The problem asks for the condition relating mass ($m$) and spring constant ($k$) for a spring-mass simple harmonic oscillator (SHO) such that its speed ($v$) versus extension ($x$) graph on the $x-v$ plane forms a circle.

SHO Energy Conservation

In simple harmonic motion, the total mechanical energy ($E$) is conserved. It is the sum of kinetic energy ($KE$) and potential energy ($PE$).

The kinetic energy is given by $KE = \frac{1}{2}mv^2$. The potential energy stored in the spring is $PE = \frac{1}{2}kx^2$. Thus, the total energy is:

$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

x-v Plane Equation Form

To visualize the path on the $x-v$ plane, we rearrange the energy equation:

$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Assuming non-zero energy ($E \neq 0$), we divide by $E$:

$1 = \frac{mv^2}{2E} + \frac{kx^2}{2E}$

Rearranging the terms, we get:

$\frac{kx^2}{2E} + \frac{mv^2}{2E} = 1$

This can be written in the standard form of an ellipse equation ($\frac{x^2}{a^2} + \frac{v^2}{b^2} = 1$):

$\frac{x^2}{\left(\sqrt{\frac{2E}{k}}\right)^2} + \frac{v^2}{\left(\sqrt{\frac{2E}{m}}\right)^2} = 1$

Here, the semi-axis along the $x$-axis is $a = \sqrt{\frac{2E}{k}}$ and the semi-axis along the $v$-axis is $b = \sqrt{\frac{2E}{m}}$.

Circle Condition Derivation

For the graph to be a circle, the semi-axes must be equal:

$a = b$

$\sqrt{\frac{2E}{k}} = \sqrt{\frac{2E}{m}}$

Squaring both sides gives:

$\frac{2E}{k} = \frac{2E}{m}$

Since $E \neq 0$, we can cancel $2E$ from both sides:

$\frac{1}{k} = \frac{1}{m}$

This leads to the condition:

$k = m$

Thus, the graph of $v$ as a function of $x$ is a circle only when the spring constant equals the mass.