A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity.…
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
1. 3 Cm
9 $\mathrm{Cm}$
6 Cm
$12 \mathrm{~Cm}$
Solution
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 $\ell=9 \mathrm{~cm}$