# Time period for simple pendulum without trigonometry

The time period for simple pendulum formula that is $t = 2\pi \sqrt{\frac{l}{g}}$ is valid only for $15^\circ$ of amplitude. Why not more than that?

How do I explain this someone who has just been introduced about these concepts and doesn't know little bit about trigonometry or free body diagrams.

• Small angle approximation. – Shahar Apr 26 '14 at 15:31
• If you don't know trigonometry and FBD then you must not be studying this. – evil999man Apr 26 '14 at 15:50

For small values of $\theta$, $\sin \theta \approx \theta$. Until he knows about Taylor expansions, take it on faith.
$$T \approx 2\pi \sqrt\frac{L}{g} \qquad \theta \ll 1 \tag{1}\,$$ For $\theta \ll 1$, $T$ is approximately the same for different $\theta$s - $T$ is independent of amplitude. For larger amplitudes, $T$ increases with amplitude, and so is longer than given by the above equation.
Why does the above equation hold for small $\theta$ only? First, explain in simple terms what $\sin\theta$ is. Then, show that $\sin\theta\approx\theta$ for small $\theta$, by, say, drawing a diagram. Since the equation of motion for a simple pendulum is $${d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0\\ \implies {d^2\theta\over dt^2}+{g\over \ell} \theta=0\qquad \theta \ll 1$$ (for him, right now, the meaning of the symbols is not needed) which has as solution $$T \approx 2\pi \sqrt\frac{L}{g} \qquad \theta \ll 1 \tag{2}\,$$