Simple pendulum, det the change of its period?

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Homework Help Overview

The discussion revolves around a simple pendulum with a specified length, examining how its period changes when moving from one gravitational field strength to another. The context involves understanding the relationship between the pendulum's period and the acceleration due to gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the formula for the period of a pendulum and question whether the change in period can be simply calculated by finding the difference between the periods at two different gravitational accelerations. Some participants express uncertainty about their calculations and the implications of small changes in gravity.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the logic behind their results. There is a recognition of the small change in period due to the slight difference in gravitational acceleration, and some participants are validating their answers against each other.

Contextual Notes

Participants note the importance of significant figures in their calculations and express varying levels of confidence in their results. There is an acknowledgment of the ideal conditions assumed for the pendulum's behavior.

hemetite
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Qn. A simple pendulum has a length of 3.00m. Det the change in it period if it is taken from a point where g=9.80ms2 to a higher elevation where the acceleration due to gravity decreases to 9.79ms2.

i cannot seem to start the question and a bit unsure here...

here is what i get...

Formula to use:
T = 2 pie sqrt L/g

i dun think the answer is just (T=2pie sqrt L/9.80) - (T=2pie sqrt L/9.70) right?

What i know, is that when the g start to reduce is at the point when it is starting going in the opposite direction, away from the equilibirium.(lowest point)

Please help to get me..i am stuck here...
 
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Why do you think it isn't just the difference of the two periods? I think it is.
 
isnt a the period pendulum is the same assuming that there isn't any damping?
 
The period of the ideal pendulum just depends on L and g. In the real world, sure, it also depends on damping and the amplitude of the oscillation. But this looks like a pretty straightforward question. Just give the easy answer.
 
Thanks for you opinion...i also think that it is a straightforward answer..but no confidence...anyway...just try...to err is a learning process..

thank you again..
 
i did the equations...working.

Change of period
= (T=2pie sqrt 3/9.80) - (T=2pie sqrt 3/9.79)
= 3.4768 - 3.4786
= - 0.0018s

is the answer logical?..
 
Well, g only changes by a fraction of a percent, so the change in the period is small. And the period should go up as g goes down. So I'd say it's logical. But is it correct? You are subtracting two numbers of similar magnitude, so if you need to, you should worry about how many significant figures are correct. Especially since I don't seem to get exactly the same numbers as you. What are you using for pi?
 
3.142
 
hemetite said:
3.142

That accounts for the discrepancy.
 
  • #10
the answer is i got after re-correction = -0.001777s
 
  • #11
I get -0.001775...s, but I'll call that good enough.
 
  • #12
thanks...see how it goes...when i submit the answers..
 

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