Question

What is the period, in seconds, of a simple pendulum of length 2 meters? Use the gravitational constant g = 9.8 m/s2 and round your answer to two decimal places.

Answer 1

@omnomnom

Answer 2

hm... why am i tagged i am of no help to you im so so sorry :(

Answer 3

aww okay

Answer 4

sorry :D

Answer 5

A physics formula sheet would come in handy right about now . . .

Answer 6

I think it's sqrt(L/g) though. . .

Answer 7

The units check out anyway
\[\sqrt{\frac{meters}{meters/seconds^2}} = \sqrt{seconds^2} = seconds\]

Answer 8

here is the equation

Answer 9

\[d(t) = acos(\omega t) \]

Answer 10

its the harmonic motion equation

Answer 11

a period is \[(2\pi)/\omega \]

Answer 12

Yes, but that is in terms of the amplitude of the swing and the frequency, which are irrelevant.

Answer 13

The only things that affect the period of a pendulum are its length and the gravitational field strength. You could probably derive the period from the wave equation, but there are too many unknowns.

Answer 14

I am not like this question lol

Answer 15

liking* lol

Answer 16

After finding the period, you could get the frequency and then determine displacement at any time, t for a particular amplitude if you wanted to, but that wasn't your original question.

If you just want the period, it's T=sqrt(L/g) like I said earlier.

Answer 17

i haven't ever used that equation before tbh... can you perhaps help me?

Answer 18

Are you actually required to derive the period from the harmonic motion equation, or did you just grab that as a possible formula to use?

Answer 19

thats what equation that goes with particular study
so i figured to use that ._.

Answer 20

Hmm, I could show you how to do it that way (it's a pain in the neck . . .), but first I want to explain the formula I gave.

Answer 21

Let us avoid pain in thy necks? :3

Answer 22

Here's how a physicist would think about this problem.
The information given is
1. sinmple pendulum
2. length of pendulum (in meters)
3. gravitational acceleration (in meters per second per second)

The information asked for is
1. period (in seconds)

So, how can we combine the given information (meters, and meters-per-second-squared) to get an answer in seconds?

Answer 23

with that equation i understand the basics of the equation it self i think ...meters over meters cancels ..and considering seconds is squared the square root will take care of it leaving the answer in seconds right?...its just solving it for seconds

Answer 24

Exactly. If you didn't happen to have the formula in front of you (I didn't), you'd have to do what I did. I thought about what sequence of arithmetic operations would transform m and m/s^2 into s.

And the process is just how you described. One has to divide m by m to cancel that, then square-root the square seconds.

The formula then gives itself as sqrt(L/g).

Answer 25

Plug in your given information and let me know what you get.

Answer 26

um so sqrt(2/9.8)?

Answer 27

Yep.

Answer 28

so that would be .45?

Answer 29

i tried 2/9.8 and to the root ..and got it wrong i guess i set it up wrong..it was 2. something anyways thnx

Answer 30

0.45 seconds is the period rounded to 2 places.

What do you mean by 2.something? What was 2.something?

Answer 31

Oh, crap, I forgot to multiply by 2π to convert the units.
Sorry about that.
Never mind, yes, the period should be 2.84