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} $