t^2 against mass graph help

Can someone tell me what I've done wrong here? I know my graph is supposed to be a straight line through the origin because t^2 is proportional to mass, but my graph isn't and I can't see what I've done wrong.

Reply 1

JustSomeGuy:/

14

Original post by kswales1

Can someone tell me what I've done wrong here? I know my graph is supposed to be a straight line through the origin because t^2 is proportional to mass, but my graph isn't and I can't see what I've done wrong.

As long as it forms a straight line you're completely okay. Even though it SHOULD go through the origin, there is bound to be some kind of systematic error when you were taking your results and hence it will be off from the origin by some amount. It's extremely unlikely that you would get the straight line going through the origin. Your practical looks fine. Now if you try change your results to make it go through the origin, then you've done something wrong.
Remember that the relationship that you use to derive that t^2 is proportional to m is dependent on the small angle approximation as well as not taking external forces such as air resistance into account (although it would probably be negligible albeit still produce some shift) so having not perfect results is actually ideal. TLDR you're in the clear your graph is fine as long as your value of g is somewhat close to the real value which it should be. When I did the practical myself me, as well as everyone else in my class, all had straight lines but they didn't go through the origin however our values of g were close.

Reply 2

Stonebridge

13

You haven't said what this experiment is - I assume it's a mass on a spring?
You also haven't explained what 'mass added' means.
T² is proportional to the total mass - not the mass added.
Was there already a mass there, to which you have added more?
If so then the graph doesn't go through the origin.

But what I think is happening is:
The spring itself has mass, so the total mass in your equation is the 'mass added' plus the mass of the spring.
For this reason, when that 'added' mass is zero, the total mass is not, there is still the mass of the spring itself.

Reply 3

kswales1

OP

8

Original post by Stonebridge

You haven't said what this experiment is - I assume it's a mass on a spring?
You also haven't explained what 'mass added' means.
T² is proportional to the total mass - not the mass added.
Was there already a mass there, to which you have added more?
If so then the graph doesn't go through the origin.

But what I think is happening is:
The spring itself has mass, so the total mass in your equation is the 'mass added' plus the mass of the spring.
For this reason, when that 'added' mass is zero, the total mass is not, there is still the mass of the spring itself.

Ah yes, sorry. This is the a-level physics core practical 16, which is to determine the value of an unknown mass using the resonant frequencies of known masses on a spring.
The question I've been given doesn't mention the mass of the spring at all though - all it gives me is a couple of tables. I've attached them below

Reply 4

kswales1

OP

8

Original post by kswales1

Ah yes, sorry. This is the a-level physics core practical 16, which is to determine the value of an unknown mass using the resonant frequencies of known masses on a spring.
The question I've been given doesn't mention the mass of the spring at all though - all it gives me is a couple of tables. I've attached them below

Never mind!! I've just reread the practical document which says it isn't supposed to go through the origin- my mistake, I must've missed it when I read it before

Reply 5

Stonebridge

13

Original post by kswales1

Never mind!! I've just reread the practical document which says it isn't supposed to go through the origin- my mistake, I must've missed it when I read it before

Yes it is as I thought. The mass of the spring can be found from the intercept as given in that last part.