Question

A particle of mass $m$ is fixed to one end of a light spring of force constant $k$ and unstretched length $l$. The system is rotated about the other end of the spring with an angular velocity $\omega$, in gravity free space. The increase in length of the spring will be:

• A. $\frac{m \omega^{2} l}{k}$
• B. $\frac{m \omega^{2} l}{k-m \omega^{2}}$
• C. $\frac{m \omega^{2} l}{k+m \omega^{2}}$
• D. None of these

Answer 1

Answer: (B)

Let increase in lengh of spring is $x$

At equilibrium,
$k x =m \omega^{2} x+m \omega^{2} l$
$k x-m \omega^{2} x =m \omega^{2} l $
$x =\frac{m \omega^{2} l}{k-m \omega^{2}}$