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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.

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  • $\begingroup$ Small angle approximation. $\endgroup$ – Shahar Apr 26 '14 at 15:31
  • $\begingroup$ If you don't know trigonometry and FBD then you must not be studying this. $\endgroup$ – evil999man Apr 26 '14 at 15:50
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For small values of $\theta$, $\sin \theta \approx \theta $. Until he knows about Taylor expansions, take it on faith.

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  • $\begingroup$ What about fundamental definition of sin and elementary of sinx/x limit? $\endgroup$ – evil999man Apr 26 '14 at 15:45
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$$T \approx 2\pi \sqrt\frac{L}{g} \qquad \theta \ll 1 \tag{1}\,$$ 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 $$T \approx 2\pi \sqrt\frac{L}{g} \qquad \theta \ll 1 \tag{2}\,$$

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