Mechanical Vibrations

A mechanical vibration is simply something that moves back and forth around a resting position, like a mass bouncing on a spring, and this lesson uses it to tie the whole unit together. All we're really asking is: where is the mass at each moment? The equation $m\,u'' + \gamma\,u' + k\,u = F(t)$ looks like a lot, but it's just forces added up: $mu''$ (mass times acceleration), $\gamma u'$ (friction, which pushes back harder the faster the mass moves), $ku$ (the spring pulling back harder the farther it's stretched) and $F(t)$ (whatever is pushing from outside). Newton's law says they have to balance, and that's the whole equation. It's not as bad as it looks, and you already have every tool you need from this unit; bear with it.

Each case below is one of the earlier lessons in disguise: the characteristic roots of $mr^2 + \gamma r + k = 0$ decide how the mass moves.

Free, undamped ($\gamma = 0$, $F = 0$): $u = R\cos(\omega_0t - \delta)$, an oscillation that lasts forever. The roots are $\pm\omega_0 i$, purely imaginary, so there's no envelope to shrink anything.

Where that form comes from. Solving gives $u = A\cos\omega_0t + B\sin\omega_0t$, two waves added together. But on a graph you only ever see one wave, so there should be a way to write it as one. Expand the single wave: $R\cos(\omega_0t - \delta) = R\cos\delta\,\cos\omega_0t + R\sin\delta\,\sin\omega_0t$. Matching the two sides gives $A = R\cos\delta$ and $B = R\sin\delta$. Square and add: $A^2 + B^2 = R^2$, so $R = \sqrt{A^2 + B^2}$. Divide: $\tan\delta = B/A$, with $\delta$ in the quadrant of the point $(A, B)$ (add $\pi$ when $A < 0$).

Reading it off a graph: $R$ is the peak height, the first peak sits at $t = \delta/\omega_0$ (the time lag), and peaks repeat every period $2\pi/\omega_0$. Notes on Diffy Qs writes the same wave as $C\cos(\omega_0t - \gamma)$, with $\gamma$ the phase and $c$ the damping.

Free, damped. The discriminant of $mr^2 + \gamma r + k$ is $\gamma^2 - 4mk$, so everything hinges on comparing $\gamma$ with $2\sqrt{mk}$:

• $\gamma < 2\sqrt{mk}$: underdamped, oscillations that die out (complex roots).
• $\gamma = 2\sqrt{mk}$: critically damped, the fastest return without oscillating (repeated root).
• $\gamma > 2\sqrt{mk}$: overdamped, a slow creep back (two real roots).

Forced. With $F = F_0\cos\omega t$, undetermined coefficients says the response settles into an oscillation at the forcing frequency. Now think about what happens when there's no damping and $\omega = \omega_0$. The guess $A\cos\omega_0t + B\sin\omega_0t$ is already in $u_c$, so the duplication rule multiplies it by $t$, and that $t$ out front makes the amplitude grow without bound: resonance. Every push lands in step with the motion. Close but not equal frequencies produce beats instead, where the swings grow and shrink in a slow rhythm.

A common mistake. It's tempting to think more damping always means the spring settles faster. It doesn't: past critical damping, an overdamped spring creeps back more slowly. Critical damping is the sweet spot.

Buildings are designed so earthquake frequencies don't match their natural frequencies, and car suspensions are tuned near critical damping. The forced oscillations lab lets you push a spring at any frequency.