Easy: given theta, uncertainity on cos(theta)

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Homework Help Overview

The discussion revolves around calculating the uncertainty in the cosine of an angle, specifically given an angle of 5 degrees with an uncertainty of 0.1 degrees. Additionally, there is a consideration of uncertainty in timing measurements related to a pendulum's oscillation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the propagation of error for the cosine function and explore the implications of timing uncertainties in pendulum measurements. Questions arise regarding the nature and sources of uncertainty in timing, particularly in relation to human reaction time.

Discussion Status

Participants are actively engaging with the problem, providing insights into the calculation of uncertainty in cosine and questioning the assumptions around timing uncertainties. There is no explicit consensus on the timing uncertainty, with various perspectives being explored.

Contextual Notes

Participants mention the subjective nature of timing errors and the influence of external factors, such as physical state and measurement techniques, on the uncertainty of time measurements.

thegame
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Hi,

I need to find the uncertainty on cos(theta) given:
theta = 5 plus/minus 0.1 degrees

what is the uncertainty on cos(theta)? [propagation of error]

2)

Suppose I measured the time taken by the pendulum to complete one oscillation. So the pendulum moves downward and I start the stopwatch. Then the pendulum completes one oscillation and I stop the stopwatch.

Is the ucertainty 0? Let's say my reaction time is 0.17 sec. So, when I see the pendulum move down, I start the clock 0.17 seconds late (which is the time taken by the brain to intrepret the signal). And at the end, I stop the pendulum 0.17 seconds later after the event has occurred...
 
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[tex]\Delta \cos\theta =\cos(\theta +\Delta\theta)-cos\theta=(\cos\theta) (\cos\Delta\theta)-(\sin\theta)(\sin\Delta\theta)-\cos\theta.[/tex]

Now,use approximations:
[tex]\sin\Delta\theta=\Delta\theta[/tex]
[tex]\cos\Delta\theta=1[/tex]

Then the uncertainty becomes:
[tex]|\Delta\cos\theta|=(\sin\theta)(\Delta\theta)[/tex]

As for the second part,what do u mean of uncertainty 0...??
 
dextercioby said:
As for the second part,what do u mean of uncertainty 0...??

Thanks for the quick reply.. By 0, I mean that there is no uncertainty on time
 
thegame said:
Thanks for the quick reply.. By 0, I mean that there is no uncertainty on time

Why wouldn't it be??Just because you estimated the time of the reflex to be be the same,that doesn't mean that your hand muscles will behave the same.Hell,if that period of oscillation is big,u might get drunk inbetween and your hand could be tremblin' like s***. :-p Then the uncertainties will depend on the amount of alcohol in your blood. :-p

Daniel.

PS.This kind of systematic and subjective errors is a bit tricky.It depends on the human body which can malfunction in between condecutive identical experiments.
 
It's worth mentioning that your brain is anticipating when the bob passes the reference point, so your actual uncertainty will most likely actually be less than your raw reaction time. I'd probably guess around .15 or .2 seconds of uncertainty, although .3s would probably be pretty safe.

Additionally, if you measure 10 or so periods at a time, your uncertainty would be less than [itex]\sigma / 10[/itex].

Also, you might try simply to repeat the measurement a large number of times and use the unadjusted standard deviation of that set of data for the uncertainty.

I'd be a little concerned with how the period is being measured, since the end up the upswing isn't a very distinguished point as far as the human eye is concerned. How well you can tell when the bob passes the reference point can add some uncertainty, especially if you're using points like at the end of the upswing.

I suppose the moral of the story is that the uncertainty on time depends on a lot of factors, not just your raw reaction time.

--J
 
Thanks Justin, dextor.
 

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