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My physics professor recently covered the concept of simple harmonic motion and touched on vertical springs. However he mentioned that the spring (with a mass attached to its end) only undergoes simple harmonic motion if it is massless. However I am unsure why making the spring have non-negligible mass would affect its simple harmonic motion hence I am seeking clarification. Thank you!

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    $\begingroup$ Effective mass (spring–mass system) $\endgroup$ – Farcher Jul 15 '20 at 8:00
  • $\begingroup$ It's a little unusual to ask questions in the answer, but in this case the answerer is waiting for some feedback from you, and they appear to be pretty flexible and willing to help. $\endgroup$ – uhoh Nov 25 '20 at 15:01
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Do you know about lumped systems? (Discrete ropes etc)

Do you know the equation $$ \omega = 2\sqrt{\frac{k'}{m}}sin(\frac{ka}{2}) $$ ?

If no, we can derive with your, i will postpone until your answer the derivation. If you know: $$ \omega = 2\sqrt{\frac{k'}{m}}sin(\frac{ka}{2}) $$ $$ a \rightarrow 0 $$ $$ \omega = 2\sqrt{\frac{k'}{m}}(\frac{ka}{2}) $$ but $$ k'a = KL $$ $$ \omega ^2 = K\frac{L}{\rho }k^2 $$

We can go on and derive a equation for oscillations which will show to us that the system actually acts like the mass on the spring was a fixed end. We will end up with oscillation of the spring than with that of the system itself.

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