Measuring period of vertical oscillations and estimating uncertainty

[MightyHound]

Here is the question:

Gently displace the mass and release it so that it performs vertical oscillations. Make measurements to determine the period T of the oscillations.

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So here is what I did:
Time for 20 oscillations = t1 (2 d.p)
Time for another 20 oscillations = t2 (2 d.p)

Average time for 20 oscillations = (t1 + t2)/2 (2 d.p)

Time for one oscillation, T = { (t1 + t2)/2 } / 20 ---> this is your answer correct to 1 decimal place.

Right. Next question:

Estimate the number of oscillations that would need to be timed in order to make the percentage uncertainty in T equal to 1%. Show your working.

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Let error in one reading be x

So we use the formula:

2x/T = 0.01
where T = { (t1 + t2)/2 } / N (N is the number of oscillations).

We use the above formula to find out N.

Here are my concerns:
Is the working above correct for the two questions?
What on Earth is the error in timers? (for protractors it is 1deg; for rulers it is 1mm).

Thanks friends.
I am getting that heartpounding, gutwrenching, vomit inducing exam jitters right about now. Exam is only a few days away.

I am not new to the forums here - I have been an avid view for some time now.
Thanks for all your help guys!

 

[member 553083]

Yes, the working above is correct for the two questions. The error in timers will depend on the accuracy of the timer you are using. Generally, if you are using a digital timer, the error should be quite small (on the order of milliseconds).