Aug 31, 2000
PHY 113 -- Additional notes for
PHY 113 - Additional notes for Chapter 1 (Problem Set # 1)
In class, we did not quite have enough time to discuss error
analysis. This will be discussed also in your laboratory work.
Some (hopefully) helpful comments follow:
Some degree of error is associated with any measurement. For
example, Suppose your ruler has centimeter and millimeter
markings. If you measured one side of your text you could say
that its length is l1 ±dl1 (for example 22.2 ±0.2) cm. Suppose the second length is measured as l2 ±dl2, while the thickness is t ±dt. If you now
wanted to compute the volume of your text, that would be
To get an idea of the error in your calculation you need to think
about the error in each length measurement. Precisely,
|
dV º (l1 ±dl1)(l2 ±dl2)(t ±dt) - l1 l2 t. |
|
If we want an estimate of
the error, then we can make the following approximations to the
above formula for dV.
- dl1 is small so that terms like dl1 ·dl2
and dl1 ·dl2 ·dt can be neglected.
- Since we want to estimate the maximum possible error, we
should replace ± with +.
Therefore,
|
dV » dl1 l2 t + l1 dl2 t + l1 l2dt. |
|
If we divide this result by V, we get the very compact result:
|
|
dV
V
|
= |
dl1
l1
|
+ |
dl2
l2
|
+ |
dt
t
|
. |
|
This shows that in this case the fractional error is equal to the
sum of the fractional errors in each of the length
measurements. Not all derived quantities will have this simple
result, but often one can estimate the error as a function of
fractional errors. Homework problem # 4 makes use of some of
these ideas.
File translated from TEX by TTH, version 2.20.
On 31 Aug 2000, 13:53.