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

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Homework Help Overview

The discussion revolves around a physics problem involving a pendulum in a grandfather clock, specifically focusing on how the local acceleration due to gravity affects the clock's period when moved to a different location. The original poster presents the problem of determining the new magnitude of gravity after observing a time discrepancy in the clock's operation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the period of a pendulum and the acceleration due to gravity, questioning how changes in gravity affect the clock's timing. There are discussions about the definitions of half period and period, as well as the implications of the time discrepancy observed.

Discussion Status

Participants are actively engaging with the problem, clarifying terms and relationships between variables. Some guidance has been provided regarding the equations involved, and there is an ongoing exploration of how to relate the observed time change to the change in gravity.

Contextual Notes

There is a focus on the specific values given in the problem, including the half period of the pendulum and the time the clock runs slow after being moved. Participants are also considering the dimensional consistency of the equations they are working with.

BrandonSW
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<< 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?
 
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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?
 
BrandonSW said:
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?
What is the period at the new location?
Yes, half is 1/2.
 
BrandonSW said:
Is a half period just T/2?
Yes.
BrandonSW said:
-(T/2g) = dT/dg
I assume you mean -T/(2g). Keep the right hand side as ΔT/Δg. What do you know about ΔT/T?
 
kuruman said:
Can you show us where you got this? How does the period depend on g?
What's the difference in your notation between "Delta T" and dT?
Yes, half a period is T/2.

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.
 
BrandonSW said:
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.
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.
 
kuruman said:
what is the dependence of the g on the period
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.
 
haruspex said:
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.
Alright awesome, so what am I solve for delta g?
 
BrandonSW said:
Alright awesome, so what am I solve for delta g?
Answer my question at the end of post #4.
 
  • #10
haruspex said:
Answer my question at the end of post #4.
ΔT/T is just the initial T?
 
  • #11
BrandonSW said:
ΔT/T is just the initial T?
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...
BrandonSW said:
found to run slow by 89.0s per day
 
  • #12
I can't figure out what deltaT/T relates to...
 
  • #13
huffy said:
I can't figure out what deltaT/T relates to...
It means the fractional change in the period.
 
  • #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?
 
  • #15
huffy said:
Does deltaT/T=deltag?
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.
 
  • #16
I rearange the problem to get ΔT/T=Δg/2g, but i get a negative on the intitial period below the ΔT.
 
  • #17
huffy said:
I rearange the problem to get ΔT/T=Δg/2g, but i get a negative on the intitial period below the ΔT.
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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