Question

A small mass $m$ attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_0$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during this rotation is

• A. $\frac{\omega^{2}L}{\omega^{2}_{0}-\omega^{2}}$
• B. $\frac{\omega^{2}_{0}L}{\omega^{2}-\omega_{0}^{2}}$
• C. $\frac{\omega^{2}L}{\omega_{0}^{2}}$
• D. $\frac{\omega_{0}^{2}L}{\omega^{2}}$

Answer 1

The Correct Option is A

The given situation can be shown as

From figure, $K x \sin \theta=m \omega^{2}(L+x) \sin \theta$
$\Rightarrow K x=m \omega^{2}(L+x)$
Also, $ \sqrt{\frac{K}{m}}=\omega_{0}$
$\Rightarrow K=m \omega_{0}^{2}$
Substituting the value in E (i)
$m \omega_{0}^{2} x=m \omega^{2}(L+x)$
$\Rightarrow x=\frac{\omega^{2} L}{\omega_{0}^{2}-\omega^{2}}$