How Can Variables l and T Be Plotted to Determine k from a Graph's Slope?

[General_Sax]

Homework Statement

How should the variables ( l and T) be plotted to obtain k from the slope of a linear graph? Identify (write out) the constants correstponding to the slope and intercept of the linear graph.

Homework Equations

l = lambda

l = (k/f)*(T/u)^0.5

The Attempt at a Solution

k^2 = [ (f^2)*(u) ] * [ (l^2)/T ]

(l^2)/T = x

(f^2)* u = m

K^2 = mx + 0

b = 0

 

[symbolipoint]

You really want to graph l as a function of (1/f)*(T/u)^0.5
The slope, if you obtain the graph of a line, will be k.

 

[General_Sax]

I truly don't understand how that function's slope represents k. Does the removal of k somehow allow it to represent the slope?

 

[symbolipoint]

A line crossing the origin can be represented with y = m*x. I used the "*" to show multiplication.

Your equation is l = (k/f)*(T/u)^0.5, which you may rewrite as l = k*(1/f)*(T/u)^0.5. In this equation, l corresponds to y, k corresponds to m, and (1/f)*(T/u)^0.5 correponds to x.

 

[General_Sax]

Oh, that's so simple, yet so elusive. Thanks so much for your help. One last question: should I square the whole expression, so that I don't get a curve from (T/u)^0.5?

 

[symbolipoint]

Oh, that's so simple, yet so elusive. Thanks so much for your help.

One last question: should I square the whole expression, so that I don't get a curve from (T/u)^0.5?

Based on the description you gave, No.
You are expecting a linear relationship for which your horizontal axis is for values of THE EXPRESSION, (T/u)^0.5 ; that expression IS your independent variable, composed of T and u.

 

[HallsofIvy]

What the others are saying is that you plot y= \lambda on one axis and the values of x= (1/f)\sqrt{T/u} on the other. That way your graph will be y= kx with k as the slope. Of course, you have to know f and u to do that.

Another thing you could do, and a fairly common way to get a linear graph from non-linear data, is to take the logarithm of both sides and plot that.

log(\lambda)= log((k/f)*(T/u)^0.5)= 0.5 log(T)+ log(k)- log(f)- 0.5log(u)[/itex]<br /> Plotting log(\lambda) against log(T) will give you a straight with slope 1/2 and log(k)- log(f)- 0.5log(u) as y-intercept. Assuming that you know f and u, that will tell you k.<br /> <br /> Does anyone remember &quot;log-log graph paper&quot; or am I showing my age here?

 

[General_Sax]

You people are so helpful. Much better than my TA, and you aren't even being paid!

 

[General_Sax]

So, in general, if I'm given two variables and I need to graph the relationship between them I do:

a) use one as the y-axis and one as the x-axis.

b) use whatever method I need to in order to have a constant infront of the x-axis variable.

c) remember that even if certain values aren't plotted onto the graph, they are still part of the expression

Right?

 

[symbolipoint]

General Sax, if your TA's are helping you with this then what they are good for is uncertain. Do you at least have the explanation and description of the process for this kind of graphing in your laboratory manual?

HallsofIvy - I also remember log-log paper; and semi-log paper; and two and three cycle log paper; and polar coordinate paper.

Those three questions, General Sax:
a) use one as the y-axis and one as the x-axis.
Yes.

b) use whatever method I need to in order to have a constant infront of the x-axis variable.
What exactly do you mean? Maybe.

c) remember that even if certain values aren't plotted onto the graph, they are still part of the expression
That means nothing. What kind of values are and are not plotted? Why?
You need to identify your chosen variables; sometimes one of your variable is composed of other variables.

 

[General_Sax]

General Sax, if your TA's are helping you with this then what they are good for is uncertain.

We have two TA's in my physics lab, but my physics lab is divided into two sections, so the TA's are also 'split'. The TA for my section is very unfriendly, I feel he has a "this is beneath me" attitude. He always makes me feel so stupid. The other TA is nice and friendly, but always has such a large queue for questions that I often don't have the time to ask him questions ( we have to hand in our lab reports at the end of the lab).

Do you at least have the explanation and description of the process for this kind of graphing in your laboratory manual?

Yes, but it reads to me like programming syntax, and I always have had difficulty learning that way. I preform much better in a class or tutorial environment.

What exactly do you mean? Maybe.

re-write the equation so that I have the form y = mx + b. where m and b are compositions of variables.

That means nothing. What kind of values are and are not plotted? Why?
You need to identify your chosen variables; sometimes one of your variable is composed of other variables.

Your response was helpful.

Let's say that the variables were l ( l = lambda ) and u.

l = (k/f)*(T/u)^0.5


I could plot the graph using:

y = l
x = 1/ (u)^0.5

thus:

m = (T/u)^0.5 * 1/f

Thanks again for any time you contribute to me. I really do appreciate it. Happy thanksgiving!

Edit: m (slope) = T^0.5 * k/f