Recent content by Physics Curiosity

Apologies for my misunderstanding. It was that simple, I just had a blind eye to the question. You have fully answered my question. I sincerely thank anyone who has taken time out of their day to guide my misunderstanding. Have a great day! :-)

Alright, I am still rather puzzled where the 2pi went, but thanks a bunch for the information. I will look into it!

So, I cant just use T = 2π/√(3g) * √(2L), the theoretical gradient being 2π/√(3g) which follows the y=mx+c format like post #6 suggested? I will be comparing the theoretical gradient to the experimental gradient when I graph √(L) in √(m) in the x-axis against Period T in (s) in the y-axis.

In that case would the theoretical gradient, or m, be √(2/3g), then x would be sqrt(L), in the y=mx+c equation?

I understand now. I had confused myself -- T = 2π/√(3g) * √(2L) was the correct "processed" equation. Cheers, many thanks for all the help guys!

I am supposed to be graphing sqrt(L), not sqrt(2L). Here's an example of "m" being "processed", I am guessing processed just means to separate the equation into y = mx+c structure, where T is period, m is the gradient, x is the independent variable (m), and c is the y-intercept. This example...

Thanks for the further clarification! I do know how to manipulate that equation in that way, except I am confused on how I can process with T = 2πsqrt(2L/3g), I am not sure how to get rid of the 2 in 2L. I came to this T = 2π/√(3g) * √(2L), but the √(2L) should be √(L) instead.

Initially I went from: T = 2π√(2L/3g) T = 2π/√(3g) * √(2L) To finally this equation: T = 2π/√(3g) * √(L) Where 2L becomes L as the 2 is lost. I am not fully sure if this is correct or how to properly get rid of the 2 in 2L.We must follow the rule of y = mx+c whereby y = T, m = the constant...

Curious. That's all.