Simple pendulum problem - having trouble with the math

[Confused20]

Simple pendulum problem - having trouble with the math!

Homework Statement

A simple pendulum is set up on Mars where g=3.72m/s^2. How long should pendulum be if period T is to be 2.00 seconds?

Homework Equations

T=2pia(3.14) times square root of l(length)/gravity

The Attempt at a Solution

I placed in 2 seconds for T and divided it by 2pia(3.14). Then I squared this number and then multiplied it by 3.72 to solve for l(length). The correct answer is supposed to be .965 but I did not get this answer. Can anyone please help. I have an exam in 2 days! Any input will be greatly appreciated.

 

[lewando]

So, what did you come up with? I think the "correct answer" is wrong.

 

[Confused20]

Thank you for your response. The answer I got was .377 meters. My professor stated in class that the answer was .965. Is this the answer that you got as well? Is the manner in which I approached the algebra correct?

 

[lewando]

Yes, that's what I got. Your algebra looks immaculate! As a check, you can always plug your result back into the equation and see it you get the right T. Good luck with your test (and welcome to PF!)

 

[Confused20]

Thank you very much for confirming that I did the algebra correctly. My professor also suggested that we do a variation of the problem where length=.965m and T=2.00 and then attempt to solve for gravity (answer should be 3.72 m/s^2). Can you assist me in solving this problem. We are of course using the same equation of T=2pia(3.14) times the square root of length over gravity. I began to attempt to solve this problem in a similar manner by dividing 2.00 by 2pia(3.14) and then squaring that number. The next step is puzzling for me because of my lack of algebra. I decided to divide that number by .965. Can you please assist me in this last step. The answer I get is .105 m/s^2. Any help on how to solve this problem will be greatly appreciated.

 

[lewando]

Notice in this equation that if you increase l (say, from 0.377 to .965), then g must also increase for T to remain constant.

[tex]T = 2\pi\sqrt{\frac{l}{g}}[/tex]

Here is what you did:

[tex]\frac{T}{2\pi} = \sqrt{\frac{l}{g}}[/tex]

[tex]\frac{T^2}{4\pi^2} = \frac{l}{g}[/tex]

[tex]\frac{T^2}{4l\pi^2} = \frac{1}{g}[/tex]

I think you solved for 1/g. Can you think of what to do to bring up g to the numerator?

 

[SteamKing]

If you use g = 9.81 m/s^2 in the formula for the period of a pendulum, then the length of 0.965 m gives a period of 2 s. An l = 0.377 m and a g = 3.72 m/s^2 also gives a period of 2 s. Your 'professor' owes you an apology and he is also due a head slap for not checking his math.

 

[dinyatta32]

Im also having problems with this simple pendulum equation if you have a 1m long pendulum with a 2kg mass is pulled back 0.1 m and then released so that it swings back and forth predict what the horizontal position x will look like as a function of time i can't figure out how this equation goes i have been on the same thing for 2 days I am stuck can someone help me please