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PHY 231
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I'm uncertain about calculating the uncertainty, or percent error. I looked back at my notes from when you were showing us how to do it but I can't make sense of what I've written. One part shows something like: (aħ `da)/(bħ `db) using opposite signs on the top and bottom to find the greatest variance. Then though I have a line connecting to this saying subtract and beside that an addition like 0.2 +0.3 =0.5. The first method really seems to add the most error by the time you go from time/distance to acceleration, but it seems to be the most logical. Is there somewhere on the site that goes through this or can you tell me in your response.
If you know a within uncertainty +-`da, then the proportional uncertainty in a is `da / a. If you multiply this by 100 you get the percent uncertainty.
If a and b are uncertain respectively by amounts +- `da and +- `db, then the product a * b in a sense really means (a +- `da) * (b +- `db). The product is a b +- a `db +- b `da +- `da * `db.
If the percent error is small then `da * `db is much smaller than a `db + b `da, and very much smaller than a * b. So the product is approximately ab +- (a * `db + b * `da).
The proportional uncertainty in ab is therefore (a * `db + b * `da) / (a b) = `db / b + `da / a, which is the sum of the proportional uncertainties in a and in b.
To you divide a +- `da by b +- `db, you get a similar result. The proportional uncertainties add up.
This is directly related to the product rule. Since the differential d(ab) is equal to da * b + a * db, the quantity ab is uncertain by proportion (da * b + a * db ) / (ab) = da / a + db / b.
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Also, I have a question about the projectile trials. The ball is going to have a slight v0y since it's coming off the tilted ramp, especially for the larger slope and I'm not sure how to figure that in right off. That's probably on the site as well but I don't know where.
If you know x, y, x0 and y0, as well as the slope, then since you can easily find the values of sin(theta) and cos(theta) the equations
x = x0 + v0 cos(theta) `dt and
y = y0 + v0 sin(theta) `dt - 1/2 g t^2
can be solved for the two unknowns v0 and `dt.
Eliminate `dt by solving the first equation for `dt, substitute into the second equation, and solve for v0.
Do this symbolically to get a formula that can be easily applied to different trials (I would set it up in Excel).
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Lastly, my `da/ `dslope is very high again when done with the pendulum and I haven't finished the last two to know about those. This is something you mentioned in my other lab report but you didn't say whether they were correct or not. So is something like 567 cm/s^2/ slope too much? It seems like it should be. I get something like `da= 25, `dslope = 0.085 and that is of course very large since the denominator is so small.
567 cm/s^2 is in the right ballpark. No problem there. of course Audacity will give you the best result.
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