Question

the end of a rod of length L and mass M are attached to two identical spring the rod is free to rotate about its Centre O .the rod is depressed slightly at and A and released the time period of the resulting oscillation will be?​

Answer 1

Answer:

(3)

Explanation:

The Energy method is an interesting method to do this

When we rotate the rod by an angle by an angle theta, so the compression in springs is l.theta/2 .We can use energy conservation equation: Here LHS is of stretched position and RHS is of when it returns to original position(mean position)

$PE_{springA} +PE_{springB} = KE\\\frac{1}{2} k x^2 + \frac{1}{2} k x^2 = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2$

Now , $v = \omega \frac{l}{2}$

$kx^2 = \frac{1}{2}(\frac{ml^2}{12})(\frac{2v}{l})^2 + \frac{1}{2}mv^2$

Now differentiate both sides with respect to the time t:

$2kx.v = \frac{2mv.a}{6}+ mv.a\\\to 2kxv = \frac{4}{3}mva \\\to ma = \frac{3}{2}kx\\\to ma = k'x \to k' = \frac{3k}{2}$

This has been converted to a spring block problem, whose time period is given by $2\pi \sqrt{\frac{m}{k}}$

So the time period here will be :

$T = 2\pi \sqrt{\frac{m}{k'}} = 2\pi \sqrt{\frac{2m}{3k}}$