A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is \(1 \mathrm{~cm} .\) If the angular speed is doubled, the elongation in the spring is \(6 \mathrm{~cm}\). The original length of the spring is

Let \(\ell\) be the original length of the spring.
Let the initial angular velocity be \(\omega\) and the corresponding elongation \(\mathbf{e}_{1}=1 \mathrm{~cm}\).
When the angular velocity is doubled the elongation \(e_{2}=6 \mathrm{~cm}\).
If \(k\) is the spring constant then we have
\(\begin{array}{l}\mathrm{m}\left(\ell+e_{1}\right) \omega^{2}=\mathrm{ke}_{1} \\\text {and } \mathrm{m}\left(\ell+e_{2}\right)(2 \omega)^{2}=\mathrm{ke}_{2} \\\text {or } \quad \mathrm{m}\left(\ell+e_{2}\right) \cdot 4 \omega^{2}=\mathrm{ke}_{2}
\end{array}\)
Dividing Eq(1) by Eq(2), we get
\(\frac{\ell+e_{1}}{4\left(\ell+e_{2}\right)}=\frac{e_{1}}{e_{2}}\)
solving we get \(\ell=9 \mathrm{~cm}\)