Where Did I make a mistake in my deviation for the Period of a Pendulum Formula

Question

Imagine a Pendulum with String Length L, and Angle θ, which forms arclength S (Will be used later in the derivation.)

First V = $\sqrt{2gh} $ (Height is measured straight down) KNOWN TO BE A CORRECT FORMULA

(From Here, Assume that nothing else is proven to be correct.)

Arclength X = $Lθ $

Time (Period ) = T

V = $(Distance) / (Time)$

V = $Lθ/T $ $\sqrt{2gh} = \frac{4Lθ}{T} $ (Four, as the pendulum covers that area four times)

Height $= L - L \cos θ$

$\sqrt{2gL(1 - cos(θ))} = \frac{4Lθ}{T} $

$T = \frac{4Lθ}{\sqrt{2gL(1 - cos(θ))}} $

This is very different from the known period of a (simple) Pendulum $T = 2\pi\sqrt{\frac{L} {g}} $, mainly where the angle $\theta $ doesn't even exist. Is there a way to get the canceled $\theta $ in my derived equation to cancel out? Did I just completely make a Mistake?

Answer 1

The original derivation takes $\theta$ to be small so $1-\cos\theta\approx{\theta^2\over 2}$ your answer becomes $$T=4\sqrt{L\over g}$$

Now your mistake is $v={s\over t}$ because this only works when velocity doesn't change (acceleration = 0) but velocity depends on height and height changes

I'll prove it here for reference a fair background in rotational mechanics would help

Taking polar coordinates we $r=L$ and $\theta$ (angle with vertical) varies

From rudimentary circular motion with fixed $r$ using torque equation about the point of contact we get $$ mr^2{d^2\theta\over dt^2}=-mgr\sin\theta $$ The gravity component is taken perpendicular to string as tension cancels the other component

Now nothing stops you from directly integration for $\theta(t)$ but this comes out as elliptic integrals and other fun stuff to avoid that we approximate $\sin\theta\approx\theta$

This is the general SHM $${d^2\theta\over dt^2}=-{g\over L}\theta$$ Which solves to $$\theta(t)=A\sin\left(\sqrt{g\over L}t+\phi\right)$$

The period of this function is $$T=2\pi\sqrt{L\over g}$$

Answer 2

1. What went wrong? The formula

$$\text{speed}=\frac{\text{distance}}{\text{time}} $$

actually defines the average speed, whereas your formula $v=\sqrt{2gh}$ is the maximum instantaneous speed of the bob (when it passes through the minimum height point). This means that you cannot directly plug $v$ to the above formula.

2. What is correct? In fact, the period of a (not-approximated) pendulum is given by

$$T = \frac{4}{\omega} K\left(\sin\frac{\theta_0}{2}\right),$$

where

• $\omega = \sqrt{\frac{g}{L}}$ is the angular frequency of corresponding simple pendulum,
• $K(k) = \int_{0}^{\frac{\pi}{2}} \frac {\mathrm{d}\theta}{\sqrt{1 - k^{2}\sin^{2} \theta}}$ is the complete elliptic integral of the first kind, and
• $\theta_0$ is the maximum angular displacement.

When $\theta_0$ is small, this is approximated by the usual formula for the simple pendulum:

$$ \theta_0 \ll 1 \qquad\implies\qquad T \approx \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{L}{g}}. $$

Check this page for the derivation.