2Free undamped vibration
With no damping and no force, the equation is $m\ddot x + kx = 0$, or $\ddot x + \omega_n^2 x = 0$. Trying $x=e^{st}$ gives the characteristic equation $s^2+\omega_n^2=0$, so $s=\pm i\omega_n$, and the general solution is a pure sinusoid:
$$ x(t) = A\cos\omega_n t + B\sin\omega_n t = X\cos(\omega_n t - \phi) $$The constants come from the initial conditions $x(0)=x_0$, $\dot x(0)=v_0$:
$$ A = x_0,\qquad B=\frac{v_0}{\omega_n},\qquad X=\sqrt{x_0^2+\left(\frac{v_0}{\omega_n}\right)^2},\qquad \tan\phi=\frac{v_0}{\omega_n x_0} $$Why a cosine? Picture a vector of length $X$ spinning at $\omega_n$. Its vertical shadow is the motion. The initial conditions set where the vector starts: $x_0$ is its vertical part and $v_0/\omega_n$ its horizontal part, so its length is $X=\sqrt{x_0^2+(v_0/\omega_n)^2}$ and its starting angle is $\phi$.
Change v₀: the vector starts tilted, which is exactly the phase angle φ. Change ωₙ: it spins faster, but the circle (amplitude) stays the same.
The motion repeats with natural frequency $f_n=\omega_n/2\pi$ (Hz) and period $\tau_n=1/f_n=2\pi/\omega_n$. Energy sloshes between the spring ($V=\tfrac12 kx^2$) and the mass ($T=\tfrac12 m\dot x^2$) with the total constant.
Since $\omega_n=\sqrt{k/m}$, quadrupling the mass halves the frequency, and quadrupling the stiffness doubles it. Race three oscillators released together:
Count the cycles: the stiff one completes two for every one of the baseline, and the heavy one only half.
Phase portrait: $\dot x/\omega_n$ vs $x$. Undamped motion traces a circle of radius $X$.
Try this
- Quadruple the mass. What happens to $f_n$? Now quadruple $k$ as well.
- Change only $x_0$. The amplitude changes but the period does not. That is a hallmark of a linear system.
- Set $x_0=0$ and give only $v_0$. Where does the phase angle $\phi$ end up?
- Watch the energy bars: when is $T$ maximal? When is $V$ maximal?
The heartbeat of a quartz watch is a tiny tuning fork cut from quartz crystal, vibrating at 32,768 Hz. That is 215, so a chain of 15 divide-by-two circuits turns it into exactly one tick per second. Its frequency is set by the same √(k/m) you just played with.