Propogation of uncertainty

[tobbles_tw93]

Hi guys Ive got really stuck trying to work out how uncertainty propogates through the formula for acceleration of gravity. I have converted units to M from cm and worked out my average periods and run it through the equation for gravity to give me:

g=4pi()^2*L/t^2

g (ms^-2)	T (mean period of T)	uncertainty in mean of T (seconds)	Uncertainty in L (metres)
9.83	0.64	0.04	0.002
10.02	0.89	0.04	0.002
9.27	1.14	0.04	0.002
9.92	1.26	0.04	0.002
9.96	1.41	0.04	0.002

I have tried and tried but cant seem to get an uncertainty that looks correct?
If anyone can help me out Id be incredibly grateful.

Many thanks,

T.

 

[topsquark]

First, a pet peeve: g is the acceleration due to gravity! Note that it's an acceleration. Gravity is a term denoting a force.

Your formula for g has two variables, L and T (You should stick to either t or T.)

$$\Delta g = \sqrt{ \left ( \dfrac{ \partial g}{ \partial L } \Delta L \right )^2 + \left ( \dfrac{ \partial g}{ \partial T} \Delta T \right )^2 }$$
which is, in this case
$$\Delta g = \sqrt{ \left ( \dfrac{4 \pi ^2}{T^2} \Delta L \right )^2 + \left ( \dfrac{4 \pi ^2L}{T^3} \Delta T \right ) ^2 }$$
-Dan

 

[Romsek]

First, a pet peeve: g is the acceleration due to gravity! Note that it's an acceleration. Gravity is a term denoting a force.

Sounds like a massive misunderstanding!

 

[topsquark]

$$\Delta g = \sqrt{ \left ( \dfrac{4 \pi ^2}{T^2} \Delta L \right )^2 + \left ( \dfrac{4 \pi ^2L}{T^3} \Delta T \right ) ^2 }$$

Correction:
$$\Delta g = \sqrt{ \left ( \dfrac{4 \pi ^2}{T^2} \Delta L \right )^2 + \left ( \dfrac{8 \pi ^2L}{T^3} \Delta T \right ) ^2 }$$
-Dan

 

[tobbles_tw93]

Thank you so much, all scoldings have been duly noted.
I am hower... a little unsure how the T becomes cubed on the right hand side of the equation? and 4pi^2 becomes 8pi^2? This may be where it all fell apart for me.

 

[Deleted member 4993]

Thank you so much, all scoldings have been duly noted.
I am hower... a little unsure how the T becomes cubed on the right hand side of the equation? and 4pi^2 becomes 8pi^2? This may be where it all fell apart for me.

$\displaystyle T = 2\pi \sqrt{\dfrac{L}{g}}$

First rewrite the above as g = f(T, L)

Then calculate........ $\displaystyle \Delta{g}$ .........using calculus and you should be able to see why do we have '8' and T³.