Solving Frequency Equation to Match y=mx+c

[l.e.d.]

How do you get to change the equation

f = 1/2L $$\sqrt{}$$ P/$$\mu$$

to match the equation y=mx+c?


I tried to solve it but only came up with

$$\mu$$ = 1P / f2L (where the 1 and the L are squared)

and still that is not in the form of y=mx +c!

Any help is greatly appreciated. Thanks

 

[chaoseverlasting]

Hi, welcome to PF. What exactly is the question? I don't think you can get a linear equation from what you have.

 

[l.e.d.]

Thanks chaoseverlasting! actually the problem is about an experiment where there's a wire carrying a current and its placed in a magnetic field. The current is AC and therefore the force of the wire will reverse direrction with the same frequency of the current. This makes the wire resonate. The equation i wrote above is given for the frequency.

It sasys that it's required to measure the mass per unit length of this wire and it says that to do this values of the frequency and wire lenth need to be taken as they are made to vary. A table is given for these values.

Then there's this question which asks to explain how you would use the tabulated data to plot a suitable graph in order to determine the mass per unit length.

Now, i thought that maybe I should get that equation in the y=mx+c form so that I can eventually plot the graph. I can't manage to do it though and that's the problem..Actually, what I really need to figure out is how I'm going to plot the graph not the equation bit.

Hope i cleared some things up here, if it's still vague just ask and maybe I can explain further!

Thanks :-)

 

[Redbelly98]

Hi,

Since f is inversely proportional to L, you won't get a nice linear relation by plotting f vs. L.

However, since f is directly proportional to (1/L), you could plot or make a table of f and (1/L), instead of L.

The graph will then be a straight line, and the slope will be related to μ somehow. Also, you will need to know what P is.

Note to others: the equation in the OP can be written (perhaps more clearly) as

$$\frac{\sqrt{P/\mu}}{2L}$$

 

[l.e.d.]

Thanks for your reply! Yes, I understand what you mean and I think that I have to draw a graph of frequency against 1/L. The problem now is how I will get to find μ because I don't know what the gradient of such a graph will give me. I think, therefore, that before I plot the graph I need to arrange the equation such that it I will be able to know beforehand what the gradient of the graph will give me.

Thanks:-)

 

[Redbelly98]

$$f = \frac{\sqrt{P/\mu}}{2}\cdot \frac{1}{L} \ + \ 0$$

That should help you see what corresponds to "x", "y", "m" and "c" in

y = m^.x + c​

 

[l.e.d.]

Thanks Redbelly98! I think I can work it out now! And sry for the late reply..Thankyou all! :)