Question

A uniform rod of length $l$ and mass $m=1kg$ is hinged at its lowest point $O$ and is connected at its highest point $A$ by means of a spring of spring constant $k$ (in fig.). When it is pushed slightly, what is the frequency of oscillation? (Take $l=2m,k=10$ units, $\sqrt{10}=3.14$ )

Answer 1

Answer: 4.71

Total $E=E_{\text{rotational }}+E_{\text{transitional }}$
$E=\frac{1}{3}m{L}^2{\left(\frac{d\theta }{dt}\right)}^2-mg\frac{L}{2}(1-\cos \theta )+\frac{1}{2}k(L\theta {)}^2$
$=\frac{1}{3}m{L}^2{\left(\frac{d\theta }{dt}\right)}^2-mg\frac{L}{2}\left[1-\left(1-\frac{{\theta }^2}{2}\right)\right]+\frac{1}{2}k{L}^2{\theta }^2$
$\dots [\because$ As $\theta \to 0^c\Rightarrow \cos \theta \approx 1-\frac{{\theta }^2}{2}]$
$=\frac{1}{3}m{L}^2{\left(\frac{d\theta }{dt}\right)}^2+\frac{1}{2}\left(k{L}^2-\frac{mgL}{2}\right){\theta }^2$
Angular frequency,
$\omega =\sqrt{\frac{\text{ Spring factor }}{\text{ Inertial factor }}}$
$=\sqrt{\frac{k{L}^2-\frac{mgL}{2}}{\frac{1}{3}m{L}^2}}$
$=\sqrt{\frac{3k}{m}-\frac{3g}{2L}}$
$=\sqrt{\frac{3\times 10}{1}-\frac{3\times 10}{2\times 2}}$
$=3\sqrt{\frac{10}{4}}$
$=\frac{3}{2}\times 3.14$
$\therefore \omega =4.71rad/s$