Question

If the frequency of a pendulum is four times greater on an unknown planet then it is on earth, then the gravitaional constant on that planet is:
a) 16 times greater
b) 4 times greater
c) 4 times lower
d) 16 times lower
e) 24 times lower
Give the correct answer and explain it.
@sarah786 @2Angel4 @Muskan @jacalneaila

Answer 1

@AllTehMaffs

Answer 2

The restoring force of a pendulum
\[-mg \sin \theta = ma\]
For small angles,
\[\sin \theta = \theta\]
so
\[a=\frac{\mathrm{d}^2v}{\mathrm{d}t^2}=\ell \frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}=-g\theta\]
giving the simple diff eq
\[\frac{\mathrm{d}^2\theta}{\mathrm{d}t^2}+\frac{g}{\ell}\theta=0\]
with the solution
\[\theta=\theta_{max} \sin(\omega t)\]
\[ \text{angular frequency} \ = \omega = \sqrt{\frac{g}{\ell}}\]

Earth
\[ \omega_{E}= \sqrt{\frac{g_{E}}{\ell}}\]
If
\[ \omega_{p} = 4\omega_{E}\]
then
\[ 4\omega_{E}= \sqrt{\frac{g_{p}}{\ell}}\]
\[4\sqrt{\frac{g_{E}}{\ell}}=\sqrt{\frac{g_{p}}{\ell}}\]
bring the 4 into the radical to solve

Answer 3

\[\sqrt{\frac{16g_{E}}{\ell}}=\sqrt{\frac{g_{p}}{\ell}}\]
\[16g_{E}=g_p\]

Answer 4

Great Master Maff @AllTehMaffs :) ^^