Stuck! Please help!

Hey guys,

Really stuck with this experiment we were doing...
We got an equation;

T^2= 4pi^2*L/g - 2pi^2*h/g

and we were asked to plot the values of T2 against h. I drew a line of best fit which was a staright line but I can't understand how I can make this equation into a y=mx + c form! In addition I was told to calculate the gradient and hence a value for g. I'm stuck can someone help me out? Bottomline is how do I make this into an y=mx + c form?

Thanks

Reply 1

AlphaNumeric

\underbrace{T^{2}}_{Y} = \underbrace{\frac{-2\pi^{2}}{g}}_{M}\underbrace{h}_{X} + \underbrace{\frac{4\pi^{2}L}{g}}_{C}

Reply 2

Stranded

AlphaNumeric

\underbrace{T^{2}}_{Y} = \underbrace{\frac{-2\pi^{2}}{g}}_{M}\underbrace{h}_{X} + \underbrace{\frac{4\pi^{2}L}{g}}_{C}

Nope that's not the equation;

it's 2pi^2*h/g :s:

Reply 3

Stranded

polynomial

Well as far as I can see if you want some help with this, you are going to have to tell us what the constants are. OK so ill assume that T^2 and h are your only variables, and therefore:

in the form y = mx + c

y = t^2

m = -2pi^2

c = 4pi^2*l/g.

However after adding my two penneth, I am contradicting someone with a degree in maths from cambridge and who is going a phd in physics, so im probably wrong.

How can the gradient be -2pi^2? :s:

Wherabouts the g?

Reply 4

Stranded

Well as far as I can see if you want some help with this, you are going to have to tell us what the constants are.

Well I was given the value of the constant L which was 0.5m. T^2 and h were the variablesand I'm asked to calculate the value of the constant g using my gradient. :s:

Reply 5

AlphaNumeric

The coefficent of h (which is equivalent to x) is

\, -\frac{2\pi^{2}}{g}

as far as the original post goes, so your reply quoting me is either wrong or your original post is wrong.

Reply 6

gorgeousguy

AlphaNumeric

The coefficent of h (which is equivalent to x) is

\, -\frac{2\pi^{2}}{g}

as far as the original post goes, so your reply quoting me is either wrong or your original post is wrong.

Iv'e changed my mind. I agree with this guy. look at what he has written and then use basic algebra and you will seehe is right.

Well I was given the value of the constant L which was 0.5m. T^2 and h were the variablesand I'm asked to calculate the value of the constant g using my gradient.

Now all oyu have to do is to takea gradient off your lovely straight line graph, then sub this in where h is, and you have yourself a simple little equation to solve.