Question

Can someone please help me to understand this?
Aim: To determine the value of 'g' with the use of the simple pendulum.

Thank you.

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

@kym02 Please note that, formula that expresses period of little oscillations of a simple pendulum, is this:
\[T=2 \pi \sqrt{\frac{ l }{ g }}\]
wher g is, as usually, the gravity, namely g=981 cm/sec.
Now squaring both sides of the above formula, we have:
\[T ^{2}=4 \pi ^{2} \frac{ l }{ g }\]
from which we can write:
\[g=4 \pi ^{2}\frac{ l }{ T ^{2} }\],
so you can measure g using measures of period T

Answer 3

@kym02 of course T is the period of little oscillations, and l is the length of your simple pendulum

Answer 4

the above formula can be derived from the equations that @kym02 posted

However - I seem to recall that it requires an assumption that sin theta = theta for small values of theta (in radians).

The above formula is ONLY valid for small amplitude of swing for the pendulum

Answer 5

Thank you very much. I now understand!😃