Simple Pendulum Amplitude Investigation: Graph and Uncertainty Analysis

[influx]

As part of a Physics experiment I have to investigate how the amplitude of a pendulum bob (attached to a string) varies with the number of oscillations it undergoes. The equation I have to work with is:

(where t = the number of swings, A = amplitude after t swings, A₀ = initial amplitude and k = the damping constant)

Am I correct in saying that ln(A/A₀) would be the label on the y-axis and t would be the label of the x-axis?

IF yes, this would suggest that at the y-intercept , ln(A/A₀) = lnA₀ which suggests that at the y-intercept (where t=0) A = (A₀^2). Is this correct?

Lastly, I am required to find the uncertainty in k by sketching the above graph and using error bars. I am unsure of how to go about this?

Briefing sheet:

http://photouploads.com/images/dzvsv.png
(PS:in the above document, 'n' is used to represent the number of swings rather than 't')

 

[Simon Bridge]

As part of a Physics experiment I have to investigate how the amplitude of a pendulum bob (attached to a string) varies with the number of oscillations it undergoes. The equation I have to work with is: [##A(t)=A_0e^{-kt}##]
(where t = the number of swings, A = amplitude after t swings, A₀ = initial amplitude and k = the damping constant)

... this would be a damped pendulum - not a simple pendulum.

[##\ln A = -kt + \ln A_0##]

Am I correct in saying that ln(A/A₀) would be the label on the y-axis and t would be the label of the x-axis?

If you plot ##\ln (A/A_0)## vs ##t## then yes.
Presumably you want to get a straight line?
Did you try it and see what you get?

(Experiments are about dealing with the data you have and not about trying to conform to an expected result: do it and see.)

IF yes, this would suggest that at the y-intercept , ln(A/A₀) = lnA₀ which suggests that at the y-intercept (where t=0) A = (A₀^2). Is this correct?

No.
Check your algebra. How does the RHS go from the ##\ln A## above to ##\ln (A/A_0)##?

You may be better to just plot ##\ln(A)## vs ##t## instead.

Lastly, I am required to find the uncertainty in k by sketching the above graph and using error bars. I am unsure of how to go about this?

You know the errors for the measurements you made - use them to calculate the errors on the numbers you are plotting.
(Hint: if the error in y is small compared to the error in x, you can eave it off.)

Your course will probably have a specific method they want you to use to convert error bars into an overall uncertainty - check your notes.

 

[influx]

... this would be a damped pendulum - not a simple pendulum.

Ah sorry. That's what I meant :)!

If you plot ##\ln (A/A_0)## vs ##t## then yes.
Presumably you want to get a straight line?
Did you try it and see what you get?

(Experiments are about dealing with the data you have and not about trying to conform to an expected result: do it and see.)

Yes I want to get a straight line. I was thinking of plotting lnA against t but then the y-axis (lnA) would have units and I thought log graphs are not meant to have units? Hence why I decided to plot ##\ln (A/A_0)## vs ##t##.

No.
Check your algebra. How does the RHS go from the ##\ln A## above to ##\ln (A/A_0)##?

You may be better to just plot ##\ln(A)## vs ##t## instead.

Well this is what I did (I can't see what I wrong?):

Thanks

 

[Simon Bridge]

when you change what you plot, you have to redo the y=mx+c step.
off y=mx+c, show me how you decided
y=
x=
m=
c=

... this is missing from your calculation pictured.

note: A/Ao is the size of A measured in units of Ao ... i.e. you still have units.
I don't think there is anything wrong with a log plot having units/dimensions on both axis, but it is tidier.

 

[HalfLostAndSearching]

Yes, you are correct in saying that ln(A/A0) would be the label on the y-axis and t would be the label of the x-axis. This is because the natural logarithm of the ratio of the amplitude after t swings and the initial amplitude is a linear function of t.

Regarding your second question, your interpretation of the y-intercept is not entirely correct. At the y-intercept, ln(A/A0) = ln(A0/A0) = ln(1) = 0, which means that A/A0 = 1 or A = A0. This suggests that at the beginning of the experiment (t=0), the amplitude of the pendulum bob is equal to the initial amplitude.

To find the uncertainty in k, you can use the error bars on the graph to estimate the range of values for k that would still fall within the error bars. This range of values would represent the uncertainty in k. Alternatively, you can use the slope of the line (which is equal to -k) to calculate the uncertainty in k using the formula for propagation of uncertainties. This would involve finding the uncertainties in A/A0 and t, and then plugging them into the formula for the uncertainty in k.