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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 :

$(a)\;\large\frac{1}{2}mv^{2} \qquad(b)\;mv^{2} \qquad(c)\;\large\frac{1}{3}mv^{2} \qquad(d)\;\large\frac{1}{6}mv^{2}$

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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:

$T =\int_m\frac{1}{2}u^2\,dm$

Since the spring is uniform, $dm=\left(\frac{dy}{L}\right)m$, where $L$ is the length of the spring. Hence,

$T = \int_0^L\frac{1}{2}u^2\left(\frac{dy}{L}\right)m\!$
$=\frac{1}{2}\frac{m}{L}\int_0^L u^2\,dy$

The velocity of each mass element of the spring is directly proportional to its length, i.e. $u=\frac{vy}{L}$, from which it follows:

$T =\frac{1}{2}\frac{m}{L}\int_0^L\left(\frac{vy}{L}\right)^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\int_0^L y^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\left[\frac{y^3}{3}\right]_0^L$
$=\frac{1}{2}\frac{m}{3}v^2$

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