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I am supposed to arrive at the following conclusion for total energy for a spring mass system with mass $M$ spring constant $k$ and spring mass $m$.

$$E = \frac{1}{2}(M + \frac{m}{3})v^2 + \frac{1}{2}kx^2$$

I consider the resting length of the spring $l_o$. An infinitesimal amount of spring will have $\frac{m}{l_0}dl$. I am having troubles relating the velocity of the mass $M$ with the velocity of this small amount of spring.... If I can do this, then the kinetic energy of the spring becomes apparent. Could someone help?

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A diagram would be easier, but assuming the mass M is at one end with a fixed wall on the other end, the spring end touching M will move the same as M, the other end will not move at all, and there should be a linear variation in between. You can integrate.
Most spring-mass systems neglect the spring mass, i.e. M >> m

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  • $\begingroup$ Why is there linear variation is my question $\endgroup$
    – A. Radek Martinez
    Commented Sep 16, 2021 at 16:49

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