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What is the magnitude of gravity of a pendulum that is moved

  1. Oct 23, 2016 #1
    << Mentor Note -- OP correctly re-posted schoolwork question in the HH forums; threads merged >>

    The full question is: The pendulum inside a grandfather clock has a half period of 1.0000s at a location where the magnitude of the local acceleration of gravity is 9.800 m/s^2. The clock carefully is moved to another location at the same temperature and is found to run slow by 89.0s per day. What is the magnitude of the local acceleration due to gravity at the new location?

    What I have so far:
    Delta T = dT/dg * delta g
    -(T/2g) * delta g = (dT/dg) * delta g
    -(T/2g) = dT/dg

    From here I am not sure what to do. Is a half period just T/2? Where should I go from here?
     
    Last edited by a moderator: Oct 23, 2016
  2. jcsd
  3. Oct 23, 2016 #2
    The full question is: The pendulum inside a grandfather clock has a half period of 1.0000s at a location where the magnitude of the local acceleration of gravity is 9.800 m/s^2. The clock carefully is moved to another location at the same temperature and is found to run slow by 89.0s per day. What is the magnitude of the local acceleration due to gravity at the new location?

    What I have so far:
    Delta T = dT/dg * delta g
    -(T/2g) * delta g = (dT/dg) * delta g
    -(T/2g) = dT/dg

    From here I am not sure what to do. Is a half period just T/2? Where should I go from here?
     
  4. Oct 23, 2016 #3
    What is the period at the new location?
    Yes, half is 1/2.
     
  5. Oct 23, 2016 #4

    haruspex

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    Yes.
    I assume you mean -T/(2g). Keep the right hand side as ΔT/Δg. What do you know about ΔT/T?
     
  6. Oct 23, 2016 #5
    Period depends on gravity because of different locations on Earth have different magnitudes of acceleration, correct? I thought that was the point to this problem. So as you change locations, acceleration due to gravity is different thus the period changes.
     
  7. Oct 23, 2016 #6

    kuruman

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    Sorry, I deleted my response because @haruspex is on it. Yes, that is the point of the problem. By "how" I meant what is the dependence of the g on the period. Anyway, you are in good hands.
     
  8. Oct 23, 2016 #7

    haruspex

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    From the equations in the OP, it looks like Brandon has dealt with that part. It remains to apply the last equation to the given data.
     
  9. Oct 23, 2016 #8
    Alright awesome, so what am I solve for delta g?
     
  10. Oct 23, 2016 #9

    haruspex

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    Answer my question at the end of post #4.
     
  11. Oct 23, 2016 #10
    ΔT/T is just the initial T?
     
  12. Oct 23, 2016 #11

    haruspex

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    No, T is the initial period, ΔT is the increase when the clock was moved. How does ΔT/T relate to the given information that it is...
     
  13. Oct 24, 2016 #12
    I cant figure out what deltaT/T relates to...
     
  14. Oct 24, 2016 #13

    haruspex

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    It means the fractional change in the period.
     
  15. Oct 24, 2016 #14
    But how does that relate to us trying to find the acceleration due to gravity in this problem? Does deltaT/T=deltag? 89.0s/2, 2 being the full period since the half period is 1.0000s, gives us 44.5s, does that mean 44.5is our deltag?
     
  16. Oct 24, 2016 #15

    haruspex

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    No, that would make no sense dimensionally.
    Rewrite Brandon's last equation in the original post as -T/(2g)=ΔT/Δg then rearrange it to find ΔT/T.
     
  17. Oct 24, 2016 #16
    I rearange the problem to get ΔT/T=Δg/2g, but i get a negative on the intitial period below the ΔT.
     
  18. Oct 24, 2016 #17

    haruspex

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    You mean, you get ΔT/(-T)? That's ok, but there is a more usual way of writing it. What is 2/(-1)?
     
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