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The tangential force exerted on a pendulum weight is $-mgsin(\theta)$. If we say that the pendulum has length L than $sin\theta$ = $\frac{x}{l}$.

Then $$F_{tangential} = \frac{-mg}{l}x$$

Then why do we need the small angle approximation at all? This relation between the force and the displacement satisfies the condition of simple harmonic motion, which is $\frac{F}{x} = c$ ; $c<0$.

My textbook uses small angle approximation and derives the same force equation from there. But to me, it seems like the relation should be linear even if the angle is large.

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    $\begingroup$ $\text{sin} \theta= x/l $can be used only for small angles. As you can see for yourself, for bigger angles, torque contains a sine term in $\theta$ and that is precisely why the EOM doesn't contain $\text{sin} \theta$ :) $\endgroup$
    – Physiker
    Commented Jan 10, 2021 at 14:20

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$F_{tangential}$ is not in the direction of the displacement $x$. The force in the $x$ direction is

$F_x = F_{tangential} \cos \theta$. Which is almost equal to your $F_{tangential}$ - if the angle is small.

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  • $\begingroup$ You should take the other answer into account too: your formula only accounts for the "remainder" of the force of gravity that applies in the horizontal direction, but the rope also adds an additional horizontal force (again, negligible at small angles) $\endgroup$
    – Kristoffer Sjöö
    Commented Jan 10, 2021 at 18:50
  • $\begingroup$ Weight has no component in the x direction. It does not make any sense to take the horizontal component of the tangential component of the weight. It is zero. $\endgroup$
    – nasu
    Commented Jan 14, 2021 at 21:08
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If you want to solve the problem along the $x$ direction, you must consider also the tension of the rope has a non-zero $x$ component which depends on the velocity of the pendulum.

As you do not know the velocity of the pendulum until you solve the differential equations, your approach does not simplify the problem.

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  • $\begingroup$ And you should also use the component of weight in the horizontal direction. Which of course is zero. So the tension is the only force with component in the x direction. I mean tension is not an additional term, is the only one. $\endgroup$
    – nasu
    Commented Jan 14, 2021 at 21:05

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