# Simple pendulum graph

1. Apr 27, 2014

### 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:

(PS:in the above document, 'n' is used to represent the number of swings rather than 't')

2. Apr 27, 2014

### Simon Bridge

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

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.)

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.

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.

3. Apr 27, 2014

### influx

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

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 $t$.

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

Thanks

4. Apr 27, 2014

### 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.