Motion along length of a spring What's relation between velocity of each part of a massive spring undergoing Simple Harmonic Motion.
 A: Let $\xi$ be an unextended length (coordinate) measured along the spring from its left end, and let u represent the displacement to the right of the particular cross section situated at unextended coordinate $\xi$ when the spring is stretched.  If the spring is stretched uniformly, then $$\frac{du}{d\xi}=\frac{u}{\xi}=\frac{U}{L}$$where U is the displacement of the right end of the spring at $\xi=L$ and $du/d\xi$ is the uniform axial strain.  Also, for uniform stretching of the spring, the tension in the spring is uniform, and given by:
$$T=kU=(kL)\frac{du}{d\xi}$$where k is the overall spring constant.
Under circumstances where the spring is stretched non-uniformly, the local tension in the spring (at unstretched coordinate $\xi$) is still determined by the local strain, and given by:$$T(\xi,t)=(kL)\frac{\partial u}{\partial \xi}$$where the above equation assumes that the displacement and tension are also changing with time.
If we perform a differential force balance on the section of the spring between $\xi$ to $\xi+\Delta \xi$, we obtain:$$\frac{M}{L}\frac{\partial^2 u}{\partial t^2}\Delta \xi=T(t,\xi+\Delta \xi)-T(t,\xi)$$Taking the limit as $\Delta \xi$ approaches zero then yields:
$$\frac{M}{L}\frac{\partial^2 u}{\partial t^2}=\frac{\partial T}{\partial \xi}=(kL)\frac{\partial ^2u}{\partial \xi^2}$$or equivalently,
$$\frac{\partial^2 u}{\partial t^2}=\frac{kL^2}{M}\frac{\partial ^2u}{\partial \xi^2}$$This is the longitudinal wave equation for the spring.
