Question

A uniform rod of length $l$ and mass $m =1 \,kg$ 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 $l=2\, m , k =10$ units, $\sqrt{10}=3.14$ )

Answer 1

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