Pendulum period [duplicate]

In plane pendulum problem, we can calculate its period using elliptic integration.

In SHO problem, we use approximation such that $\theta\ll 1$ and get the period, $2\pi\sqrt{l/g}$.

Is there another way of explaining that the period of pendulum (no approximation) is larger than the period of SHO without considering elliptic integration?

That the true period of a pendulum is longer than the period found using the harmonic oscillator (HO) approximation can actually be deduced just from the fact that $\sin\theta<\theta$ for $\theta>0$. That means that the restoring torque on the pendulum bob is always less than in the approximate HO system. Weakening the restoring force automatically means that the pendulum's excursions away from the equilibrium position are going to be longer, which is the desired result.
• For small $\theta$ in an experimental setup, $sin\theta \approx \theta$, $T\pm\Delta T=2\pi\sqrt{\frac{l\pm\Delta l}{g}}$, and the difference from the theoretical period is negligible.