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 M_1 to the spring. For the mass configuration #2, the attached mass will be M_2.
For each mass configuration, find the natural period of oscillations and the corresponding natural frequency.
For the first mass configuration, we have:
<br />
f_1=\frac{1}{2 \pi} \sqrt{\frac{k}{m_{eq1}}}<br />
and
<br />
m_{eq1} = M_1 + \frac{1}{3} M_{spring}<br />
We don't know the stiffness and the mass of the spring. So let us perform another test with the second mass configuration:
<br />
f_2=\frac{1}{2 \pi} \sqrt{\frac{k}{m_{eq2}}}<br />
with
<br />
m_{eq2} = M_2 + \frac{1}{3} M_{spring}<br />From the natural frequencies equations, we can find:
<br />
{f_1}^2 {m_{eq1}} = {f_2}^2 {m_{eq2}}<br />
This last equation will allow us to find out the value of the mass of the spring.
Please let me know if it works
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… you're obviously intended to time something …
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