# Standard deviation question.

## Homework Statement

Sally, Bob and Charlie each measure the period of the same pendulum to determine the
acceleration of gravity, g. The lab instructions say that you should determine the period by
timing the time of 100 swings (complete cycles) of the pendulum. Sally is the first to do the
experiment and she times 100 swings of the pendulum. Bob does the experiment next and
decides to make two such measurements of 100 swings each and averages the values to get a
better result while Charlie decides to make 10 sets of measurements and average them. The
final data set consists of three times (TS, TB, and TC) for 100 swings. Assume that the dominant
uncertainty in timing the 100 swings is random and that all three students have the same
reaction time.
If Sally obtains a standard deviation $$\sigma$$S, what are the standard deviations calculated by
Bob and Charlie ($$\sigma$$B and $$\sigma$$C) expressed in units or multiples of $$\sigma$$S?

## Homework Equations

$$\sigma$$ = sqrt[(sum(x-xavg)2)/(N-1)]
where N is number of trials, x is each measured value and xavg is the mean of the measured values.

## The Attempt at a Solution

I'm not quite sure where to start with this because I don't know how sally could have gotten a standard deviation with only one measurement.
I feel like this should be pretty simple, but I must be overlooking something easy or misreading the question. I would like to figure this out on my own, but I cant seem to even get out of the batters box, so if maybe if someone could just get me started, I think i could finish on my own.

Thanks.

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ehild
Homework Helper
Bob gets two data for the time of 100 swings t1 and t2, with standard deviations $$\sigma_S_ {t1}[/itex] [tex]\sigma_S_ {t2}[/itex] both supposedly equal to Sally's [tex]\sigma_S [/itex] and calculates both the average time and the standard deviation of the final value TB. The final time is the average of t1 and t2. TB=0.5*(t1+t2). Then he uses this formula to get the standard variation of TB: This is [tex]\sqrt {(\frac{\partial TB}{\partial t1})^2*\sigma_{t1}^2+(\frac{\partial TB}{\partial t1})^2*\sigma_{t1}^2}=\sqrt{2(0.5)^2\sigma_S^2}=\sigma_S/\sqrt 2$$

Charlie gets his standard derivation with the same method, but calculates the average from three measurements.

ehild

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