Question

A light wooden rod fixed at one end is kept horizontal. A load of $0.4\text{kg}$ tied to the free end of the rod causes that end to be depressed by $2.8\text{cm}$ . If this load is set into up and down vibrations, with what frequency(in $\text{Hz}$) will it oscillate?

Answer 1

$2.98$
As plank is massless netforce on it should be zero. Let $K=kx$ be spring force

$\Rightarrow K=\frac{2.8}{100}\times k=mg$
$\Rightarrow K=\frac{{10}^3}{7}\frac{N}{m}$.
Now,

$\Rightarrow dx=\frac{ld\theta }{\cos \theta }$

Torque equation about fixed point
$-\left(K\right(dx\left)\cos \theta \right)l$= $I_0\alpha$

$\Rightarrow -k{l}^{2}d\theta =m{l}^2\alpha$
$\frac{k}{m}d\theta ={\omega }^{2}d\theta \Rightarrow \omega =\sqrt{\frac{k}{m}}=\frac{50}{\sqrt{7}}$
Therefore frequency
$\Rightarrow f=\frac{\omega }{2\pi }=2.968\text{H}z.$