3
$\begingroup$

I want to conduct an experiment. I want to observe time spent after a certain oscillation. I have actually did that and get some data but I want to know error margin.

I know the equation for osciallation time until rest. It should be T=2π√(L/g), but I don't know how to formulate for specific oscillation amount like time spent for 10 oscillation.

$\endgroup$
  • 2
    $\begingroup$ $\text{time spent for 10 oscillations}=10\,\left(\text{time spent for 1 oscillation}\right)$ $\endgroup$ – AccidentalFourierTransform Dec 19 '16 at 19:26
2
$\begingroup$

If, as you correctly noted, the period of a pendulum is $$T=2\pi \sqrt{\frac{L}{g}}$$

then the time for one oscillation is just ten times that (be careful with units!).

As for an error calculation, there are a number of ways in which this might be approached. One simple way to do so is to perform the experiment a number of times. We can presume that the mean of these is the "true" value and that the error is $\pm \sigma$, where $\sigma$ is the standard deviation of the collected measurements.

Note, however, this only accounts for experimental error and not the precision or accuracy of the instrument(s) used for the measurement or any problems with the experimental apparatus itself. Search the internet for "experimental error" to find more resources that describe this important concept in more detail.

$\endgroup$
  • $\begingroup$ Oh, I wrongly thought that this formulation is for time spent until rest but I have understood it now. It is only for one oscillation. $\endgroup$ – Samet Dec 19 '16 at 19:44
  • $\begingroup$ Yes, this is the period time, the time for one oscillation and in this model/approximation the pendulum is never coming to rest as there is no friction. $\endgroup$ – user1583209 Dec 19 '16 at 20:22

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.