Question

A particle of mass is fixed to one end of a light spring having force constant $k$ and unstretched length $l$. The other end is fixed. The system is given an angular speed about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is

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

Answer 1

The correct option is A

$\frac{m{\omega }^{2}l}{k-m{\omega }^2}$

Step 1: Given data.

Fixed mass of the particle$=m$

Spring force constant$=k$

Unstretched length of spring$=l$

Step 2: Finding the stretch length in the spring.

We have,

In the given condition elastic force will provides the required centripetal force show in figure.

$kx=m{\omega }^{2}r$

Where, $x$ is increase in spring length, $\omega$ angular speed and $r$ total length of spring.

$kx=m{\omega }^2\left(l+x\right)$ $\left[\because r=\left(l+x\right)\right]$

$\Rightarrow$$kx-m{\omega }^{2}x=m{\omega }^{2}l$

$\Rightarrow$ $x=\frac{m{\omega }^{2}l}{k-m{\omega }^2}$

Hence, option A is correct.