Question

A particle of mass $m$ is fixed to one end of a light of a light spring having force constant $k$ and unstretched length $l$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring, such that it rotates in a circle in gravity free space. Then the stretch in the spring is

• A. $\frac{ml{\omega }^2}{k-\omega m}$
• B. $\frac{ml{\omega }^2}{k-m{\omega }^2}$
• C. $\frac{ml{\omega }^2}{k+m{\omega }^2}$
• D. $\frac{ml{\omega }^2}{k+m\omega }$

Answer 1

The correct option is B $\frac{ml{\omega }^2}{k-m{\omega }^2}$



At elongated position $\left(x\right)$,

$F_{\text{radial}}=\frac{m{v}^2}{r}=mr{\omega }^2$
$\therefore kx=m(l+x){\omega }^2$
$(\because r=l+x\text{here})$
$kx=ml{\omega }^2+mx{\omega }^2$
$\therefore x=\frac{ml{\omega }^2}{k-m{\omega }^2}$

Hence, option $\left(B\right)$ is correct.