A pendulum is 1.20 m long. (a) Find its period for small swings…

(a) Small swings. $T_0 = 2\pi\sqrt{\dfrac{1.20\ \text{m}}{9.8\ \text{m/s}^2}} = 2.20\ \text{s}$.

(b) A wide swing. $\theta_0 = 40 \times \dfrac{\pi}{180} = 0.698$ rad. Then $\dfrac{\theta_0^2}{16} = \dfrac{0.487}{16} = 0.0305$. So

(c) The increase. $0.0305$ of $T_0$, which is about $3\ \%$.

Why? For a wide swing, $\sin\theta$ is smaller than $\theta$. The pull back is weaker than the simple formula assumes, so the bob takes longer. The program follows the true motion step by step and gives $1.0313$ times $T_0$, very close.