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Since the question allows us to assume simple harmonic motion:

$\theta (t)=\theta _{\max } \sin (t)$

We look for when $\theta (t)=\pm\frac{\theta _{\max }}{3}$ .

Case 1:

$\theta (t)=\frac{\theta _{\max }}{3}\\ t=\pi -\sin ^{-1}\left(\frac{1}{3}\right)\lor t=\sin ^{-1}\left(\frac{1}{3}\right)$

Case 2:

$\theta (t)=-\frac{\theta _{\max }}{3}\\ t=2 \pi -\sin ^{-1}\left(\frac{1}{3}\right)\lor t=\pi +\sin ^{-1}\left(\frac{1}{3}\right)$

Graphically, we are looking at the shaded areas:

Graph

To find the fraction of time, we take the valid intervals divided by $2 \pi$ :

Case 1:

$\frac{\left(\pi -\sin ^{-1}\left(\frac{1}{3}\right)\right)-\sin ^{-1}\left(\frac{1}{3}\right)}{2 \pi }$

Case 2:

$\frac{\left(2 \pi -\sin ^{-1}\left(\frac{1}{3}\right)\right)-\left(\pi +\sin ^{-1}\left(\frac{1}{3}\right)\right)}{2 \pi }$

Summing both gives a numerical answer of $0.783653$ .

The forumla for the angular displacement $\theta$ of a pendulum is $\theta = \theta_{max} sin(\sqrt{\frac{g} {L}} t)$ , where $g$ refers to the gravitational acceleration and $L$ refers to the length of the string. Since these parameters are irrelevant, we will set $\sqrt{\frac{g} {L}}$ to unity.

Now we are left with $\theta = \theta_{max} sin(t)$ . For convenience's sake, we can set $\theta_{max}$ to $3$ , leaving us with the even simpler $\theta = 3sin(t)$ .

Now, the question asks for what fraction of time is $|\theta(t)| > \theta_{max}/3$ . Since $\theta(t)$ is $\theta = 3sin(t)$ , $\theta(t)$ has a period of $2\pi$ . Thus we only need to consider the length of $t$ for which $\theta(t)$ is above $\theta_{max} / 3 = 3 / 3 = 1$ for the first $2\pi$ , and divide it by $2\pi$ .

So we solve the inequality $|3sin(t)| > 1$ , to give the basic angle $\alpha = 0.33984\text{ rad (to 5 sf)}$ . Therefore, the answer is $\frac{((\pi - \alpha) - \alpha) + ((2\pi - \alpha) - (\pi + \alpha))}{2\pi} = 0.78365\text{ (to 5 sf)} = \boxed{0.784}\text{ (to 3 sf)}$ .

$\theta$ " + (g/l) sin $\theta$ = 0 x = sqrt(g/l) t => d² $\theta$ /dx² + sin $\theta$ = 0 linearisation : d²(theta)/dx² + theta = 0 => theta = A cos(x) + B sin(x) take theta(0) = 0 => A = 0 => theta = B sin(x) => theta max = B, because sin(x) has maximum value 1 => theta(x) = theta max * sin(x) |theta| = theta max / 3 => sin(x) = +- 1/3 => x = +- arcsin(1/3) = +- 0.3398 x has period 2 pi so the amount of time for which |theta(t)|>(theta max)/3 is (2 pi - 2 * 0.3398)/(2 pi) = 1 - 0.3398/pi = 0.892 or 89.2 % I am now thinking that the pendulum swings from left to right and then from right to left again, so that we have to take 4 times 0.3398 which would yield = 1 - 2*0.3398/pi = 0.7837 = 78.37 %