Driven oscillation and resonance

What is it about?

Pohl’s wheel is a torsion pendulum: a copper wheel on a spiral spring that, once pushed, oscillates at its natural frequency $f_0 = 0.5\,\text{Hz}$ . An eddy-current brake damps it with decay constant $\delta$ .

A motor moves the outer end of the spring (the driver) back and forth: $\varphi_E(t) = \hat\varphi_E \sin(\omega_E t)$ . After a transient, in which the natural oscillation dies away with $e^{-\delta t}$ , the wheel oscillates at the driving frequency $f_E$ :

\varphi(t) = \hat\varphi \sin(\omega_E t - \Delta\varphi), \qquad \hat\varphi = \hat\varphi_E \cdot \frac{\omega_0^2}{\sqrt{(\omega_0^2 - \omega_E^2)^2 + (2\delta\,\omega_E)^2}} .

The resonance curve $\hat\varphi(f_E)$ peaks at $\omega_R = \sqrt{\omega_0^2 - 2\delta^2}$ , slightly below $f_0$ ; weak damping makes it tall and narrow. The wheel lags behind the driver: $\Delta\varphi \approx 0°$ for slow driving (in phase), $90°$ at $f_E = f_0$ , $\approx 180°$ for fast driving (antiphase). With very weak damping the amplitude at resonance, $\hat\varphi \approx \hat\varphi_E\,\omega_0/(2\delta)$ , exceeds the load limit of $120°$ : a resonance disaster.

Try it

Start from rest at $f_E = 0.3\,\text{Hz}$ and watch the amplitude settle. At what frequency does the wheel oscillate afterwards?
Set resonance and compare the amplitudes and the phase shift in the phasor diagram.
Fast driving: does the wheel move with the driver lever or against it?
Record the resonance curve, then repeat with strong damping and weak damping.
Try the resonance disaster. How can you save the wheel?

Tasks

Task 1: Match $f_E = 0.1$ , $0.5$ , $1.5\,\text{Hz}$ (with $f_0 = 0.5\,\text{Hz}$ ) to $\Delta\varphi \approx 0°$ , $90°$ , $180°$ .

Show solution

$0.1\,\text{Hz} \to 0°$ , $0.5\,\text{Hz} \to 90°$ , $1.5\,\text{Hz} \to 180°$ . The amplitude is smallest at $1.5\,\text{Hz}$ (about $0.6°$ for $\hat\varphi_E = 5°$ ).

Task 2: Calculate the amplitude at resonance for $\hat\varphi_E = 5°$ and $\delta = 0.25\,\text{s}^{-1}$ .

Show solution

$\hat\varphi = \hat\varphi_E \cdot \omega_0/(2\delta) = 5° \cdot \pi / 0.5 \approx 31°$ . Check

Task 3: Which damping keeps the wheel below $120°$ at resonance with $\hat\varphi_E = 10°$ ?

Show solution

$\delta > \hat\varphi_E\,\omega_0 / (2 \cdot 120°) = 10° \cdot \pi / 240° \approx 0.13\,\text{s}^{-1}$ . Check

Ideas for the classroom

Model: $\ddot\varphi = -\omega_0^2(\varphi - \varphi_E) - 2\delta\dot\varphi$ , solved with Runge–Kutta; $f_0 = 0.5\,\text{Hz}$ fixed; load limit $120°$ .
“Record resonance curve” skips the transient at each frequency and measures amplitude and phase shift; theory curves can be hidden.
Common stumbling blocks: the wheel keeps its natural frequency (no – it follows $f_E$ ); at resonance driver and wheel are in phase (no – $\Delta\varphi = 90°$ ); faster driving always means larger amplitude (no).