Calculating g on an Unfamiliar Planet Using a Simple Pendulum

[lollypop]

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 47.0 cm. She finds that the pendulum makes 97.0 complete swings in a time of 141 s .
What is the value of g on this planet?

a complete swing is a cycle? so the period would be 97/ 141 = 0.688??is this right?
i was thinking of using the formula frequency = (1/2pi)*sqrt(g/ L), and get g from there. I tried using the period i said before but it is not right. What am i doing wrong?

 

[cookiemonster]

Post your work. Your approach is correct.

cookiemonster

 

[lollypop]

so g = (f*2pi)^2 * L

f = 1/period = 1.453 Hz

which gives me g= 39.2 and this is wrong .

 

[cookiemonster]

Check your algebra. I'm getting a different answer.

cookiemonster

 

[lollypop]

i have checked the algebra and i keep on getting 39.2 m/s^2 as the gravity, is my formula right?

 

[cookiemonster]

T = 2pi*Sqrt[L/g]

g = L*T^2/(4pi^2)

Yes?

cookiemonster

 

[lollypop]

by your formula u get 0.0082? how did u get T^2 on the numerator, when I solve for g using your formula I get (L*4pi^2)/(T^2) which i think is right.

 

[cookiemonster]

Eh, heh, heh... I suck at typing.

g = 4pi^2*L/T^2

cookiemonster

 

[lollypop]

which gives 39.2 for g. what other number u get that u said was different??

 

[cookiemonster]

4pi^2*(.47m)/(141s/97)^2 = ?

cookiemonster

 

[HallsofIvy]

You have an error back in your first post:

"a complete swing is a cycle? so the period would be 97/ 141 = 0.688??is this right? "

No, it's not right. It would be a good idea to carry the units along with your calculation. The problem tells you that the pendulum "makes 97.0 complete swings in a time of 141 s ." so that 97/141 would be 97/141 swings per second which is not "period". It is, in fact, the frequency in cycles per second.

The frequency is 0.688 cycles per second, not the period.

 

[Chen]

(And the period is one over frequency.)