"An impulse is then applied to the mass giving it an initial velocity."
If the impulse gets mass moving in a circular motion, then at least some of the impulse is in the tangential direction. You may soon realise that the type of circular motion is not simple, and that it depends on the initial magnitude of tangential velocity. Illustration at the bottom accompanies explanation.
A spring-less scenario (grey trajectory) would mean that the mass finds its equilibrium rotational velocity and keeps rotating at that velocity if there are no energy losses. In the case of a spring, circular motion will cause radial motion as the centrifugal force experienced by the mass stretches the spring (see diagram).
With spring (blue trajectory in figure): It is convenient to divide the motion into circular and radial for analysis. The total energy of the system $E_{tot}$ is:
$E_{tot} = E_{tan} + E_{rad}\tag1$
where $E_{tan}$ is the energy due to circular motion only (if there were no spring) and $E_{rad}$ is the radial motion energy (associated with the spring oscillations). In turn, the radial energy has two components, kinetic ($K_{rad}$) and potential ($U_{rad}$), so that:
$E_{tot} = E_{tan} + (K_{rad} + U_{rad})\tag{1.1}$
For the circular motion, the applicable energy equation is the usual:
$E_{tan} = 0.5mv^2 = 0.5mr_a^2\omega_{ta}^2\tag2$
where for the mass, $\omega_{ta}$ is the average rotational angular velocity, and $r_a$ is the average distance from the centre of rotation.
For the radial oscillations, we have simple harmonic motion with a spring, position (ref) is expressed by:
$r_i(t) = Acos(\omega_r t + \varphi) \tag3$
where $A$ is the maximum amplitude of the radial oscillation, $\omega_r$ is the radial (spring induced) oscillation 'angular' velocity, and $\varphi$ is the phase. The radial direction instantaneous potential energy is:
$E_{rad} = K_{rad}+U_{rad} = 0.5m\dot r_i^2 + 0.5kr_i^2= 0.5m\omega_r^2 A^2sin^2(\omega_r t + \varphi) + 0.5kA^2cos^2(\omega_r t + \varphi) \tag4$
As $\omega_r t$ varies, the kinetic energy $K_{rad}$ oscillates, achieving a max when $U_{rad}$ is at a minimum, and vice-versa. We can express instantaneous values of $E_{tan}$ in equation $2$ by replacing average rotational radius $r_a$ by the instantaneous rotational radius $r_i$, and $\omega_{ta}$ with $\omega_t$. $\omega_t$ in $E_{tan}$ must also oscillate to conserve angular momentum as the rotational radius of the mass $r_i$ changes. See this related question.
Equations $2$ and $4$ are therefore the total energy expression. Note that the (tangential) rotational angular velocity $\omega_{t}$ is different from $\omega_{r}$, the radial (spring induced) oscillation 'angular' velocity. There is an important caveat:
If the tangential velocity is high enough, there will be no
radial direction oscillations (orange trajectory in figure). Why is this? The tangential velocity results in an outward centrifugal force: $F_{out} = v^2/r_i$, and at this max radius $r_i$ the spring is pulling the mass towards the centre with maximum force possible for the given spring = $F_{inward} = kr_i$. However:
$F_{out} > F_{inward} \implies v^2/r_i > kr_i$
at all times, so there are no radial oscillations and:
$E_{rad} = K_{rad}+U_{rad} = 0 + 0.5kA^2 \tag5$
and by substituting into equation 1.1:
$E_{tot} = 0.5mr_a^2\omega_{ta}^2 + 0.5kA^2 \tag6$
![rotating mass with spring](https://cdn.statically.io/img/i.sstatic.net/5QjuS.png)