Question regarding Pendulum on Fictional Planet

Question

This one is a long as its all related. so confusing ...

1. What would the time for 20 oscillations of a 1.0m long pendulum be on Planet X?

Given :

L = 1.0m

Acceleration due to gravity on Planet X = 6.91m/s²

T=?

How do I solve this and which formula do I use?

1. Contrast the time for 20 oscillations on Planet X and that of a small Moon of Plant X?

1. Contrast the time for 20 oscillations on Moon A , which is twice as big as Moon B , but with the same radius?

The most confusing one here:

1. Sketch two lines on a graph to compare the results the astronomers would obtain for the relationship between the square of the period and the length of the pendulum on Moon A and Moon B.

This was very long so thank you in advance for taking the time to help me!

Answer 1

Hi, Tyler!

Let's do one step at a time. You a little bit confused me, because I prefer to see something like this: question No. 1, question No. 2 and so on. Anyway...

Time of N = 20 oscillations is t = N x T , where T - period for 1 oscillation. Now we use straightforward the formula

T = 2π√ L / g

You know everything and easily find time of 20 oscillations, which is

t = 2π N√ L / g

Now I believe we have a different problem or another part of the problem.

From the formula for the period of oscillations of a pendulum after squaring it, we can get

g = 4π² L / T² = 4π² / (T² / L)

We can plot the graph T² vs L and find its slope; it will be T² / L (tangent or gradient of the line). So I can write that

g = 4π² / Slope

This is how you can find the acceleration due to gravity of any body, where gravitation exists.

Now about the acceleration due to gravity on moons A and B.

Acceleration due to gravity on any body with mass M and Radius R could be found with the use of the formula

g = G M / R²,

where G = 6.67 x 10 ^{- 11} Nm²/kg² - universal gravitational constant.

" twice as big, but with the same radius" - here you confused me. When we say "big" we talk about size.

Hope that you wanted to say that the mass was as twice as big. Then it makes sense.

If moon A has double mass , but the same radius as moon B, then acceleration due to gravity on it, g, will be as twice as big then on moon B (they have the same radii but one mass is doubled).

It means that for the moon with greater mass the slope of the graph T² vs L will be steeper. Tangent of the graph will be 2 times greater for moon A compare to the moon B. That is it!

Hope that I help you to clarify the situation. Free to ask in comments any additional questions. Good luck!