Question

The assembly shown in the figure is composed of two massless rods of length $l$ with two particles, each of mass m. The natural frequency of this assembly for small oscillations is

• A. $\sqrt{g/l}$
• B. $\sqrt{2g/l\left(cos\alpha \right)}$
• C. $\sqrt{g/l\left(cos\alpha \right)}$
• D. $\sqrt{\left(gcos\alpha \right)/l}$

Answer 1

The correct option is D $\sqrt{\left(gcos\alpha \right)/l}$

$I=m{l}^2+m{l}^2=2m{l}^2$

Considering moment equilibrium conditions about the axis of rotation.

$2m{l}^2\ddot{\theta }+mglsin(\alpha +\theta )-mglsin(\alpha -\theta )=0$

$\Rightarrow 2m{l}^2\ddot{\theta }+mgl[sin\alpha cos\theta +cos\alpha sin\theta -sin\alpha cos\theta +cos\alpha sin\theta ]$

$\Rightarrow 2m{l}^2\ddot{\theta }+2mglcos\alpha sin\theta =0$

$\text{For very small}\theta ,sin\theta =\theta$

$2m{l}^2\ddot{\theta }+2mglcos\alpha =0$

$\omega =\sqrt{\frac{2mglcos\alpha }{2m{l}^2}}=\sqrt{\frac{gcos\alpha }{l}}$