Error Analysis in Physics?

Question

You are determining the period of oscillation of a pendulum. One procedure would be to measure the time for 20 oscillations, t20, and repeat the measurement 5 times. Another procedure would be to measure the time for 5 oscillations, t5, and repeat the measurement 20 times. Assume, reasonably, that the error in the determination of the time for 20 oscillations is the same as the error in the determination of the time for 5 oscillations. Calculate the error in the period for both procedures to determine which will give the smallest error in the value of the period?

Answer 1

If you assume the errors are independent (no systematic error), then the error for the average of N measurements is the error for one measurement over sqrt (N). So as you take more measurements, your error goes down like 1/sqrt(N). If you take 4 times as many measurements, your error is half as great.

Answer 2

The time period of oscillation in a pendulum is a function of the acceleration as per Newtonian theory.
The pendulum period depends on what location of the gravity field which is defined as accelleration of Gravity force per Unit
mass.At the surface of the earth the acceleration is not Uniforn and varies depending on the location .
This is where difference of time period of oscilation occurs. If the pendulum is taken to the space station where the gavity field is relatively low the time period of the pendulum would be a lot longer.
Concerning error in the calculation, it is dependent on the friction of the pendullum is subject to. The enegy of the swing is max. at the begining and slow down exponentially with time.
There fore to compare the initial condition to the final condition is the error of the swing time.
That means the measure 5 oscilation would have less friction losses than 20 oscilation.

Answer 3

the superb extra healthful line could pass with the aid of one hundred% of the favored blunders bars if attainable. while you're using a know-how analysis application or maybe Microsoft Excel, there's a trendline function which provides you with an R^2 cost. The R^2 cost could be as on the factor of a million.00 as attainable, ususally 0.ninety 9 or 0.ninety 8 is acceptable.