Question

Considering massless rigid rod small oscillations, the natural frequency (in rad/s) of vibration of the system is shown in the figure.

• A. $\sqrt{\frac{400}{1}}$
• B. $\sqrt{\frac{400}{2}}$
• C. $\sqrt{\frac{400}{3}}$
• D. $\sqrt{\frac{400}{4}}$

Answer 1

The correct option is D $\sqrt{\frac{400}{4}}$

Method 1:


Q $\Rightarrow$ changes in angular positions of the rod in an anticlockwise direction.

Restoring torque = $(kr\theta$)r will act in clockwise direction

So, $\tau =-k{r}^2\theta$

$I\alpha =-k{r}^2\theta$

$\alpha =-\frac{k{r}^2}{I}\theta$

${\omega }_n=\sqrt{\frac{k{r}^2}{I}}$

$I=m(2r{)}^2=1(4r{)}^2$

${\omega }_n=\sqrt{\frac{400\times r^2}{1\times 4r^2}}=\sqrt{\frac{400}{4}}$

$=10\text{rad/s}$

Method II:



Given: k= 400, m = 1 kg

When the system is disturbed from its equilibrium position

$m\frac{d^2{x}_2}{d{t}^2}2r+k{x}_{1}r=0$

$2mr\frac{d^2{x}_2}{d{t}^2}+k\left(r\theta \right)r=0$

$2mr\frac{d^2\left(2r\theta \right)}{d{i}^2}+k{r}^2\theta =0$

$4m{r}^2\frac{d^2\theta }{d{i}^2}+k{r}^2\theta =0$

$\frac{d^2\theta }{d{t}^2}+\frac{k{r}^2}{4m{r}^2}\theta =0$

${\omega }_n=\sqrt{\frac{k{r}^2}{4m{r}^2}}=\sqrt{\frac{k}{4m}}=\sqrt{\frac{400}{4}}$