Question

A particle of mass m is moving with a constant speed v in a circular path in a smooth horizontal plane (plane of paper) by a spring force as shown in Fig. If the natural length of the spring is $l_0$ and stiffness of the spring is k, find the elongation of the spring.

Answer 1

If the particle executes uniform circular motion, its speed must be uniform because there is no tangential force to speed it up. Here, the spring force $kx$acting on the particle is the centripetal force caused by the elongation x of the spring.
Equation of motion:
$\sum F_r=F_{sp}m{a}_r$ (where R= radius of circular path)
But centripetal acceleration
$a_r=\frac{v^2}{R}$ (where R = radius of circular path)
Radius of rotation: $R=l_o+x$
$|F_{sp}|=kx$
Put in (i), $kx=\frac{m{v}^2}{(l_0+x)}\Rightarrow k{x}^2+k{l}_{0}x-m{v}^2=0$

This gives $x=\frac{\sqrt{k^2{l}_0^2+4m{v}^{2}k}-k{l}_0}{2k}$