HELP Mass per unit length equation

[mac b]

Homework Statement

Ok so using the standard equation v = Sqrt(T/Mu)

I want to find Mu as a gradient for the first fundamental.

I've made it into F x lambda = sqrt(T/Mu)

Then f x 2l = sqrt(T/Mu)

Then f= sqrt(T/4l^2 Mu)

As I'm having the tension held by a mass suspended over a pulley.

f= sqrt(mg/4l^2 Mu)

Note: Mu is used to represent mass per unit length.

T is TENSIONMY QUESTION IS:

How do I convert f^2= mg / (4l^2 Mu) into a form where I can get Mu as the gradient?

Thanks.

 

[mac b]

Any help would be appreciated.

 

[mac b]

Does it usually take this long to get a reply? I thought this forum was active.
If someone can PLEASE help it would really be appreciated.
Thanks.

 

[mac b]

I have concluded that this forum is dead.

 

[Doc Al]

How do I convert f^2= mg / (4l^2 Mu) into a form where I can get Mu as the gradient?

If you had an equation A = B/(Cx), could you solve for x? Same thing.

(μ is the mass/length. Not sure what you mean when you say you want it 'as the gradient'.)

 

[mac b]

If you had an equation A = B/(Cx), could you solve for x? Same thing.

(μ is the mass/length. Not sure what you mean when you say you want it 'as the gradient'.)

I want to make a graph with mass per unit length as a gradient, how do I manipulate that equation to do so?

 

[truesearch]

First of all there are several approaches to do this.
I would start by arranging the expression to have no square root.
If you start with v^2 =T/m (m = mass per unit length) can you arrange this to be
f^2 = T/(m4L^2)
There are only 2 experimental variables in your case I think...f and L
Do you knopw what to plot here that would have m as a part of the gradient? It can be tricky depending on your experience with graph plotting and equations of lines !