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Is there any limitation in acceleration and frequency of a spring. Please, imagine a horizontal spring with an object of mass $m$ attached in the free side and the friction is neglected. $x''=\frac{k}{m}x$ and $f=\frac{1}{2\pi}\sqrt{k/m}$. Imagine now that $m$ is very small (even a dust particle). Will the frequency be very very high? What is the physical explanation of any limitation (atomic bonds, crystal properties)?

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    $\begingroup$ If the mass m is very small then the mass of the spring itself will start playing a role in determining the dynamics and you will not be able to write the simple expression for frequency (since that assumes an ideal massless spring). If you also keep decreasing the mass of the spring to keep it much smaller than the mass then the k will keep changing. If this change is such that the k decreases for lighter springs(which seems plausible) then (since m is also decreasing at the same time) the ratio $\sqrt(k/m)$ might not change too much keeping you from obtaining infinite frequency. $\endgroup$
    – gautam1168
    Apr 27, 2015 at 9:37
  • $\begingroup$ If the spring could be a simple beam, the stiffness would be: $k=\frac{S_{cross-section}.E_{Hooke}}{L}$. It seems that it could be independent to the mass in a particular geometry. $\endgroup$
    – aayyachi
    Apr 28, 2015 at 23:20
  • $\begingroup$ For a rod you will not be able to keep the mass of the spring small and the mass of the load even smaller than that beyond a limit. But for this case you can easily find how high the frequency can go, because we make vibrating rods all the time. Every stringed instrument has this, a piano, a guitar, a violin... Just look for the highest frequency producible. $\endgroup$
    – gautam1168
    Apr 29, 2015 at 16:28

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