Question

Can someone please help me to understand what this means?
Topic: Calculating the value of g with the use of the simple pendulum.

This is from the marking scheme and I have no idea what it means. The textbooks that I have has something completely different from what I'm about to write.

Answer 1

From the second law, \[mg \sin \theta = \frac{ mdv }{ dt }\] where the velocity, v, can be written in terms of \[v= \frac{ Ld \theta }{ DT }\]

Answer 2

From Newtons second Law in circular accelerated motion we get\[T-mg \cos \theta=m \frac{ v^2 }{ l }\]\[-mg \sin \theta =m \frac{ dv }{ dt }=ml \frac{ d \theta }{ dt } \rightarrow \dot{\theta}=-\frac{ g }{ l }\sin \theta \approx -\frac{ g }{ l } \theta \rightarrow \theta = \theta _0 \cos (\sqrt{\frac{ g }{ l }} t + \phi)\]From the last equation we can see that the period of oscillation will be \[T=2 \pi \sqrt{\frac{ l }{ g }}\]Therefore, if you measure the period, you can deduce what the gravitational acceleration is through the equation\[g=4 \pi ^2 \frac{ l }{ T^2 }\]