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Experiment: determine the mass of a spring

by moonkey
Tags: determine, experiment, mass, spring
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moonkey
#1
Nov26-10, 01:22 PM
P: 26
1. The problem statement, all variables and given/known data
Find the mass of the spring, given the following apparatus:
1. a spring, attached to a stand,
2. a set of graduated 10g masses,
3. a stop-clock


2. Relevant equations



3. The attempt at a solution
I just can't really get my head around how to do it without being able to measure the displacement from equilibrium
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tiny-tim
#2
Nov26-10, 05:19 PM
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hi moonkey! welcome to pf!

(what are "graduated 10g masses" ? )

hmm you're obviously intended to time something

how about hanging weights on the spring and timing the bounces?
moonkey
#3
Nov26-10, 05:49 PM
P: 26
Hi tiny-tim. Thanks for the reply. using the timer I would be able to work out the period of the oscillations which would in turn allow me to work out w(omega) using the formula

T=2Pi/w

From there I know that omega squared is equal to k/M(mass of spring plus any additional mass) but I don't know k or the mass of the spring. So i'm thinking that there must be some way of either calculating k on its own or else I'm going about it all wrong.

tiny-tim
#4
Nov27-10, 02:43 AM
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Experiment: determine the mass of a spring

hi moonkey!

(just got up )
Quote Quote by moonkey View Post
From there I know that omega squared is equal to k/M(mass of spring plus any additional mass) but I don't know k or the mass of the spring.
(have an omega: ω and a square-root: √ )

so do it for different values of M + m, and draw a graph
moonkey
#5
Nov27-10, 09:59 AM
P: 26
Ok, just had an idea. So,

T=4π(m/k) and k is constant so

k=4πm/T

So where m is (Ms+Mn) I came up with this:

k=4π(Ms+Mn-1)/(Tn-1) , k=4π(Ms+Mn)/(Tn)

k=k => (Ms+Mn-1)/(Tn-1)=(Ms+Mn)/(Tn)

(Tn)(Ms+Mn-1)=(Tn-1)(Ms+Mn), which turns out to be

Ms=[(Tn-1)Mn-(Tn)Mn-1]/[(Tn)-(Tn-1)]

If I do this for a couple of different values of Mn using the 10g masses I'm given then take the average for Ms.

Does this sound like it would work? I haven't made any drastic mistakes or assumptions or anything?
JolileChat
#6
Nov27-10, 10:34 AM
P: 32
When we consider the mass of the spring, the equivalent mass is equal to the particle mass plus one third of the spring mass.

Let us perform two tests, with two mass configurations.

For the mass configuration #1, you will attach a mass [tex]M_1[/tex] to the spring. For the mass configuration #2, the attached mass will be [tex]M_2[/tex].

For each mass configuration, find the natural period of oscillations and the corresponding natural frequency.

For the first mass configuration, we have:

[tex]
f_1=\frac{1}{2 \pi} \sqrt{\frac{k}{m_{eq1}}}
[/tex]

and

[tex]
m_{eq1} = M_1 + \frac{1}{3} M_{spring}
[/tex]

We don't know the stiffness and the mass of the spring. So let us perform another test with the second mass configuration:

[tex]
f_2=\frac{1}{2 \pi} \sqrt{\frac{k}{m_{eq2}}}
[/tex]

with

[tex]
m_{eq2} = M_2 + \frac{1}{3} M_{spring}
[/tex]


From the natural frequencies equations, we can find:

[tex]
{f_1}^2 {m_{eq1}} = {f_2}^2 {m_{eq2}}
[/tex]

This last equation will allow us to find out the value of the mass of the spring.

Please let me know if it works
moonkey
#7
Nov27-10, 11:39 AM
P: 26
Thanks alot JolileChat.

I'm sure that will work perfectly. That's really put me on the right track. I'll go and look it up so that I fully understand it. No point doing an experiment if I'm not fully aware of why I'm doing what I'm doing.

Thanks again


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