# 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 :

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

i was having problem with the smae question since yesterday.... i found it here...

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:

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

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

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