Relation between two second derivatives

Question

I got two equations $\ (1) \ $ and $\ (2) \ $ of which the $\ (2) \ $ is said to be related in some relation with $\ (1)$.

$$\ F_t= ma_t \rightarrow - mg \sin \theta = m \frac{d^2 s}{dt^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

$\ s$ is defined to be $\ s=L\theta$, therefore the equation below is obtained:

$$\ \frac{d^2 \theta}{dt^2} = -\frac{g}{L} \sin\theta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$

I don't know how to substitute $\ \theta = \frac{s}{L} \ $ into $\ (1)$ to get $\ (2)$.

p/s: This is the equation of Restoring Force for Simple Pendulum.

Answer 1

Given that, $$ F=-mg\sin \theta=m \frac{d^2 s}{dt^2}$$ so, $$-mg\sin \theta=m \frac{d^2 s}{dt^2}.........(1)$$ $$\text{Again,}~\theta =\dfrac{s}{L}$$ $$\implies s=L\theta$$ substituting the value of s in (1), $$-mg\sin \theta=m \frac{d^2 (L\theta)}{dt^2}$$ $$\implies -g\sin \theta=L\frac{d^2 (\theta)}{dt^2}~\text{[dividing both side by m]}$$ $$\implies \boxed{-\dfrac{g}{L}\sin \theta=\frac{d^2 \theta}{dt^2}}$$

Answer 2

Yep, so the equation you're talking about is done using the small angle assumption. This is a differential equation, and the only thing you can do to solve it explicitly is to make the substitution:

$$ \theta \approx \sin \theta $$

If you do that, then your equation will simplify quite a lot. So you will get:

$$ {d^2 \theta \over dt^2} = -{g\over L} \theta $$

Which you can now solve explicitly if you want to (hint: the second derivative of a sine wave or a complex exponential is proportional to itself).