Question

The system shown in the figure consists of block A of mass 5 kg connected to a spring through a massless rope passing over pulley B of radius r and mass 20 kg. The spring constatn k is 1500 N/m. If there is no slipping of the rope over the pully, the natural frequency of the system is rad/s

Answer 1

Total energy of system
$=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}m{(r\times \frac{d\theta }{dt})}^2+\frac{1}{2}k\times (r\theta {)}^2+\frac{1}{2}\times \frac{m{r}^2}{2}\times {\left[\frac{d\theta }{dt}\right]}^2$
$=\frac{1}{2}m{r}^2.(\theta {)}^2+\frac{1}{2}k\times r^2{\theta }^2+\frac{m{r}^2}{4}(\dot{\theta }{)}^2$
Since total energy remains constant with respect to time.
$\frac{dE}{dt}=0$
$\frac{dE}{dt}=\frac{1}{2}m{r}^2.2.\dot{\theta }\ddot{\theta }+\frac{k{r}^2}{2}.2\theta .\dot{\theta }+\frac{M{r}^2}{4}2\dot{\theta }\ddot{\theta }$
$0=m{r}^2\dot{\theta }\ddot{\theta }+k{r}^2\theta .\dot{\theta }+\frac{M{r}^2}{2}\dot{\theta }\ddot{\theta }$
Divid both sides by $\dot{\theta }$
$0=m{r}^2\ddot{\theta }+k{r}^2\theta +\frac{M{r}^2}{2}\ddot{\theta }$
Divide both sides by $r^2$
$0=m\ddot{\theta }+k\theta +\frac{M\ddot{\theta }}{2}$
$0=(m+\frac{M}{2})\ddot{\theta }+k\theta$
$0=\ddot{\theta }+\left[\frac{k}{m+\frac{M}{2}}\right]\theta$
${\omega }_n^2=\frac{k}{m+\frac{m}{2}}$
${\omega }_n=\sqrt{\frac{2k}{M+2m}}=\sqrt{\frac{2\times 1500}{20+(2\times 5)}}$
$=\sqrt{\frac{3000}{20+10}}=\sqrt{100}=10$ rad/s