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A spring of unstretched length $l$ has a mass $m$ with one end fixed to a rigid support .Assuming spring to be made of a uniform wire, the kinetic energy possessed by it if its free end is pulled with uniform velocity $v$ is
$\frac {1}{2}\,mv^2$
$mv^2$
$\frac {1}{3}\,mv^2$
$\frac {1}{6}\,mv^2$
Solution
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$:$
${T=\frac{1}{2} \frac{m}{L} \int_{0}^{L}\left(\frac{v y}{L}\right)^{2} d y}$
${=\frac{1}{2} \frac{m}{L^{3}} v^{2} \int_{0}^{L} y^{2} d y}$
${=\frac{1}{2} \frac{m}{L^{3}} v^{2}\left[\frac{y^{3}}{3}\right]_{0}^{L}}$
${=\frac{1}{2} \frac{m}{3} v^{2}}$
