The period of a pendulum oscillating on the surface of the planet Ceti-Alpha-5 is 1.4 times as...

Question:

The period of a pendulum oscillating on the surface of the planet Ceti-Alpha-5 is 1.4 times as great as its period on the surface of the earth. If the mass of Ceti-Alpha-5 is 0.6 times that of the earth, what is its radius?

Simple Pendulum

We have to use the concept and equation of time period of simple pendulum to solve this question. We know that the time period of pendulum depends upon the gravity of the planet. And from the gravitational law we know that the acceleration due to gravity depends upon the mass and radius of the planet. Therefore by using time period and acceleration due to gravity equation we can solve this question.

Given:

Time period on planet {eq}\displaystyle T_{C} = 1.4 T_{e} {/eq}

Mass of the ceti Alpha 5 {eq}\displaystyle M_{C} = 0.6 M_{e} {/eq}

Now we know that the time period of pendulum is given by

{eq}\displaystyle T_{e} = 2\pi \sqrt{\frac{L}{g}} {/eq}

where g is acceleration due to gravity on earth

Now comparing the time period on the given planet , we get

{eq}\displaystyle g_{C} = \frac{g_{e}}{1.4^{2}} \\ {/eq}

Now we know that the acceleration due to gravity is given by

{eq}\displaystyle g = \frac{GM}{R^{2}} {/eq}

Now putting the value in above equation we get

{eq}\displaystyle \frac{GM_{c}}{R_{C}^{2}} = \frac{GM_{e}}{R_{e}^{2}1.4^{2}} \\ \frac{0.6M_{e}}{R_{c}^{2}} = \frac{M_{e}}{R_{e}^{2}1.4^{2}} \\ R_{c} = \sqrt{0.6}R_{e}*1.4 \\ R_{c} = \sqrt{0.6}*6378.1*10^{3}*1.4 = 6916.637 \ km {/eq}

Pendulums in Physics: Definition & Equations

from

Chapter 11 / Lesson 6
30K

After watching this lesson, you will be able to explain what a pendulum is, why it is an example of simple harmonic motion, and use equations to solve pendulum problems. A short quiz will follow.