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Let's say you have a pendulum moving in SHM with a length $L$ and an amplitude $\theta$. Suppose you wanted to find the linear velocity $v$ at it's lowest point. The way that gets the right answer involves using the energy equations and a ton of trigonometry, but I end up with $$v = 2sin(\frac\theta2)\sqrt{Lg}.$$

I thought my approach would be much easier, but it doesn't work. It involves using the simple harmonic equations, namely: $$\omega=\theta\sqrt\frac{g}{L}cos(\sqrt\frac{g}{L}t+\phi).$$ Since the angular velocity is at it's max at that point, the cosine term will equal 1, and $\omega=\frac{v}{L}$. So you end up with: $$\frac{v}{L}=\theta\sqrt\frac{g}{L}.$$ Solving for $v$ I ended up with $v=\theta\sqrt{Lg}$. This is so close to the solution. What is wrong with my approach?

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    $\begingroup$ Actually they are equivalent ... well at least for small $\theta$ i.e expand $\sin ( \theta/2) $ to first order using a Taylor series. $\endgroup$ – Rumplestillskin Aug 25 at 1:50
  • $\begingroup$ A simple pendulum is only approximately (for small angles $\theta$) a simple harmonic oscillator. Your first equation is correct while the rest are approximations. $\endgroup$ – Puk Aug 25 at 4:05
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For the pendulum case, equation of torque is$$\tau = mgL\sin\theta$$ thus the angular acceleration is proportional to sine of $\theta$, not $\theta$.

But this motion becomes angular SHM if the amplitude of its motion is small. Then using Taylor series we take an approximation, $$\sin\theta \approx \theta$$ Using energy equations and trigonometry, you might have solved it for a general case, thus your answer has a sin term. Because it's SHM, you might take the above mentioned approximation. The SHM equation, however, already included it.

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For small angles, $\sin(\theta/2)\approx\theta/2$. Therefore, your two equations are equivalent.

The first equation has not yet applied this small angle approximation. The second one already applied it, since you went from the equation $\ddot\theta=-(g/L)\sin\theta$ to $\ddot\theta=-(g/L)\theta$ using this approximation.

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