One end of a massless spring of spring constant 100 N/m and natural length 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 (in cm).

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2/r.The horizontal force on the particle is due to the spring and equals kl, where l is the elongation and k is the spring constant. Thus, kl=mv2r=mω2r=mω2l0+lHere co is the angular velocity l0 is the natural length (0.5 m) and l0+ I is the total length of the spring which is also the radius of the circle along which the particle moves. Thus k−mω2l=mω2l0or, l=mω2l0k−mω2Putting the values, l=0.5×4×0.5100−0.5×4m=1100m=1cm