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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
Options:
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
Solution:
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Verified Answer
The correct answer is:
9 $\mathrm{Cm}$
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 $\ell=9 \mathrm{~cm}$
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 $\ell=9 \mathrm{~cm}$
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