_{0}\). The term \(\dot{x}\) is the velocity of the mass and is equal to \(y\). 2. Substitute \(\dot{x} = y\): We eliminate \(\dot{x}\) from the equation using the substitution, \(\dot{x}=y\), and rewrite the equation as $$\frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{m_{0}\dot{x}}{\sqrt{1 - \dot{x}^{2} / c^{2}}}\right)+k x=0$$ 3. Integrate both sides over time: Integrating both sides of the equation with respect to time, \(t\), we get: $$\frac{m_{0} c^{2}}{\sqrt{1 - \dot{x}^{2} / c^{2}}} + \frac{1}{2} k x^{2} = const$$ 4. Replace \(\dot{x}\) with \(y\): Finally, we replace \(\dot{x}\) back to \(y\): $$\frac{m_{0} c^{2}}{\sqrt{1 - y^{2} / c^{2}}} + \frac{1}{2} k x^{2} = const$$ This expression represents the phase trajectories for the given relativistic spring equation.