A particle of mass `m` is fixed to one end of a massless spring of spring constant `k` and natural length `l_(0)`. The system is rotated about the other end of the spring with an angular velocity `omega` ub gravity-free space. The final length of spring is

Options

A. `(momega^(2)t)/(k)`
B. `(momega^(2)l)/(k+momega^(2))`
C. `(momega^(2)l)/(k-momega^(2))`
D. none of the above

Answer

C

Explanation

Let `dl` be the increase in length of spring. It means the particle will move on a circular path of radius `(l+dl)` and the restoring force due to spring `=(kdl)` will provide the required centripetal force
`:. m omega^(2)(l+dl)=k dl` or `dl=(momega^(2)l)/(k-momega^(2))`