d
We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral:

\(T=\int_{m} \frac{1}{2} u^{2} d m\)

since the spring is uniform, \(d m=\left(\frac{d y}{L}\right) m,\) where \(L\) is the length of the spring. Hence,

\(T =\int_{0}^{L} \frac{1}{2} u^{2}\left(\frac{d y}{L}\right) m\)

\(=\frac{1}{2} \frac{m}{L} \int_{0}^{L} u^{2} d y\)

The velocity of each mass element of the spring is directly proportional to its length, i.e.

\(u=\frac{v y}{L},\) from which it follows\(:\)

\({T=\frac{1}{2} \frac{m}{L} \int_{0}^{L}\left(\frac{v y}{L}\right)^{2} d y}\)

\({=\frac{1}{2} \frac{m}{L^{3}} v^{2} \int_{0}^{L} y^{2} d y}\)

\({=\frac{1}{2} \frac{m}{L^{3}} v^{2}\left[\frac{y^{3}}{3}\right]_{0}^{L}}\)

\({=\frac{1}{2} \frac{m}{3} v^{2}}\)