Question

A particle of mass $m$ is attached to a spring of natural length $L$ and spring constant $k$. If it is rotated in a horizontal plane with an angular speed $\omega$, then what will be the new length of the spring ?

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

Answer 1

The correct option is A $\frac{kL}{k-m{\omega }^2}$

Spring will start to elongate, so that required centripetal force can be provided by the spring force.


Let the elongation in the spring be $x$, so the radius of the particle for the circular path becomes $r=L+x$

$\therefore F_{\text{spring}}=kx$

Applying the equation of circular dynamics along radial direction:
$m{\omega }^{2}r=F_{\text{spring}}...\left(i\right)$
$\Rightarrow$ $m{\omega }^2\left(L+x\right)$$=kx$
$\Rightarrow$$m{\omega }^{2}L+m{\omega }^{2}x=kx$
$\therefore x=\frac{m{\omega }^{2}L}{k-m{\omega }^2}$

New length of the spring is:
$r=L+x=\frac{kL}{k-m{\omega }^2}$