Simple pendulum acceleration

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  • #1
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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?
 

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  • #2
mgb_phys
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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?
 
  • #3
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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?
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:
  • #4
mgb_phys
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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'
 
  • #5
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thank you for your help, i been confuse for whiles.
 
  • #6
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What if you square a quantity between -1 and 1? Doesn't the error decrease?
 
  • #7
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damn! i cannot understand it, what does between -1 and 1 mean
 

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