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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 its axis with an angular velocity $\omega$ , $(k> > m\omega ^{2})$ the relative change in the length of the spring is best given by the option :
Option: 1 $\frac{m\omega ^{2}}{3k}$
Option: 2 $\frac{m\omega ^{2}}{k}$
Option: 3 $\sqrt{\frac{2}{3}}\left ( \frac{m\omega ^{2}}{k} \right )$
Option: 4 $\frac{2m\omega ^{2}}{k}$

Answers (1)

best_answer

As natural lentgh=l₀

Let elongation=x

Mass m is moving with angular velocity $\omega$ in a radius r

where $r=l_{0}+x$

Due to elongation x spring force is given by $F_{s}=Kx$

And $F_{C}=m\omega ^{2}r=m\omega ^{2}(l_{0}+x)$

as $F_{C}=F_{s}$

$Kx=m\omega ^{2}(l_{0}+x)$

$\Rightarrow x=\frac{m\omega ^{2}l_{0}}{K-m\omega ^{2}}$

Using $k> > m\omega ^{2}$

So, $\frac{x}{l_{0}}$ is equal to $\frac{m\omega ^{2}}{k}$

Hence the correct option is (2).

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