Problem 86 - A Complete Resource Book in Physics for JEE Main

Question

A particle of mass \(m\) is fixed to one end of a light spring of force constant \(k\) and unstretched length \(l\). The other end of the spring is fixed and it is rotated in horizontal circle with an angular velocity \(\omega\), in gravity free space. The increase in length of the spring will be (A) \(\frac{m \omega^{2} l}{k}\) (B) \(\frac{m \omega^{2} l}{k-m \omega^{2}}\) (C) \(\frac{m \omega^{2} l}{k+m \omega^{2}}\) (D) none of these

Step-by-Step Solution

Answer

The increase in length of the spring is \(x = \frac{m \omega^{2} l}{k}\).

Step 1: Identify the centripetal force

The centripetal force acting on the particle towards the center of the circle is given by \(F_c = m \omega^{2} x\), where x is the extension in the length of the spring.

Step 2: Identify the spring force

According to Hooke's law, the force exerted by a spring is given by \(F_s = kx\), where k is the spring constant and x is the increase in length of the spring.

Step 3: Equate the centripetal force and spring force

Since the centripetal force is provided by the spring force, we can equate the two forces: \(m \omega^{2} x = kx\)

Step 4: Solve for the increase in length of the spring

We need to solve the equation for x, the increase in length of the spring. Divide both sides of the equation by x: \(\frac{m \omega^{2} x}{x} = \frac{kx}{x}\) \(m \omega^{2} = k\) Now divide both sides by k: \(\frac{m \omega^{2}}{k} = 1\) And finally, multiply both sides by l: \(\frac{m \omega^{2} l}{k} = l\) So the increase in length of the spring is: \(x = \frac{m \omega^{2} l}{k}\) This matches option (A).