Q: A spring mass system (mass m , spring constant k , natural length l) rests in equilibrium on a horizontal disc . The free end of the spring is fixed at the centre of disc . If the disc together with spring mass system , rotates about its axis with an angular velocity ω , (k >> m ω2 ) the relative change in the length of the spring is best given by option

(a) $\displaystyle \frac{2 m \omega^2}{k} $

(b) $\displaystyle \frac{ m \omega^2}{3 k} $

(c) $\displaystyle \frac{ m \omega^2}{k} $

(d) $\displaystyle \sqrt{\frac{2}{3}} \frac{ m \omega^2}{k} $

Ans: (c)

Solution: Let Δl be the extension in the spring

Therefore , k Δl = m ω2 (l + Δl )

k Δl – m ω2 Δl = m ω2 l

Δl ( k – m ω2) = m ω2 l

$\displaystyle \frac{\Delta l}{l} = \frac{m \omega^2 }{k-m \omega^2} $

Using , k >> m ω2

$\displaystyle \frac{\Delta l}{l} = \frac{m \omega^2 }{k} $