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


$$g=\frac{4\pi^2 l}{T^2}$$ $$\implies \log g=\log(4\pi^2)+\log l-2\ log T$$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.