Can the formula for G be used to measure gravity using a mass and spring system?

[sharpy]

ok, I am trying to measure thevalue of g using a mass attatched to a spring. i have the period of bounce, the mass of the..well...mass, and the elastic constant of the spring. i have been informed its relatively easy to find g using this setup but I am struggling. iv spent the last 3 nights either rearranging formulae or serching the internet and my brains dead and i really need a break from this. anyone? any help like atall is massivley appreciated. though as usual the answers probably pretty obvious and I am jst being blind. but thanks anyway.

 

[HallsofIvy]

The period of motion is not dependent on g and so cannot give a value for it.
The motion of a weight on a spring, assuming it is hanging, is given by the differential equation
m\frac{d^2 x}{dt^2}= -kx- mg
or
\frac{d^2 x}{dt^2}= -\frac{k}{m}- g

The general solution to that is
x(t)= C_1 cos(t\sqrt{\frac{k}{m}})+ C_2 sin(t\sqrt{\frac{k}{m})- frac{g}{k}
The period is given by
2\pi\sqrt{\frac{m}{k}}
with no dependence on g.

Of course, you have mg= kx where x is the "stretch" when you hang the weight on the spring initially so g= kx/m. That's done with the spring motionless.

 

[sharpy]

...but the period is dependant on the mass which can be affected by altering gravity?
i know I am grasping at nothing but I am pretty sure my friend managed this and I am pretty stubborn about not failing when I've wasted so much on this

 

[FredGarvin]

How about

T = 2 \pi \sqrt{\frac{L}{g}

EDIT: Whoops. Never mind. If I passed basic reading comprehension I would have noticed that you said a mass on a spring, not a string. Don't mind me.

 

[sharpy]

would this work? what could i use for l?
i still have mass and spring elastic to use up.
i didnt really measure anything else

 

[Doc Al]

How about

T = 2 \pi \sqrt{\frac{L}{g}

Which is for a simple pendulum, not a mass on a spring (as FredGarvin certainly knows).

 

[sharpy]

simple pendulum indeed, however ill use anything. i am definetley that desperate.

 

[sharpy]

surley it would be possible somehow to use the elastic constant of the spring to find the restoring force which would be somewhat the same as the force of gravity acting on the mass? no? anyone?

 

[lpfr]

Maybe the catch is to use the period Tof the oscillation and the lengthening \ell of the string:
g = \ell\left({2\pi\over T}\right)^2
(if I didn't did a mistake)

 

[sharpy]

omg thankyou I am getting aroundabout 9 for my value of g, and since my technique can be considered fairly crude i will have to assume that this is the best formulae ever concived. thankyou much, and my mum thanks you as i hae stopped sulking about not being able to work this out. thanks to you all though, i realize your time is important so thanks for even reading the post, but to the guy who posted that formulae i owe my life. :!)
... any chance of a proof for this?
but really thankyou
im so happy
my easter holidays might not be ruined afterall

 

[Cyrus]

......no. What he told you is wrong and you got a close answer by dumb luck. The period of oscillations has nothing to do with g. Read the very first post and what HallsofIvy told you.

 

[sharpy]

:( i feared this day would come
are you sure?
i know someone who has done this before and I am convinced that there's a way. sure it was dumb luck? i tried it on all of my readings and they were all pretty much right. its late where i am and I am tired and i can't work out if they all would be right from just luck
could this not work then since the mass in the spring constant cancels with the mass in the shm equation so that i can match the extension to the period o the same masses? i really don't know. ill be quiet now.

 

[Cyrus]

Read HallsofIvys post. Gravity does not play a role in SHM of a spring. Are you talking about a spring or a string?

 

[sharpy]

...yeh i read that and was disheartened, however
t=2pi root m/k
and
w^2 = k/m = g/e
so can't we change this to
t=2pi root e/g
...which means that the period can be dependant on gravity?
no?
i know hallsofivy said that it didnt but why couldn't this work?
thanks for your patience
and the fast response that was awesome cheers

and yeh sorry definately a sPring

 

[Cyrus]

w^2 = k/m = g/e

where did this come from?

The answer is no. The spring and mass does not depend on gravity, ever. You can take it to the moon and you will get the same answer. You can turn gravity off and still get the same answer.

 

[sharpy]

...ok... erm...
i have it written in a book, and i think its because the whole hookes law thing gives f = kx and f = ma gives xk = ma
a as the acceleration can substitute for g
and e for extension because its easier that way
so we have
mg = ke
and then we rearrange or something to get
k/m = g/e
...maybe?

 

[lpfr]

I do not know if I make it clear.
When you hang a mass m the spring lengthens of \ell:
mg=k\ell
If you put the mass and the spring in oscillation the period is:
T=2\pi\sqrt{m\over k}
then:
{k\over m}= \left({2\pi\over T}\right)^2
From the first formula:
g=\ell{k\over m}
g = \ell \left({2\pi\over T}\right)^2
OK?

 

[Cyrus]

This is not a pendulum lpfr. (And please delete one of your double posts.)

 

[lpfr]

Do I say it was?
It is a mass at the extremity of a spring and it oscillates VERTICALLY!

 

[Cyrus]

I see what your doing now, but its still rather meaningless because your calling \ell, what should really be called \Delta L and that is a STATIC measurement of the physical system. Its exactly measuring the displacement of the mass. You don't need SHM to do that. All you have to do is put the mass on and measure the deflection. Its a static experiment.

see what I mean?

 

[sharpy]

...well i don't know how to do a fancy l anyhow
so your agreeing it will work?
or do the physiics wars continue?

 

[Cyrus]

Although...that is a very clever way of finding g by eliminating the need to find the spring constant k altogether!


(Though it would be much more inaccurate)

 

[sharpy]

...so its all good?

 

[Cyrus]

Yes, that's fine, but l is the static deflection of the mass on the spring.

NOT the entire length of the spring.

 

[sharpy]

yeh cool l is the extension when loaded yesss.
i was going to try the static one but then the forces and stuff would require me changing mass to weight apparently? and id have to use g for that which is why we went for the difficult one
well anyway thankyou very much guys youve been wonderfull.

 

[Cyrus]

Say that again?

make sure you use the static deflection.

 

[sharpy]

hahaha never mind I am babaling now.
thanks for the help.
goodnight.

 

[DyslexicHobo]

Are you looking for g or G? I'm assuming you mean gravity, not the gravitational constant, by reading some posts.

Anyway, it's pretty much just avoiding the situation, but you could just put a mass on a spring and measure how much the spring has changed in size (assuming you know the k for that spring).

 

[sharpy]

i was trying to find the gravitational field strength.
i did however make a mistake stating that i had the elastic constant , as i only had a table of mass against extension, and I am now under the impression that the elastic constant requires weight against extension? so to find the elastic constant id need to know g which I am trying to find.
so i cant.

 

[lpfr]

If you use the formula that I posted, the only thing you need to measure is the stretching (the lengthening) of the spring when you hang the mass.
To know the weight you need to know g and this is what you are looking for.
You do not need to have a table of extensions against masses. You just need the extension for the mass you use to measure the period of oscillation.

Do no expect to obtain g with high precision. The measure of the extension is not very precise and the measure on period is not much better.

 

[sharpy]

cheers lpfr i allready have and its doing great.
i do need my table of mass against extension as I am trying repeating with a range of mass so the extension and period change with the mass and i need to know what the extension is with a certain mass.
but that's awesome cheers.