Over dampened Pendulum System: $$ mL^{2}\ddot{\theta } +b\dot{\theta } +mgL\sin \theta =\Gamma $$ First order approximation:
$$ b\dot{}+mgL\sin{}=Γ $$
Nondimensionalize, diving through by mgL:
$$ \frac{b\dot{\theta }}{mgL} =\frac{\Gamma }{mgL} -\sin \theta $$
Let $$\tau =\frac{mgL}{b} t$$ and $$\gamma =\frac{\Gamma }{mgL}$$
Then
$$ \frac{d\theta }{d\tau } =\gamma -\sin \theta $$
The discussion in section 4.4 of Strogatz describes $\gamma$ as the dimensionless group that represents the applied torque to the gravitational torque. As $\gamma$ decreases to 1 a fixed point at pi/2 appears, as it goes lower stable and unstable pair appear from this fixed point.
I get what he's saying. I understand the system, but then he asks the following questions:
The solution manual only presents the answers to odd questions. How are you supposed to represent $\frac{d\theta }{d\tau } =\gamma -\sin \theta$ against time when its a function of $\tau$?
I do not have a digital copy of the solution manual. However it shows a graph of $\dot{}$($\frac{\dot{d\theta }}{dt}$) vs $t$ notated at $\gamma = 100$. The graph is a straight line.
Unsure how he came to this result as $\dot{}$ is never a function of $\gamma$.