mω2(l0+x)=kx
(xl0+1)=mω2k
x=k−mω2l0mω2
k≫mω2
So, l0x is equal to kmω2
JEE Main 2020 — Physics Mechanics
A spring mass system (mass m, spring constant k and natural length l ) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system rotates about it's axis with an angular velocity ω,(k>>mω2) the relative change in the length of the spring is best given by the option:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
32(kmω2)
k2mω2
kmω2
3kmω2
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