Question

Common Data Questions:
A uniform rigid slender bar of mass 10 kg, hinged at the left end is suspended with the help of spring and damper arrangement as shown in the figure where K=2 kN/m, C = 500 Ns/m and the stiffness of the torsional spring $K_{\theta }$ is 1 kN/m/rad. Ignored the hinge dimensions.


The undamped natural frequency of oscillations of the bar about the hinge point is

• A. 42.43 rad/s
• B. 30 rad/s
• C. 17.32 rad/s
• D. 14.14 rad/s

Answer 1

The correct option is A 42.43 rad/s

For small deflection, after equilibrium

Now,
$\theta =\frac{x_1}{0.4}=\frac{x_2}{0.5}$

$\therefore$ $x_1=0.4\theta$

$\text{and}{x}_2=0.5\theta$

Moment of inertia

$=\frac{m{l}_2^2}{3}=\frac{10\times (0.5{)}^2}{3}=0.833kg{m}^2$

Now, from Newton's law of motion
$c{\dot{x}}_1{l}_1+k{x}_2{l}_2+k_{\theta }+I\ddot{\theta }=0$

$c{\dot{x}}_2^1\theta +k{x}_2^2+k_{\theta }.\theta +I\ddot{\theta }=0$

$\Rightarrow 500\times (0.4{)}^2\dot{\theta }+(2000\times (0.5{)}^2+1000)\theta +0.833\ddot{\theta }$

$\Rightarrow 0.833\ddot{\theta }+30\dot{\theta }+1500\theta =0$

Equation can be written as :

$I\ddot{\theta }+C_T\dot{\theta }+K_T\theta =0$

$\therefore I=0.833$

$C_T=80$

$K_T=1500$

Now the natural frequency ,

${\omega }_n=\sqrt{\frac{K_T}{I}}=\sqrt{\frac{1500}{0.833}}=42.43\text{rad/s}$