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The gravitational acceleration,$g$ can be determined by using a pendulum. If a pendulum of length $l$ has a period of $T$ s. A 2m pendulum is timed to take 57 s for 20 swings.

A)calculate the value of $g$ from the data. B) find an expression for the approximate error in $g$ for an error of $\delta t$ in the timing of $20$ swings. C)calculate the possible error in $g$ if the timing was made to the nearest second.

I solved question a. But I've no idea for the question b and c. Can anyone explain it to me? Thanks

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$$g=\frac{4\pi^2 l}{T^2}$$ $$\implies \log g=\log(4\pi^2)+\log l-2\ log T$$ Differentiate both sides to get: $$\frac{\delta g}{g}=\frac{\delta l}{l}+\frac{2\delta T}{T}$$ (It changed to plus as errors are always added)

As length is given to be constant $\delta l$=0, hence $$\delta g=\frac{2 g \delta T}{T}$$ put $T= 57$, and the value of $g$ from part a, and you're done.

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  • $\begingroup$ The answer is $\delta g=0.348t$ I really stuck at here. I couldn't get it $\endgroup$ – Mathxx Feb 24 '16 at 15:25
  • $\begingroup$ Is length assumed to be constant?( also give the value of g from part a) $\endgroup$ – Nikunj Feb 24 '16 at 15:26
  • $\begingroup$ Yes. 2meter for the length $\endgroup$ – Mathxx Feb 24 '16 at 15:26
  • $\begingroup$ I'll complete my answer. $\endgroup$ – Nikunj Feb 24 '16 at 15:28
  • $\begingroup$ Why should I put 0.5 for part 3? $\endgroup$ – Mathxx Feb 24 '16 at 15:35

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