1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple pendulum swing problem

  1. Apr 1, 2004 #1
    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?
     
  2. jcsd
  3. Apr 1, 2004 #2
    Post your work. Your approach is correct.

    cookiemonster
     
  4. Apr 1, 2004 #3
    so g = (f*2pi)^2 * L

    f = 1/period = 1.453 Hz

    which gives me g= 39.2 and this is wrong . :confused:
     
  5. Apr 1, 2004 #4
    Check your algebra. I'm getting a different answer.

    cookiemonster
     
  6. Apr 1, 2004 #5
    i have checked the algebra and i keep on getting 39.2 m/s^2 as the gravity, is my formula right???
     
  7. Apr 1, 2004 #6
    T = 2pi*Sqrt[L/g]

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

    Yes?

    cookiemonster
     
  8. Apr 1, 2004 #7
    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.
     
  9. Apr 1, 2004 #8
    Eh, heh, heh... I suck at typing.

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

    cookiemonster
     
  10. Apr 1, 2004 #9
    which gives 39.2 for g. what other number u get that u said was different??
     
  11. Apr 1, 2004 #10
    4pi^2*(.47m)/(141s/97)^2 = ?

    cookiemonster
     
  12. Apr 2, 2004 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  13. Apr 2, 2004 #12
    (And the period is one over frequency.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Simple pendulum swing problem
  1. Simple Pendulum Problem (Replies: 10)

Loading...