Determining acceleration due to gravity using a spring

Odrevan

Hey, so this seems like the most relevant place to post this. I'm a first year Physics student studying in UCC and we received the task of finding he acceleration due to gravity at the surface of the earth using a set of eight masses (whose values are unknown, all equal mass though), a stand with scale graduated in cm, a spring with attached holder for mass, and a stop clock. We have to make up the method ourselves and we were told you are allowed to consult friends or look up how to do it on the internet.

Unfortunately, I can't find any method how to do it, the closest I came was when I found an experiment that showed how to get the gravitational intensity using this apparatus, but that's not what I need and it required known masses for the calculations.

Could anybody point me in the right direction with a relevant link, or if anyone's int he know could they explain how to do it? Thanks in advance!

dlouth15

Well I think you have two equations to play around with.

Hooke's Law:
[latex]\displaystyle{x=kF} [/latex],

where [latex]x[/latex] is the extension from the equilibrium position when the force [latex]F[/latex] is applied.
And
[latex]\displaystyle{\omega^{2}=k/m} [/latex],

where [latex]\omega[/latex] is the angular frequency of the oscillations.

We know that the force [latex]F=mg[/latex] so Hooke's law becomes

[latex]\displaystyle{x=mgk} [/latex] and so [latex]\displaystyle{g=\frac{x}{mk}} [/latex]

Rearranging the other equation

[latex]\displaystyle{k=\frac{\omega^{2}}{m}
} [/latex]

Combining these two

[latex]\displaystyle{g=\frac{x}{m}\,\frac{m}{\omega^{2}}=\frac{x}{\omega^{2}}} [/latex]

Both [latex]x[/latex] and [latex]\omega^{2}[/latex] are are measurable. So what you could do is plot a graph of [latex]x[/latex] against [latex]\omega^{2}[/latex] for various quantities of weights. The slope should then give you your value for [latex]g[/latex].

Odrevan

Thanks for the reply! I'm just wondering though, I think you made a slgiht mistake in one of your equations. Shouldn't it be F=-kx and not F=-x/k as yours turns out to be?

dlouth15

Odrevan wrote: »

Thanks for the reply! I'm just wondering though, I think you made a slgiht mistake in one of your equations. Shouldn't it be F=-kx and not F=-x/k as yours turns out to be?

You are right. it is actually a big mistake. Thanks for pointing it out.

So

[latex]\displaystyle{F=-kx}[/latex]
implies
[latex]\displaystyle{mg=-kx}[/latex]
and so
[latex]\displaystyle{g=\frac{-kx}{m}}[/latex]

Then we have
[latex]\displaystyle{\omega^{2}=\frac{k}{m}}[/latex]

and so
[latex]\displaystyle{m=\frac{k}{\omega^{2}}}[/latex]

Substituting m into the expression for g

[latex]\displaystyle{g=-kx\frac{\omega^{2}}{k}=-x\omega^{2}}[/latex]

So instead of plotting x against [latex]\omega^2[/latex], plot x against [latex]1/\omega^2[/latex]. Slope should give g.

I believe that should work.

[latex]\displayvalue{x=g\left(\frac{1}{\omega^{2}}\right)+C}[/latex] is in the form y=mx+c with the slope m being the acceleration due to gravity [latex]g[/latex]. The [latex]1/\omega^2[/latex] values will be along the bottom of the chart and the displacements will be on the vertical axis.