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Oscillation question...

  1. Sep 19, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    To increase the period of oscillation of a pendulum in 1 second, it is needed to increase the length of it in 2 meters. Calculate, in seconds, of the initial period of oscillation of the pendulum.

    I found this question online a few minutes ago. I have not learned this in physics class yet so bare with me :)
    Because it is asking for the period of the oscillation I figured I would need the equation: T = 2π√(l/g)

    To set up the equation, I wrote it like this

    2π√(l/g) + 1 = 2π√(l+2/g)

    as the period of the oscillation increased by one when the length increased by 2 meters. When I solve this equation though, everything except cancels itself out, so I can't find the initial length to plug back into the equation to find the initial period... Again, I am 1 month in of my first year of physics so don't crucify me for my mistakes please (lol), and the wording on this question is quite confusing too, so I may have it all wrong but hopefully you can see where I'm going with this ;)

    Thanks in advance!
     
  2. jcsd
  3. Sep 19, 2016 #2

    Charles Link

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    I think you set it up correctly. Try squaring both sides, cancelling the terms that cancel, and then grouping terms and squaring once more. If you are careful with the algebra, you should get an answer for "l". (You do need a parentheses around your (l+2).)
     
  4. Sep 19, 2016 #3
    I only put the parentheses around the l/g to show that it was included in the square root. Anyway, you can take a look at my work:
    0cfc21fbc2e419d27f0295d07110d6f3.jpg
    Unless I messed it up, the lengths do in fact cancel each other out. :/
     
  5. Sep 19, 2016 #4

    Charles Link

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    How about the 2ab term when you square the left side? ## (a+b)^2=a^2+2ab+b^2 ##.
     
  6. Sep 19, 2016 #5
    ohhhhh.... I completely forgot. So now when I solve the equation, I get about 3.09047 for the initial length. When I put this back into the equation T = 2π√(l/g) I got about 3.525678 seconds. Im not sure if it is correct yet but thanks a lot for the help!
     
  7. Sep 19, 2016 #6

    Charles Link

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    I didn't compute an exact answer either, but that's approximately what I got. Good work !!!
     
  8. Sep 20, 2016 #7

    CWatters

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    "in 1 second" = "by 1 second" or "to 1 second"?
     
  9. Sep 20, 2016 #8
    The person who asked this problem meant "by." I just copied the problem down verbatim in case that wasn't the case.
     
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