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A small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes virtual oscillation with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is
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Verified Answer
The correct answer is:
\(\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}\)
The given situation can be shown as 
From figure, $K\left(x \sin \theta=m \omega^{2}(L+x) \sin \theta\right.$
$\Rightarrow \quad K x=\operatorname{m\omega}^{2}(L+x)$...(i)
Also, $\quad \sqrt{\frac{K}{m}}=\omega_{0}$
$\Rightarrow \quad K=m \omega_{0}^{2}$
Substituting the value in Eq. (i)
$$
\begin{aligned}
m \omega_{0}^{2} x &=m \omega^{2}(L+x) \\
\Rightarrow \quad x &=\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}
\end{aligned}
$$
From figure, $K\left(x \sin \theta=m \omega^{2}(L+x) \sin \theta\right.$
$\Rightarrow \quad K x=\operatorname{m\omega}^{2}(L+x)$...(i)
Also, $\quad \sqrt{\frac{K}{m}}=\omega_{0}$
$\Rightarrow \quad K=m \omega_{0}^{2}$
Substituting the value in Eq. (i)
$$
\begin{aligned}
m \omega_{0}^{2} x &=m \omega^{2}(L+x) \\
\Rightarrow \quad x &=\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}
\end{aligned}
$$
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