Error calculations involving gradients

[acidburner]

I'm trying to work out the percentage error in working out the value of gravity,g, from a pendulums motion.
i know that percentage error is (possible error/value used)*100 however I am using multiple values multiple times and its getting a little confusing.
In the investigation g=k(∆T²/∆L), my problem is that i have the error for working out T which is 0.001 seconds. The error in L is 0.001m. As I'm using a difference of two values for each of the ∆'s would i double each error and for the error of T² would i square 0.001 and then double it.
any explanations would be helpful
Thanks
∆T²=2.112 ∆L=0.51

 

[rock.freak667]

You know that

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

so that

$$T^2=4\pi^2 \frac{L}{g}$$

and that means that

$$g=\frac{4\pi^2}{T^2}L$$

To find the error you do this.

$$\frac{\delta g}{g}=2\frac{\delta T}{T} + \frac{\delta L}{L}$$

$\delta T$ would be the error in T and similarly for $\delta L$ is the error in L.

 

[Moetasim]

I understand your confusion with calculating the percentage error in your investigation. Let me break it down for you.

Firstly, when calculating the percentage error, it is important to consider the possible error in each individual measurement. In your case, the possible error in T is 0.001 seconds and in L is 0.001 meters. It is important to note that these errors are independent of each other.

Now, when calculating the value of g using the equation g=k(∆T²/∆L), you are essentially taking the ratio of two measurements, ∆T² and ∆L. This means that the percentage error in g will be the sum of the percentage errors in ∆T² and ∆L.

To calculate the percentage error in ∆T², you will need to square the error in T, which is 0.001 seconds. This gives you an error of 0.000001 seconds. Then you will need to double this error, as you are using the difference of two values for ∆T². This gives you a final error of 0.000002 seconds for ∆T².

Similarly, for ∆L, you will need to double the error of 0.001 meters, giving you a final error of 0.002 meters for ∆L.

Now, to calculate the percentage error in g, you will need to use the formula you mentioned, which is (possible error/value used) * 100. So, for ∆T², the percentage error will be (0.000002/2.112) * 100 = 0.000095%. And for ∆L, the percentage error will be (0.002/0.51) * 100 = 0.392%.

Finally, to calculate the overall percentage error in g, you will need to add the percentage errors of ∆T² and ∆L. In this case, it will be 0.000095% + 0.392% = 0.392095%. This means that the value of g you have calculated may have an error of 0.392095%.

I hope this explanation helps you in understanding how to calculate the percentage error in your investigation. It is always important to consider the possible error in each measurement and to combine them appropriately when calculating the overall percentage error. If you have any further questions, please do not