Understanding Error in Measurement for Simple Pendulum Experiment

[AStaunton]

Hi there

when doing an experiment with my university associated with my university related to a simple pendulum, I was not confident when calculating the error there was in my measurement, the relevant equation is:

$$T=2\pi\sqrt{\frac{L}{g}}$$

where T=period L=length of pendulum and g=accel due to grav

my problem is with deciding the error in the period, I measure the length L which was 50cm+.5cm and took g as 9.8+.1 as this is a product (L/g) the general idea is we add the percentage error and of course if we are adding two quantities for example (distance1 + distance2) we add the absolute error.
But what I am not sure of is; does taking the square root of the expression (L/g) have any bearing on the error...also does multiplying by 2pi change the error, I think the multiplying by 2pi might for the following reason:
when you want the width of a piece of paper, it is better to measure 50 pieces of paper so we can then divide by 50 and reduce the error by a factor of 50...so I think it is possible that multiplying by 2pi might increase error by a factor of 2pi..

another example is if I measure the angle theta to accuracy of +1degree and then take the sine of theta, does taking the sine have an effect on the error?

I would be grateful if someone could answer these questions and also suggest rules of thumb when calculating error in more complicated expressions, as I am a third year university physics student now and am too embarassed to ask any of my teachers for advice on something I should have learned years ago!

 

[HallsofIvy]

The simplest way to handle that is to use the differential. From $T= 2\pi L^{1/2}g^{-1/2}$, we have $$dT=\pi g^{-1/2}L^{-1/2}dL- \pi L^{1/2}g^{-3/2}dg$$. You are saying that L= 50 cm.= 0.5 m, dL= 0.5 cm= .005 m, g= 9.81 m/s^2, and dg= .1 m/s^2.

 

[Andy Resnick]

I would be grateful if someone could answer these questions and also suggest rules of thumb when calculating error in more complicated expressions, as I am a third year university physics student now and am too embarassed to ask any of my teachers for advice on something I should have learned years ago!

As HallsofIvy mentioned, the way to determine error propagation is to take the differential. When the different measurements are independent, the errors add in quadrature.

This is the essential book to read:

https://www.amazon.com/dp/093570275X/?tag=pfamazon01-20

 

[Vanadium 50]

The Taylor book is excellent. One case where you can judge a book by its cover.

 

[pmacias]

Hello,

Thank you for sharing your concerns about understanding error in measurement for a simple pendulum experiment. As a scientist, it is important to accurately measure and analyze data in order to draw meaningful conclusions. I will do my best to address your questions and provide some general guidelines for calculating error in more complex expressions.

Firstly, it is important to note that the equation you provided, T=2\pi\sqrt{\frac{L}{g}}, is the formula for the theoretical period of a simple pendulum. This means that it assumes certain ideal conditions such as a perfectly rigid and massless string, and a small amplitude of motion. In a real experiment, there will always be some degree of error due to various factors such as human error, equipment limitations, and environmental conditions. Therefore, it is important to consider and account for this error in your calculations.

In terms of determining the error in the period, you are correct in adding the percentage errors of the individual measurements. However, when taking the square root of the expression (L/g), the error does not change. This is because when we square a number, the percentage error also gets squared, resulting in the same error value. Similarly, multiplying by 2pi does not change the error. The same principle applies here - when we multiply a number by a constant, the percentage error remains the same. Therefore, the error in the period will be the same as the error in the individual measurements of length and acceleration due to gravity.

To address your example of measuring the angle theta and taking the sine, taking the sine does not introduce any additional error. The error in the angle will remain the same when taking the sine, as long as the angle is measured consistently in degrees or radians.

In general, when calculating error in more complicated expressions, it is important to consider the individual errors of each measurement and how they affect the overall expression. As a rule of thumb, if the values are being multiplied or divided, the percentage errors should also be multiplied or divided. If the values are being added or subtracted, the absolute errors should be added. Additionally, it is always a good practice to round off your final answer to the same number of significant figures as the measurement with the least number of significant figures.

I hope this helps address your concerns and provides some guidance for calculating error in more complex expressions. It is always important to double check your calculations and ask for help if needed. Remember, asking questions and seeking clarification