Natural frequency by energy method

For a conservative system with no damping, the total energy stays constant:

$$T + U = \text{constant}$$

Spring-mass (massless spring)

Let the mass $m$ be displaced by $x$ from the static equilibrium position and have velocity $\dot{x}$.

$$T = \tfrac12 m\dot{x}^2, \qquad U = \tfrac12 k x^2$$

(The change of potential energy of gravity is cancelled by the static spring force, so only the extra spring energy $\tfrac12kx^2$ remains.)

Differentiating $T + U =$ constant:

$$m\dot{x}\ddot{x} + kx\dot{x} = 0 \;\Rightarrow\; m\ddot{x} + kx = 0$$ $$\omega_n = \sqrt{\frac{k}{m}}, \qquad f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{\frac{g}{\delta_{st}}}$$

Alternatively, with $x = X\sin\omega_n t$, the maximum kinetic energy $\tfrac12 m\omega_n^2X^2$ equals the maximum potential energy $\tfrac12 kX^2$, which gives the same result.

Effect of the spring's own mass

Let the spring have mass $m_s$ and length $l$. A small element $dy$ at distance $y$ from the fixed end moves with velocity $\dfrac{y}{l}\dot{x}$ (assuming a linear variation).

$$T_{spring} = \int_0^l \tfrac12\left(\frac{m_s}{l}dy\right)\left(\frac{y\dot{x}}{l}\right)^2 = \tfrac12\cdot\frac{m_s}{3}\dot{x}^2$$

Total kinetic energy $T = \tfrac12\left(m + \dfrac{m_s}{3}\right)\dot{x}^2$. Hence

$$\omega_n = \sqrt{\frac{k}{m + m_s/3}}$$

One-third of the spring mass is added to the suspended mass as an effective mass. The natural frequency decreases slightly; the correction is negligible when $m_s \ll m$.