Why Is My Calculated Error for Resonant Frequency So High?

In summary, the speaker is trying to calculate the error for a resonant frequency using values for an inductor and capacitor. They get a value for f_0 that is close to their experimental observations, but the final error value is too high. They discuss the equations they used and make corrections to their calculations. Eventually, they come to a final error value and thank the other person for their help.
  • #1
ma18
93
1

Homework Statement



So I am calculating the error for something and I am getting really weird values.
So I know that the value for the Inductor is 24.97 +- 0.005 mH and that for the capacitor is 105.7+-0.0005 nf.

So I am finding the value for the resonant frequency

Homework Equations



f_0 = 1/(2*pi*sqrt(LC))


The Attempt at a Solution



So for the f_0 I get 3097 Hz which is very close to my experimental observations. But for the error I get:

error in LC = (2.64e-9) * sqrt((0.005/24.97)^2+(0.0005/105.7)^2) = 5.29e-13

error in LC^-.5: (19464.95)*0.5*2.64e-8/5.29e-13 = 37994

final error: 37994 * 1/(2*pi) = 6039.

Now this final value is too high, what am I doing wrong? Thanks
 
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  • #2
ma18 said:
error in LC = (1.03e-8) * sqrt((0.005/24.97)^2+(0.0005/105.7)^2) = 5.29e-13
I don't see where the 1.03e-8 comes from. LC is about 2.6e-6, no?
error in LC^-.5: (19464.95)*0.5*1.03e-8/5.29e-13 = 37994
What's the reason for dividing by 5.29e-13? It would be clearer if you were to write the equation in purely symbolic form, not plugging in numbers.
 
  • #3
Sure, sorry. The 1.03e-8 is the LC value and the following values in the equation are the error divided by the value. For the second equation 19464.95 is the LC^-.5 is the value and 0.5 is the exponent and 5.29e-13 is the error in LC^-0.5.

The eqn's are:

for multiplication (z=xy): dz = z *sqrt((dx/x)^2+(dy/y)^2)

for exponents (z=x^y): dz = abs(y)*z*dx/x
 
  • #4
ma18 said:
Sure, sorry. The 1.03e-8 is the LC value
Isn't LC = 24.97 * 105.7e-9?
for exponents (z=x^y): dz = abs(y)*z*dx/x
Sure, but when you plugged in the numbers you seem to have used x/dx instead of dx/x.
 
  • #5
L is in mH so there is an extra 10^-3 factor I forgot note here. I wrote down the exponent thing wrong here but the number is right, it should be:

error in LC^-.5:
value*exponent*error_LC/LC
= (19464.95)*0.5*2.64e-8/5.29e-13 = 37994

ah there are so many numbers in my sheet I get mixed up
 
  • #6
Alright I think I've got it, my data was just too ugly, I cleaned it up and did it another sheet.

I got:

Error in LC: 5.29E-13
LC: 2.64E-09
LC^-0.5: 1.95E+04
Error in LC^-.5: 1.95E+00
Error in final : 3.10E-01

which ironically seems a little small but whatever.

Thanks for the help!
 
  • #7
I find it easier to think in terms of fractional errors. (Mistakes in the calculation usually become more obvious.)
The fractional error in LC is 5.29E-13/2.64E-09 ~ 2E-4 (almost entirely owing to the error in L).
The fractional error in sqrt(LC) will be half that: 1E-4.
The fractional error in f_0 will also be 1E-4, giving an absolute error of ~ 3E3 * 1E-4 = 3E-1.
That confirms your answer.
 
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1. What is error propagation when dividing?

Error propagation when dividing is the process of estimating the uncertainty or error in the result of a division operation. It takes into account the uncertainties or errors in the values being divided and uses mathematical formulas to calculate the overall uncertainty in the result.

2. Why is error propagation important in scientific calculations?

Error propagation is important in scientific calculations because it allows us to determine the reliability and accuracy of our results. By understanding the uncertainty in our calculations, we can make informed decisions about the validity of our conclusions and the significance of our findings.

3. How is error propagation calculated when dividing two values with uncertainties?

Error propagation when dividing two values with uncertainties is calculated using the formula:

δz = z * √((δx/x)^2 + (δy/y)^2)

Where δz is the uncertainty in the result, z is the result of the division, δx is the uncertainty in the numerator, and δy is the uncertainty in the denominator.

4. Can error propagation be avoided in scientific calculations?

Error propagation cannot be completely avoided in scientific calculations as all measurements and values have some degree of uncertainty. However, careful experimental design and use of precise instruments can minimize the uncertainties and reduce the impact of error propagation on the final result.

5. How does error propagation differ from error in a single measurement?

Error propagation takes into account the uncertainties in multiple measurements or values that are used in a calculation, while error in a single measurement only considers the uncertainty in one measurement. Error propagation provides a more comprehensive understanding of the overall uncertainty in the final result, while error in a single measurement only reflects the uncertainty in that particular measurement.

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