A body of mass m is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity and elongation in the spring is 1 cm.If the angular velocity is doubled,the elongation in the spring becomes 5 cm.The original length of spring is

Let the length of the spring is l. When the system is whirled round in a horizontal circle the centripetal force is given by F=mv2r=m(rω)2r=mrω2 Then, r=l+elongation Given: elongation =1 cm (in the first case) For angular velocity ω the force required is F1=m(l+1)ω2=kx=k×1=k or k=m(l+1)ω2 ?(i) For second case, ω=2ω,elongation =5cm=x radius, r=l+5 So, F2=m(l+5)(2ω)2=kx=k×5=5k or 5k=4m(l+5)ω2 ?(ii) Now, dividing Eq. (i) by Eq.(ii), we get k5k=m(l+1)ω24m(l+5)ω2 ⇒ 5(l+1)=4(l+5) ⇒ 5l+5=4l+20 l=20−5=15cm