Question

A concentrated mass m is attached at the centre of a rod of length 2L as shown in the figure. The rod is kept in a horizontal equilibrium position by a spring of stiffness k. For very small amplitude of vibration, neglecting the weights of the rod and spring, the undamped natural frequency of the system is

• A. $\sqrt{\frac{k}{m}}$
• B. $\sqrt{\frac{2k}{m}}$
• C. $\sqrt{\frac{k}{2m}}$
• D. $\sqrt{\frac{4k}{m}}$

Answer 1

The correct option is D $\sqrt{\frac{4k}{m}}$

Method I:
Taking moments about O

$kx\times 2L=mg\times L$

$x=\frac{mg}{2k}$

$\text{Now}\frac{x}{2L}=\frac{\delta }{L}\text{(from similar triangles)}$

$\delta =\frac{x}{2}$

Using static deflection of mass
${\omega }_n=\sqrt{\frac{g}{\delta }}=\sqrt{\frac{4k}{m}}$

Method II:

By equation of motion,

$m{L}^2\ddot{\theta }+k(2L{)}^2\theta =0$

$m{L}^2\ddot{\theta }+4L^2\theta =0$

$\ddot{\theta }+\frac{4k{L}^2}{m{L}^2}\theta =0$

We know that,

${\omega }_n=\frac{4k{L}^2}{m{L}^2}=\frac{4k}{m}$

${\omega }_n=\sqrt{\frac{4k}{m}}\text{rad/s}$