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. $\frac{{\omega }^{2}L}{{\omega }_0^2-{\omega }^2}$
• B. $\frac{{\omega }_0^{2}L}{{\omega }^2-{\omega }_0^2}$
• C. $\frac{{\omega }^{2}L}{{\omega }_0^2}$
• D. $\frac{{\omega }_0^{2}L}{{\omega }^2}$

Answer 1

Answer: (A)

The given situation can be shown as

From figure, $Kx\sin \theta =m{\omega }^2(L+x)\sin \theta$
$\Rightarrow Kx=m{\omega }^2(L+x)$
Also, $\sqrt{\frac{K}{m}}={\omega }_0$
$\Rightarrow K=m{\omega }_0^2$
Substituting the value in Eq. (i)
$m{\omega }_0^{2}x=m{\omega }^2(L+x)$
$\Rightarrow x=\frac{{\omega }^{2}L}{{\omega }_0^2-{\omega }^2}$