$v^2 = \omega^2(A^2 - x^2) \Rightarrow \frac{v^2}{\omega^2 A^2} + \frac{x^2}{A^2} = 1$
Compare with $\frac{v^2}{b^2} + \frac{x^2}{a^2} = 1$: $A = a$, $\omega^2 a^2 = b^2 \Rightarrow \omega^2 = b^2/a^2$
$k = m\omega^2 = m \cdot \frac{b^2}{a^2} = \dfrac{mb^2}{a^2}$
Answer: $\boxed{\dfrac{mb^2}{a^2}}$
SHM: restoring force $F = -kx$. Equation of motion: $\ddot{x} = -\omega^2 x$ where $\omega^2 = k/m$. General solution: $x = A\sin(\omega t + \phi)$. Velocity: $v = A\omega\cos(\omega t + \phi) = \omega\sqrt{A^2 - x^2}$. Acceleration: $a = -\omega^2 x$. At equilibrium $(x=0)$: $v = v_{max} = A\omega$, $a = 0$. At extreme $(x=\pm A)$: $v = 0$, $|a| = a_{max} = A\omega^2$.
Total mechanical energy $E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$ (constant). PE: $U = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$. KE: $K = E - U = \frac{1}{2}m\omega^2(A^2 - x^2)$. At $x = 0$: KE is max, PE = 0. At $x = \pm A$: PE is max, KE = 0. Average KE = Average PE = $E/2$.
The trajectory in v-x (phase space) is an ellipse: $\frac{v^2}{(\omega A)^2} + \frac{x^2}{A^2} = 1$. Semi-major axis on x-axis: $A$ (amplitude). Semi-major axis on v-axis: $\omega A$ (max speed). If $\omega = 1$: circle. If $\omega \neq 1$: ellipse. Clockwise for $x = A\sin(\omega t)$, counter-clockwise for $x = A\cos(\omega t)$.
Spring-mass: $\omega = \sqrt{k/m}$, $T = 2\pi\sqrt{m/k}$. Simple pendulum: $\omega = \sqrt{g/l}$, $T = 2\pi\sqrt{l/g}$ (valid for small $\theta$). Compound pendulum: $T = 2\pi\sqrt{I/mgl}$. LC circuit: $\omega = 1/\sqrt{LC}$. Spring in series: $1/k_{eff} = 1/k_1 + 1/k_2$. Spring in parallel: $k_{eff} = k_1 + k_2$.
Kinetic energy is $\frac{1}{2}m\omega^2(A^2 - x^2)$ and potential energy is $\frac{1}{2}m\omega^2x^2$, so their sum $\frac{1}{2}m\omega^2A^2$ stays constant throughout the motion. Both energies vary with position as a squared term, which means each completes two full cycles for every one cycle of displacement — so the frequency of the energy variation is twice the frequency of the oscillation. Averaged over a complete cycle, kinetic and potential energy are each exactly half the total.
A real oscillator loses energy, and with a damping force proportional to velocity the amplitude decays exponentially as $A = A_0e^{-bt/2m}$ while the frequency shifts slightly below the natural value. When a periodic driving force is applied instead, the system eventually oscillates at the driving frequency, not its own. Resonance occurs when the two coincide, and the amplitude then becomes very large — limited only by damping, which is why a lightly damped system resonates violently and a heavily damped one barely responds.
Two springs in parallel share the load, so their stiffnesses add: $k_{eff} = k_1 + k_2$, and the combination oscillates faster than either alone. In series each spring carries the full load and the extensions add, so the reciprocals combine: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$, giving a softer system and a longer period. Note this is the opposite of how resistors behave, which is a frequent source of confusion. Cutting a spring of constant k into n equal pieces gives each piece a constant of nk, because a shorter spring is stiffer.
Assuming the pendulum formula works for large angles. $T = 2\pi\sqrt{L/g}$ holds only for small oscillations, where $\sin\theta \approx \theta$ makes the restoring force proportional to displacement.
Mixing up where velocity and acceleration peak. Velocity is maximum at the mean position where acceleration is zero, and acceleration is maximum at the extremes where velocity is zero.