# Quick Question ( y = mx + c )

1. Feb 26, 2008

### _Mayday_

[SOLVED] Quick Question ( y = mx + c )

Okay, this shouldn't take long. Would it be okay if I did not show any working as I want to know how one equation related to another, which will then aid me in trying to solve the actual question. $$T^2$$ is the y-axis and $$d^2$$ is along the x-axis.

The equation of the straight line I have drawn is...

$$T^2=(0.10K)d^2+0.042Km$$

I am asked to find a value for for K, and then to calculate a value for m.

m = the effecitve mass
K = the constant for the system

Now I would have thought I would need to consturct a similtaneous equation, but I have a feeling that this equation can be linked to...

$$y=mx+c$$

Would x and y in this equation relate to my axis? What would K be? If I put the intial equation in the form $$y=mx+c$$ will it work? If so here is how I would have thought it should look.

$$T^2=m\times d^2 + c$$

Now I can put in the values I have for $$T^2$$ and $$d^2$$ so...

$$3.13=m\times 0.026 + c$$

I am still not sure what c is....is it K?

Any help would be great. If this does not make sense please ask for anything that isn't there and I will go and find it :tongue:

2. Feb 26, 2008

### e(ho0n3

If you know T2 and d2, you can rewrite the first equation you wrote as:

K = T2 / (0.10d2 + 0.042m)

This equation has the form y = a / (b + x), not y = mx + b.

This assumes that K depends on m.

3. Feb 26, 2008

### _Mayday_

The problem is that I do not know the values for m or k, so I cannot calculate either of them from that.

4. Feb 26, 2008

### cristo

Staff Emeritus
You can calculate K from the gradient. Your equation is $T^2=(0.1K)d^2+0.042Km$ which is, on letting y=T^2 and x=d^2 in the form $y=nx+c$ where n=0.1K and c=0.042Km. Now, what does n represent? Can you obtain this from the graph? If so, then you have K and can use your y intercept to find m.

5. Feb 26, 2008

### _Mayday_

What is n? n isn't in any of my calculations? I know what y and x represent but I do not know what y stands for. m is the gradient...

6. Feb 26, 2008

### cristo

Staff Emeritus
I changed n to be the gradient when I wrote the equation y=nx+c. You've got two m's and so that's bound to be confusing. Anyway, the important point is that you have the equation in the form y=(gradient)x + (intercept). Compare this to $T^2=(0.1K)d^2+0.042Km,$ with y=T^2 and x=d^2. What is the gradient of your graph? What is the y-intercept?

7. Feb 26, 2008

### _Mayday_

$$Gradient = \frac{T^2}{d^2}$$

I calculated this at 3.3, I did this by drawing a triangle onto the graph to determine this value.

8. Feb 26, 2008

### cristo

Staff Emeritus
Ok good, but I think you mean gradient= delta(T^2)/delta(d^2). Now you have that the gradient=3.3=0.1K. From this you should be able to find K.

9. Feb 26, 2008

### _Mayday_

Thanks for your time Cristo, much appreciated.

10. Feb 26, 2008

### cristo

Staff Emeritus
You're welcome!!