Pendulum experiment (1 Viewer)

[adomad]

when you draw you line of best fit, can you use the gradient to find the value of 'g' becuase when the formula is re-arranged, you get T^2 =((4*pi)l)/g

in terms of x and y, T^2 =y and l=x therefore, from y=mx, the gradiaent will equal 4*pi on g. when i do this i get g to be 11.25ms^-2??? or even if i pick a random point on the line of best fit!!!


help!

cheers,
ADomad

 

[LouiseFaz]

thats the idea of it, the fact that you can rearrange the formulae into the form y=mx + b and get the gradient from your line of best fit in your graph. if you got 11. something it could just be experimental errors which you could discuss in your discussion.

I'm not sure if this helps if you need anything else i have my whole formal write up of this experiment, with correct values and formulaes.

 

[youngminii]

Yes, often you are actually required to find the value of g by using the gradient.
This means that you have to draw your graph to scale, otherwise your value may not be correct.
In a practical assessment task, there is usually a range of answers that will get you the mark. ie. a 10% difference from 9.8ms^-2 may earn you full marks, a 20% difference may earn you half marks and any more will earn you nothing.
In my school, a 11.25ms^-2 would have gotten us nothing, so make sure the experiment is done accurately.

That said, later you talk about the factors that could have led to the error: human/parallax error, accuracy of apparatus used, air resistance etc.

 

[adomad]

Yes, often you are actually required to find the value of g by using the gradient.
This means that you have to draw your graph to scale, otherwise your value may not be correct.
In a practical assessment task, there is usually a range of answers that will get you the mark. ie. a 10% difference from 9.8ms^-2 may earn you full marks, a 20% difference may earn you half marks and any more will earn you nothing.
In my school, a 11.25ms^-2 would have gotten us nothing, so make sure the experiment is done accurately.

That said, later you talk about the factors that could have led to the error: human/parallax error, accuracy of apparatus used, air resistance etc.

Im pretty sure that would also apply at my school. this is how i did the experiment. i change the length of the string, i put the mass in motion and time the period of 10 complete swings from the top of the motion. i then get the average period of ONE swing .i repeated it like 4 times (and removing outliers). when i plug in these separate values, i get 'g' being 9.7,9.65 (pretty close to 9.8) i then draw the graph and the line of best fit

i then measure the gradient using rise over run and then re-arrange the formula for pendulums and as i said, when i use the gradient to find 'g' i get something like 11. does there have to be something special about the graph?

 

[youngminii]

You are not meant to use T^2 as the subject. The gradient of the line is rise/run, or T^2/L, not 4pi/g.

Right now you're using T^2 = m(L)
But m = T^2/L
So you're actually using T^2 = T^2
But you're substituting the right hand side as 4pi/g
So you're saying T^2 = 4pi/g
g = 4pi/T^2
This whole process is inaccurate. Let me guess, your T was somewhere around 1.14?

This is how you're meant to use the gradient:
Okay if we rearrange the equation, it becomes g = 4pi(L/T^2)
Now, this means g = k(L/T^2) (where k is the constant, 4pi).

Now as you said, by convention, L is on the x-axis and T^2 is on the y-axis. But now we have to use the gradient (often required). But the gradient is rise over run, or y-axis/x-axis. Hence, the gradient turns out to be T^2/L, which is not what we're looking for.

As such, you can swap around the axes but you have to state why you did it.
Or you can just say g = k(1/m), where m is the gradient.

Now, just sub in k = 4pi, and m as the gradient and you're done.

 

[adomad]

i have T^2 as my y-axis and L as my y axis. i then do rise over run to get the gradient.

Right now you're using T^2 = m(L)
But m = T^2/L

these are the same thing but you just cross multiplied. hmm ill try graphing it again

 

[untouchablecuz]

 

[youngminii]

k's meant to be the constant.. m is usually the gradient <.<

 

[Aerath]

Doesn't really matter though.

 

[untouchablecuz]

 

[js992]

thats the idea of it, the fact that you can rearrange the formulae into the form y=mx + b and get the gradient from your line of best fit in your graph. if you got 11. something it could just be experimental errors which you could discuss in your discussion.

I'm not sure if this helps if you need anything else i have my whole formal write up of this experiment, with correct values and formulaes.

May not always be the case. If it's for a practical exam try and get it as accurate as possible. In our school, teachers only gave marks for those who got a value within 0.1 of 9.8ms^-2 ==..... so 9.7-9.9 was correct.