One end of a massless spring of spring constant 100 N/m and natural 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.

Explanation

The restoring force in the spring provides the centripetal force to the mass.
If .x. is the extension in the spring, then
`Kx=m(L+x)omega^(2)`
Putting the values,
we get
`x=(0.5xx4xx0.5)/(100-0.5xx4)`
`x=1 cm`