Question

A simple pendulum on Earth has a period of 6.0 seconds. What would the period of this pendulum be on Jupiter, if the acceleration of gravity on the surface of Jupiter is 26 m/s ?
A. 2.3 sec
B. 3.7 sec
C. 6.0 sec
D. 15.6 sec

Answer 1

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
\[\theta=\theta_{max} \sin(\omega t)\]
\[ \text{angular frequency} \ = \omega = \sqrt{\frac{g}{\ell}}\]

\[ \text{Period}=T=\frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\ell}{g}}\]

Solve for l - the length of the pendulum - by plugging in the period on Earth along with earth's gravity, then use the found length and the gravity of Jupiter to find the period on Jupiter

^_^

Answer 2

The period should be shorter on Jupiter.

Answer 3

3.7 sec? @AllTehMaffs

Answer 4

yup yup ^_^

Answer 5

@Partycool nice!