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)l_(0))/(k)`
B. `(momega^(2)l_(0))/(k-momega^(2))`
C. `(kl_(0))/(k-momega^(2))`
D. `(momega^(2)l_(0))/(k+momega^(2))`

Answer

C

Explanation

`kx=momega^(2)(l_(0)+x)=momega^(2)l_(0)+momega^(2)x`
`(k-momega^(2))x=momega^(2)l_(0)`
`x=(momega^(2)l_(0))/(k-momega^(2))`
Final length=`l_(0)+x=l_(0)+(momega^(2)l_(0))/(k-momega^(2))=(kl_(0))/(k-momega^(2))`