Simple Pendulum Amplitude Investigation: Graph and Uncertainty Analysis

Click For Summary
SUMMARY

The discussion focuses on investigating the amplitude of a damped pendulum bob as it relates to the number of oscillations. The relevant equation is A(t) = A0e^(-kt), where t represents the number of swings, A is the amplitude after t swings, A0 is the initial amplitude, and k is the damping constant. Participants confirm that plotting ln(A) against t is preferable to plotting ln(A/A0) against t, as the latter introduces unnecessary complexity. Additionally, uncertainty in k can be determined by applying error bars to the graph based on measurement errors.

PREREQUISITES
  • Understanding of damped oscillations in physics
  • Familiarity with logarithmic functions and their properties
  • Basic skills in graphing and data analysis
  • Knowledge of error propagation and uncertainty calculations
NEXT STEPS
  • Learn about error propagation techniques in experimental physics
  • Study the characteristics of damped oscillations and their mathematical representations
  • Explore graphing software tools for plotting and analyzing data
  • Investigate the significance of the damping constant (k) in oscillatory systems
USEFUL FOR

Students and educators in physics, particularly those involved in experimental mechanics and data analysis, will benefit from this discussion.

influx
Messages
162
Reaction score
1
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:

cvpv.png



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

sssssqcq.png


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')
 
Physics news on Phys.org
influx said:
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.
 
Simon Bridge said:
... this would be a damped pendulum - not a simple pendulum.

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

Simon Bridge said:
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##.

Simon Bridge said:
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?):

SwoPKD2l.jpg



Thanks
 
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.
 

Similar threads

  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K