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Homework Help: Simple pendulum acceleration

  1. Jan 13, 2009 #1
    For a simple pendulum practice with clamp, boss head and resort stand assembled, also a string attached to the end of clamp and a bob, then swings it and measure the time for it to complete a period.

    At the end use the information collected to calculate acceleration due to the gravity by the formula g = 4π2l/T2.

    What could the accuracy and reliability be effected? (Equipments, angles or the error should avoid) And what kind of string or bob will most suit? Why?
  2. jcsd
  3. Jan 13, 2009 #2


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    Welcome to PF.
    For HW questions you have to make some sort of attempt before we can help.

    To get you started, if a quantity is squared would that make any error in that quantity worse or better?
  4. Jan 14, 2009 #3
    i think that will make it worse. so i measured the period for it complete ten oscillations and repeat it with different length of strings, due that both of them are the only variables in the equation. Even L and T are the only variables, but the angle should some how involved, would it?

    and also for the string and 'bob', only thing i can think of was the air resistance, anything else that i should think about?

    sorry for didn't show enough effort, i thought that could make the question simple. i just something else that can increase the accuracy and reliability in that practise. just looking for some hint
    Last edited: Jan 14, 2009
  5. Jan 14, 2009 #4


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    There are two types of error to consider .

    1, You are trying to use an ideal law for the period - no friction, all the weight acts at L etc.
    So you want your experiment to be as close to the imaginary ideal as possible and reduce effects like.
    Air friction on the string (do you want a big swing or small ?)
    All the mass to be at distance L (so the string must be much lighter than the bob)
    No friction where the string pivots.

    2, There is also propagation of errors - which is what I hinted at.
    The answer is proportional to L so an error in L will give the same percentage error in 'g'
    But 't' is squared so an error in 't'gives a bigger error in 'g' - so you have to put more effort into measuring 't' accurately than you do into measuring 'L'
  6. Jan 15, 2009 #5
    thank you for your help, i been confuse for whiles.
  7. Jan 15, 2009 #6
    What if you square a quantity between -1 and 1? Doesn't the error decrease?
  8. Jan 15, 2009 #7
    damn! i cannot understand it, what does between -1 and 1 mean
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