Measurement Help: Homework Statement and Equations

[kakab00]

Homework Statement

http://img487.imageshack.us/img487/5702/untitledna1.png

Homework Equations

v square = u square + 2as

The Attempt at a Solution

a) Let v square - u square be x
delta x/x = 2(delta v/v) + 2(delta u/u)
delta x/300 = 0.03
delta x = 9

b) i) a = (v square - u square)/2s
delta a/a = 2(delta v/v) + 2(delta u/u) + deltas/s
delta a/3 = 0.03 + 0.002
delta a = 0.096 = 0.1 (round off to 1 sig fig)
ans : 3.0 plus minus 0.1

Homework Statement

I think I did it correctly, but my answers seem to be wrong , because this is a question on my school website and I have to key in the answers and it'll tell me whether it is right or wrong. I thought maybe I did something wrong, so maybe somebody could help please?

 

[malawi_glenn]

a) well if we make a function:

$$f(u,v) = v^{2} - u^{2}$$

then, from error propagation:

$$\delta f = \sqrt{\left( \dfrac{\partial f}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial f}{\partial v} \delta v\right)^{2}}$$

So:


$$\delta f = \sqrt{\left( -2*10 * 0.1 \right)^{2} + \left( 2*20 * 0.1 \right)^{2}} = 4.4721$$

the forumula you used is when f(u,v) is a product and quoitent only.

 

[malawi_glenn]

b) the expression for the acceleration is:

$$a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}$$

then the uncertainty in a is:

$$\delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}$$

 

[kakab00]

I don't get your equations, wouldn't the delta u and delta u or other symbols cancel each other out?

 

[malawi_glenn]

it is not deltas, it is the partial derivative of f with respect to u, and so on..

what is the answers for these problems?

Here you can read more:
http://www.rit.edu/~uphysics/uncertainties/Uncertaintiespart2.html#mixtures

see section 5(f) , formulas Eq. 4a and Eq. 4b

 

[kakab00]

I don't know the exact answers, I only know whether the answers would be right or wrong because we have to key in them inside a system after we found them.

I've not learned partial derivatives yet though, so I think there should be another easier way to do it using delta/fractional uncertainty?

 

[malawi_glenn]

partial derivatives are very easy.. you just do usual derivatives and treat the rest as constans. for example: if f(x,y) = 2xy. the f 'x =2y and f 'y = 2x

According to my knowledges in science of errors, the functions we deal with here we must use error propagation as the formulas I have given.

you can read a lot of it on the web page I gave you.

 

[kakab00]

it's not easy for me XD

for the 2nd part I tried using your method:
so delta a = square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square + (0.1 x 50 x 2)square)] = root(150) which would give you a value of 12.2, much bigger than a=3 calculated. doesn't make sense, or did I do something wrongly?

 

[malawi_glenn]

well doing the partial derivative is just doing the usual treating the rest as constants.

Well you did not to the derivatie for a vs s right.. try again. Remeber to trea't v and u just as constant (values)

 

[kakab00]

I have no idea how to do it..

upon further thought, even without finding the derivative for a vs s, the answer already seems to be wrong from the 1st and 2nd part of the equation,

square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square] would give you a value of around 4 which is still greater than a=3.

 

[malawi_glenn]

are you telling me that you can not differentiate functions like:

f(x) = 1 / x ?

if you wait a minute I can do the function for you and show how you do this..

 

[malawi_glenn]

$$\delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}$$

$$a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}$$

$$\dfrac{\partial a}{\partial v} = \dfrac{2v-0}{2s} = \dfrac{v}{s}$$

$$\dfrac{\partial a}{\partial u} = \dfrac{0-2u}{2s} = \dfrac{-u}{s}$$

$$\dfrac{\partial a}{\partial s} = \dfrac{-(v^{2} - u^{2})}{2s^{2}}$$

so we get:
$$\delta a = \sqrt{\left( \dfrac{-u}{s} \delta u\right)^{2} + \left( \dfrac{v}{s} \delta v\right)^{2} + \left( \dfrac{(u^{2} - v^{2})}{2s^{2}} \delta s\right)^{2}}$$

 

[kakab00]

The answers are still wrong though, I used

a) 4 (round off to one significant figure)
b) 3.0 + - 0.2 (this was the value I calculated from the equation you gave me)

 

[malawi_glenn]

Well I got (b) -> error in a = 0.045211

are you sure you have the right formula for the acceleration?

Sorry I can't help you more kid.

And what kinda sucky school has an electronic answer machine lol

it is also the way you get to the answer that is important, not the result itself. So you can basically try every number from 0 to 20 til it gets right :)

 

[kakab00]

Well I got (b) -> error in a = 0.045211

are you sure you have the right formula for the acceleration?

Sorry I can't help you more kid.

And what kinda sucky school has an electronic answer machine lol

it is also the way you get to the answer that is important, not the result itself. So you can basically try every number from 0 to 20 til it gets right :)

hehe

but the answers are still wrong I don't really get your method either, because I haven't learned them yet. I think the school wouldn't have set something they hadn't taught yet, so I believe there's still another (easier( way ) of doing, maybe some other members might know?

Thanks for your help and time anyway! at least I learned something from you

 

[malawi_glenn]

then you show me the simpler way to do it. It is not the right one I will promise you.

we have functions that are both multiplcation, addition and qouitent. And the errors are independent of each other, hence we must use the error propgation formula.

 

[kakab00]

then you show me the simpler way to do it. It is not the right one I will promise you.

we have functions that are both multiplcation, addition and qouitent. And the errors are independent of each other, hence we must use the error propgation formula.

If I knew the easier method I wouldn't be here asking :shy:

Maybe I calculated something wrongly up there? Or maybe the school system just sucks

 

[malawi_glenn]

If I knew the easier method I wouldn't be here asking :shy:

Maybe I calculated something wrongly up there? Or maybe the school system just sucks

I bet your school sucks hehe :)

Can´t you write donw all that we have discussed here, and maybe print out the relevant section from the www-page i showed you. Then go to the professor and ask "what is wrong?" :P