If you stand on one leg and let the other dangle freely back and forth starting at an initial amplitude of, say, 20° or 30°, the amplitude will decay to one-half of the initial amplitude after about four swings. Regarding the dangling leg as a damped oscillator, what value of "Q" can you deduce from this?
the amplitude is no longer constant but a function where y is a constant that represents the frictional effects. $$A(t) = A_oe^{-yt/2}$$
at a characteristic time $t^* = 2/y$ the pendulum reach equilibrium in our case taking 25 as $A_0$: $$A(t^*) = A_0e^{-1} = 9.196 $$
The quality factor is: $$Q = \frac{\omega}{y}$$ where Q is $$Q = 2\pi (\frac{E}{\Delta E})$$
also:
$$\frac{t^*}{T} = \frac{Q}{\pi}$$
but $\frac{t^*}{T}$ is just the number of swings in time
now we are given the number of swings when the amplitude is half but we want the number of swings when the amplitude is at its equilibrium(9.16) let's build a function relating the amplitude after s swings
$$A(s) = 25 - \frac{12.5}{4} S$$ $$9.16 = 25 -\frac{12.5}{4}S$$ $$S = 5.06$$
now S is the number of swings such that amplitude reaches equilibrium we plug S in the formula for quality factor:
$$S = \frac{Q}{\pi}$$ $$Q = 5.06*\pi = 15.95$$
which is not the desired answer what did I get wrong?
the true answer must be around 18
For a damped oscillator we have $A(t) = A_0 e^{-\lambda t}\cos(\omega t)$ and then
with
$$A_0 e^{-8\pi\frac{\lambda}{\omega}}=\frac12 A_0 \implies e^{8\pi\frac{\lambda}{\omega}}=2\implies \zeta=\frac{\lambda}{\omega}=\frac{\log 2}{8\pi}$$
and then
$$Q=\frac1{2\zeta}=\frac{4\pi}{\log 2}\approx 18.13$$