How do I use the number of oscillations given?

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The discussion revolves around determining the accuracy of the gravitational constant (g) using a simple pendulum's oscillation data. The pendulum's length is 100 cm, with a period of 2 seconds measured over 100 oscillations using a clock with a 0.1-second resolution. The key confusion lies in how to apply the number of oscillations to calculate the period accurately, which is clarified by recognizing that the total time for 100 oscillations affects the uncertainty in the period. The correct approach involves dividing the uncertainty in the total time by the number of oscillations to find the uncertainty in the period. Ultimately, the accurate percentage error in g is determined to be 0.2%, aligning with the answer provided in the textbook.
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Homework Statement


The length of a simple pendulum is about 100 cm known to have an accuracy of 1 mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g?
(a) 0.2% (b) 0.5%
(c) 0.1% (d) 2%

Homework Equations



3. The Attempt at a Solution [/B]
→Now, l=100 cm=1 m, Δl=1 mm= 10-3m, t=2 s Δt=0.1 s
Then, T=2π✓(l/g)
Therefore, g=4π2l/T2
Hence, max error is
Δg/g = Δl/l + 2(Δt/t)
i.e., max percentage error is
(Δg/g)100 = (Δl/l)100 + 2(Δt/t)100
=10.1%

Solving this, I'm getting 10.1%, which is obviously not matching with any of the choices.
Where am I going wrong?
How do I use the number of oscillations given?
 
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Hi HPPAS,

Welcome to Physics Forums!

HPPAS said:
How do i use the number of oscillations given?
Consider how the pendulum period was determined. How was the clock measurement used to determine the value of T?
 
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gneill said:
Hi HPPAS,

Welcome to Physics Forums!Consider how the pendulum period was determined. How was the clock measurement used to determine the value of T?
I have considered the resolution of the clock, but I can't figure out how to use the number of oscillations
 
HPPAS said:
I have considered the resolution of the clock, but I can't figure out how to use the number of oscillations
What procedure was used to find the period of the pendulum? If you were to replicate it, what would be your steps?
 
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Oh so are you saying that I have to consider the time as 200 s (period for 100 oscillations) ?
 
HPPAS said:
Oh so are you saying that I have to consider the time as 200 s (period for 100 oscillations) ?
No, not quite.

Suppose you were given the task of determining the period of some pendulum. To begin with you only know that the period is close to 2 seconds, but you want an accurate value. You have available a timer with a resolution of 0.1 seconds. How would you go about it?
 
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gneill said:
No, not quite.

Suppose you were given the task of determining the period of some pendulum. To begin with you only know that the period is close to 2 seconds, but you want an accurate value. You have available a timer with a resolution of 0.1 seconds. How would you go about it?
No, I'm not finding anything else than what I've already done
 
From the problem statement:
HPPAS said:
Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution.

Can you write out the calculation performed to find the period?
 
HPPAS said:
No, I'm not finding anything else than what I've already done
To make it a bit more explicit, the piece you are missing is to relate the uncertainty due to the clock's precision to the uncertainty of the calculated period, T. That relationship is tied to what was actually measured and how the value of T was obtained from it.
 
  • #10
gneill said:
To make it a bit more explicit, the piece you are missing is to relate the uncertainty due to the clock's precision to the uncertainty of the calculated period, T. That relationship is tied to what was actually measured and how the value of T was obtained from it.
Actually, contrary to what you said, I got the answer (0.2%) on substituting t=200 seconds. That is the answer given at the back of the book. But I still don't get how else can the uncertainty due to the clock's precision be related to the uncertainty in time period. The uncertainty is 0.1 second, so the actual period might range from 1.9-2.1, right?
 
  • #11
All right, wait, I'm getting something.
The resolution is 0.1 second but the given period is 2 sec, not 2.0 sec. So, is the uncertainty 1 sec?
 
  • #12
HPPAS said:
Actually, contrary to what you said, I got the answer (0.2%) on substituting t=200 seconds. That is the answer given at the back of the book. But I still don't get how else can the uncertainty due to the clock's precision be related to the uncertainty in time period. The uncertainty is 0.1 second, so the actual period might range from 1.9-2.1, right?
The only measurement performed was for the total time it took for 100 oscillations. The measurement uncertainty applies to that value. But you don't know what that measured value was, you're only given the result of the calculation (2 s). Likely that 2 s value is the result of rounding appropriate to the measurement uncertainty and the calculation performed that transformed the measured value to the period.

So how do you determine what uncertainty should be associated with the 2 s? Suppose the measured time for the 100 oscillations is t. What calculation do you perform on t to arrive at T, the period? What's the uncertainty calculation associated with it?
 
  • #13
gneill said:
The only measurement performed was for the total time it took for 100 oscillations. The measurement uncertainty applies to that value. But you don't know what that measured value was, you're only given the result of the calculation (2 s). Likely that 2 s value is the result of rounding appropriate to the measurement uncertainty and the calculation performed that transformed the measured value to the period.

So how do you determine what uncertainty should be associated with the 2 s? Suppose the measured time for the 100 oscillations is t. What calculation do you perform on t to arrive at T, the period? What's the uncertainty calculation associated with it?
No idea.
By the way, can you explain why I got the right answer (0.2%, given at the back) after substituting T=200 seconds (period for 100 oscillations)?
 
  • #14
HPPAS said:
No idea.
You have, but you just don't realize it :smile:
By the way, can you explain why I got the right answer (0.2%, given at the back) after substituting T=200 seconds (period for 100 oscillations)?
Yes. Because you unwittingly applied the required expression to convert the uncertainty in the measurement of the 100 oscillations to the 2 s derived period.

You have a measurement t with uncertainty δt. You then calculated the period by writing T = t/100. What's the uncertainty in T?
 
  • #15
gneill said:
You have a measurement t with uncertainty δt. You then calculated the period by writing T = t/100. What's the uncertainty in T?
δΤ=δt
 
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  • #16
HPPAS said:
δΤ=δt
No. t is being divided by a constant (100). How does that affect the uncertainty in the result?
 
  • #17
gneill said:
No. t is being divided by a constant (100). How does that affect the uncertainty in the result?
Oh I thought 100 was just a constant.
Then is it δt/100 ?
 
  • #18
HPPAS said:
Oh I thought was just a constant.
Then is it δt/100 ?
Yes!
 
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  • #19
gneill said:
Yes!
Thanks :)
 

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