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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Mick
    Commented Aug 23, 2018 at 8:50
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For an SHO, the magnitude of the force pulling it back to the neutral position has to be proportional to the distance or the angle relative to the neutral position.

In a pendulum, this force is proportional to the sine of the angle and, therefore, as the angle increases, the magnitude of the force increases slower than it would if it was proportional to the angle.

This allows a pendulum to keep rising for a longer time and, thus, increases its period.

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