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 center of the disc. If the disc together with spring mass system, rotates about it’s axis with an angular velocity $ω, (k > > m ω^{2})$ the relative change in the length of the spring is best given by the option

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$\frac{m ω^{2}}{k}$

$\sqrt{\frac{2}{3}} (\frac{m ω^{2}}{k})$

$\frac{2 m ω^{2}}{k}$

$\frac{m ω^{2}}{3 k}$

answer is B.

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Detailed Solution

Spring force=centrifugal force

$k x = m (l + x) ω^{2}$
$\Rightarrow k x - m x ω^{2} = m l ω^{2} \Rightarrow x = \frac{m l ω^{2}}{k - m ω^{2}} \Rightarrow x = \frac{m l ω^{2}}{k} [k > > m ω^{2}]$
Relative change is $\frac{x}{l} = \frac{m ω^{2}}{k}$

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