Question

A very small mass $m$ is fixed to one end of a massless spring of constant $k$ and normal length $l$. The spring and the mass are rotated about the other end of the spring with angular speed $\omega$. Neglect the effect of gravity. Extension in the spring is:

• A. zero
• B. $\frac{m{l\omega }^2}{k+m{\omega }^2}$
• C. $ml{\omega }^2$
• D. $\frac{m{\omega }^{2}l}{k-m{\omega }^2}$

Answer 1

The correct option is D $\frac{m{\omega }^{2}l}{k-m{\omega }^2}$

Let the extension of spring $=x$
here, the spring force $=$ the centripetal force.
i.e, $kx=m{w}^{2}r$ where $r=l+x$
or $kx=m{w}^2(l+x)$
or $x(k-m{w}^2)=m{w}^{2}l\Rightarrow x=\frac{m{w}^{2}l}{k-m{w}^2}$