Question

A body of mass $m$ is tied to one end of spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one centimeter. If the angular velocity is doubled, the elongation in spring is 5 $cm$. The original length $L$ of spring (in $cm$ ) is

Answer 1

Let $K$ be the spring constant and $L$be the initial length in centimeter.
$m(L+x_1){w_1}^2=K{x}_1$
and $m(L+x_2){w_2}^2=K{x}_2$
Taking ratio
$\left(\frac{L+x_1}{L+x_2}\right)\times {\left(\frac{{\omega }_1}{{\omega }_2}\right)}^2=\frac{x_1}{x_2}$
Now,
${\left(\frac{{\omega }_1}{{\omega }_2}\right)}^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
$x_1=1$
&
$x_2=5$
Putting these value in above equation, we get,
$\left(\frac{L+1}{L+5}\right)\times \frac{1}{4}=\frac{1}{5}$$\Rightarrow 5L+5=4L+20\Rightarrow L=15cm.$