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Pendulum Equation Question

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Hello,

I'm given the following from a lab and there are two unknown variables and I'm not sure how to get the values for a,b and I was wondering if someone could help me solve for a,b. Thank you.

Period = 2pi( length/a)^b

where a & b are unknown

Y-intercept = log(2pi) + blog(1/a)

1. What is value of a? What physical constant does this value suggest?

2. What is the value of b?
 

Answers and Replies

Kurdt
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This should be in the homework section. Since this is for a lab you will most likely have to plot a graph of the results taken for period versus pendulum length.

EDIT: Well in this experiment one normally plots the log of the period versus the log of the pendulum length. The equation is then that of a straight line and you can find the values from the intercept and gradient.
 
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Hi, Thanks for the quick reply.

No there is no plotting at all. It just ask for find the values of "a" and "b" and they give two equations. I'm not sure what "a" or "b" is and what physical constant suggest on the value of "a".
 
Kurdt
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What values have you been given and what two specific equations have you been given?
 
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These are the only two equations given:

1. Period = 2pi( length/a)^b

2. Y-intercept = log(2pi) + blog(1/a)

There are no additional values and we're suppose to find what "a" and "b" are.
 
Kurdt
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From the first equation it is easy to get equation 2.

[tex] T=2\pi \left( \frac{L}{a} \right)^b [/tex]
[tex] \ln(T) = \ln(2\pi) + b\ln \left(\frac{L}{a}\right) [/tex]
[tex] \ln(T) = \ln(2\pi) +b\ln(L) +b\ln(1/a) [/tex]

The final line is what is normally plotted during an experiment and to be honest I can't see any way of solving without plotting a graph.
 
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Well it seems that if everyone says you have to graph this to get the answer, maybe that is what is required. When we graph these functions, are the 'a' and 'b' values clear? What are we looking for, could you walk me through this problem. Thank you
 
Kurdt
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Its basically the form of a straight line [tex] y = mx+c [/tex]

Perhaps since this was given to you in a lab you are to perform an experiment and graph the results to determine the answer.
 
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The lab has already ended, this is for the post lab report. I've answered all the other questions, this was to be a thought question at the end. I'm not really sure how to find the values of 'a' and 'b'. I've tried to email my TA, but haven't gotten a response.
 
Kurdt
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Right well I've basically given you the answer. What do you know of straight line graphs? Compare the straight line graph formula with [tex] \ln(T) = \ln(2\pi) +b\ln(L) +b\ln(1/a) [/tex].
 
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y = mx + c

m = slope

c = bln(1/a)

I know there should be a relationship between the line graph formula and the primary equation.
 
Kurdt
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You're almost there. You need to work out which part of this [tex] \ln(T) = \ln(2\pi) +b\ln(L) +b\ln(1/a) [/tex] is y, x, m and c.
 
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Well I originally defined

Y-intercept = log(2pi) + blog(1/a)

Which looks like

y = ln(T)
mx = log(2pi) + blog(1/a)
c = bln(1/a)
 
Kurdt
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Still not there but keep trying. Theres not much more I could do other than give you the answer which is not allowed.
 
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Can you at least say whats right or wrong, like parts of the variables?
 
Kurdt
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Well you have the equation for the y intercept which in the straight line equation is equivalent to c. What do you think mx will be?
 
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So are you saying log(2pi) + blog(1/a) + blog(1/a) is = C and mx = 1?
 
Do you know the value of y intercept? If yes,
since, you know Y-intercept = log(2pi) + blog(1/a)
You have 2 unknown for one equation... you need one more equation...
(Hints: the last equation is related to the slope of the graph)
 
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I'll keep looking at it, but I'm really lost on what to do.
 
Kurdt
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Since you said this was a question of thought afterwards and was in a lab class I think all you will be expected to do is recognise that if you take the log of both sides of the original period equation you can graph it and obtain a and b from that. I really don't want to give you any more information because if you look through all the posts i have made it should be enough to work out what goes where.
 

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