Help deriving expression for the error in velocity.

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Homework Help Overview

The discussion revolves around deriving the error expression for velocity based on the equation \( v_{x}=\frac{s}{\sqrt{\frac{2h}{g}}} \). Participants are exploring the rules for calculating error propagation in this context, particularly focusing on the uncertainties associated with division and square roots.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss their previous experiences with deriving error expressions and express uncertainty about their methods. They question the application of error propagation rules, particularly for division and square roots. Some participants suggest that the error in a square root might be treated similarly to exponentiation.

Discussion Status

The conversation is ongoing, with participants sharing their thoughts on the correct application of error propagation techniques. Some guidance has been offered regarding the treatment of uncertainties, but there is no explicit consensus on the correctness of the original poster's approach.

Contextual Notes

One participant notes that the gravitational constant \( g \) typically does not carry an error unless measured, while another mentions that lab equations usually provide gravity along with its uncertainty. There is a discussion about the conservative approach to calculating resultant uncertainty.

Vasili
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Homework Statement


Derive the error expression [tex]\delta v_{}x[/tex] from the equation [tex]v _{x}[/tex]=[tex]\frac{s}{sqrt(\frac{2h}{g})}[/tex]

The Attempt at a Solution



I've derived error expressions before, but I have a history of getting the calculations right and the error equations wrong. So, if possible I'd like to have someone tell me what I am doing wrong (if anything) when I derive this.

For any two numbers divided by one another (2h/g), the error z is:

[tex]\delta[/tex]z=|z|([tex]\frac{\delta x}{|x|} + \frac{\delta y}{|y|}[/tex])
So:
[tex]\delta[/tex]z=|z|([tex]\frac{\delta h}{|h|} + \frac{\delta g}{|g|}[/tex])

At this point intuition tells me that I should just multiply that by two to get the error in 2(h/g), but I think I've done that before and got it wrong.

I have no idea what the error of a square root is. I'm going to guess it's the same as the error in squaring something, so the error in the initial equation would be:

[tex]\delta[/tex]v=|v|([tex]\frac{2 \delta h}{|h|} + \frac{2 \delta g}{|g|} + \frac{\delta s}{|s|}[/tex])

So, am I right? If not, what am I doing wrong? It's fairly crucial to understand these rules, I think, so any help would be wonderful.

Thanks.
 
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Vasili said:

Homework Statement


Derive the error expression [tex]\delta v_{}x[/tex] from the equation [tex]v _{x}[/tex]=[tex]\frac{s}{sqrt(\frac{2h}{g})}[/tex]

The Attempt at a Solution



I've derived error expressions before, but I have a history of getting the calculations right and the error equations wrong. So, if possible I'd like to have someone tell me what I am doing wrong (if anything) when I derive this.

For any two numbers divided by one another (2h/g), the error z is:

[tex]\delta[/tex]z=|z|([tex]\frac{\delta x}{|x|} + \frac{\delta y}{|y|}[/tex])
So:
[tex]\delta[/tex]z=|z|([tex]\frac{\delta h}{|h|} + \frac{\delta g}{|g|}[/tex])

At this point intuition tells me that I should just multiply that by two to get the error in 2(h/g), but I think I've done that before and got it wrong.

I have no idea what the error of a square root is. I'm going to guess it's the same as the error in squaring something, so the error in the initial equation would be:

[tex]\delta[/tex]v=|v|([tex]\frac{2 \delta h}{|h|} + \frac{2 \delta g}{|g|} + \frac{\delta s}{|s|}[/tex])

So, am I right? If not, what am I doing wrong? It's fairly crucial to understand these rules, I think, so any help would be wonderful.

Thanks.

One thing I would note is that g is a gravitational constant that unless you are measuring it and using it generally doesn't carry an error or if it does carries a relative error of 0.
 
Well, the equation is a lab equation and they always give gravity along with its uncertainty. But is the rest of it right?
 
Vasili said:
Well, the equation is a lab equation and they always give gravity along with its uncertainty. But is the rest of it right?

Some treatments of uncertainty for independently measured variables take the Root Sum of the Squares of the relative uncertainties for multiplication division and the absolute uncertainties for addition subtraction. You will have to be the judge of what your lab may be wanting from you.

Your answer would be the more conservative approach as it would yield a greater resultant uncertainty.
 
As for exponents I think you take the exponent and multiply it times the relative uncertainty of the quantity being taken to the exponent. So for sqrt I think you take 1/2 the relative error quantity.
 
Oh, right. Square root is the same as taking it to the 1/2 exponent. I keep forgetting that, thanks.
 

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