anonymous
  • anonymous
Consider measuring the period of a pendulum with a stopwatch. Suppose that the stopwatch is running slow. Which one of the following statements is correct?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
A)This will vary our results in both the positive and negative direction, allowing for a varied estimation. We would need to average these errors to obtain our true systematic error B)This will lead to overestimation of most our time results. Systematic errors, unlike random errors, shift the results always in the positive direction. C)This will not change our estimation because all of our time results will be nearly the same. We would just need to consider random error, not systematic error. D)This will lead to underestimation of all our time results. Systematic errors, unlike random errors, shift the results in one direction.
anonymous
  • anonymous
I believe its C am I right
theEric
  • theEric
I think that the period will not be measured correctly. This error isn't one that will cancel, I think. If the stopwatch is slower than actual time, that the period would be measured to be smaller than it actually is.

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theEric
  • theEric
Also, this is a consistent error, and not random. It is systematic!
anonymous
  • anonymous
Please help!!!!!!!
theEric
  • theEric
What do you think about my reply?
anonymous
  • anonymous
I think it could be d then
theEric
  • theEric
That's what I'm leaning towards, too!
anonymous
  • anonymous
Thank you so much!
theEric
  • theEric
Here's the equation for a simple pendulum : \(T=2\pi\sqrt{\dfrac{L}{g}}\) If the actual \(T\) is greater than the measured \(T\) (I'll say \(T_m\)), then the actual \(2\pi\sqrt{\dfrac{L}{g}}\) is greater than what you would think it is as well. Solving for \(L\), as if you measured the \(T\) calculate \(L\).... \(T_m\lt2\pi\sqrt{\dfrac{L}{g}}\\\implies T_m(2\pi)\sqrt g\lt\sqrt L\\\implies L\gt T_m^2(4\pi^2)g\qquad\text{assuming all variables are positive.}\) The real \(L\) value would be greater than you would calculate. This would be pretty consistent since \(T_m = T(1-\epsilon_\text s)\) (I think...) where \(\epsilon_\text s\) is the percentage that the stopwatch is inaccurate by.

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