Problem Statement

Consider an undamped linear oscillator governed by the equation:
\( m x'' + c x' + k x = 0 \)
where: Assume the system is **undamped** (\(c = 0\)) and let \(m = 1\), \(k = 1\). Introduce a new variable \(y = x'\) to rewrite the second-order equation as a system of first-order equations:
\[ \begin{aligned} x' &= y, \\ y' &= -\frac{k}{m} x - \frac{c}{m} y \end{aligned} \]
Determine the trajectory of the system in the phase plane and show why it forms a circle.

Phase-Plane Trajectory of an Undamped Oscillator

Step 1: Write down the system

We consider the oscillator with \( m = 1, \quad k = 1, \quad c = 0 \). The phase-plane system is:
\( x' = y, \quad y' = -x \)

Step 2: Define a candidate conserved quantity

We define the total energy:
\( E(x,y) = \frac{1}{2} (x^2 + y^2) \)

Step 3: Differentiate \(E\) with respect to time

Using the chain rule:
\(\frac{dE}{dt} = \frac{\partial E}{\partial x} x' + \frac{\partial E}{\partial y} y'\)
Compute the partial derivatives:
\(\frac{\partial E}{\partial x} = x, \quad \frac{\partial E}{\partial y} = y\)
Substitute:
\(\frac{dE}{dt} = x x' + y y'\)

Step 4: Substitute the system equations

From the system:
\( x' = y, \quad y' = -x \)
Substitute into \(dE/dt\):
\(\frac{dE}{dt} = x(y) + y(-x)\)

Step 5: Simplify

\(\frac{dE}{dt} = xy - xy = 0\)

Step 6: Conclude conservation

Since \(\frac{dE}{dt} = 0\), we have
\(E = \text{constant} \quad \Rightarrow \quad \frac{1}{2}(x^2 + y^2) = C\)
Multiply both sides by 2:
\(x^2 + y^2 = 2C\)
This is the equation of a circle centered at the origin.

Step 7: Physical interpretation

Step 8: Takeaway

\(x^2 + y^2 = \text{constant} \quad \Rightarrow \quad \text{phase-plane trajectory is a circle}\)

For an undamped oscillator, the motion is perfectly periodic, and the phase-plane trajectory is a circle centered at the origin.