Question

A body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity and elongation in the spring is $1cm$. If the angular velocity is doubled, the elongation in the spring becomes $5cm$. The original length of spring is:

• A. $13cm$
• B. $14cm$
• C. $15cm$
• D. $16cm$

Answer 1

The correct option is C $15cm$

For the first case we have
$m{\omega }^2(l+1)=k\times 1$
For the second case we have
$m\left(2\omega {)}^2\right(l+5)=k\times 5$

Dividing, we get:
$\frac{l+1}{4(l+5)}=\frac{1}{5}$
$5l+5=4l+20$
$l=15cm.$