Question

Two springs of spring constant $\frac{mg}{l}$ each are attached to the end of a uniform rod of a mass m as which is hinged as shown, As the rod is rotated slightly about the hinge, it undergoes SHM with angular frequency $\omega$. Find $\omega ($in rad/s$)$.

• A. 18
• B. 9
• C. 12
• D. 6

Answer 1

The correct option is D 6

When the rod is displaced by an angular displacement $\theta$
Net torque about axis:
$\tau =(-kxl\cos \theta )+(-kxl\cos \theta )+mg\frac{l}{2}\sin \theta$
$\tau =(-kxl)+(-kxl)+mg\frac{l}{2}\theta$
Since displacement is small:
$\tau =(-k{l}^2\theta )+(-k{l}^2\theta )+mg\frac{l}{2}\theta$
Here
$K_{eff}=2k{l}^2-mg\frac{l}{2}$
Thus
$\omega =\sqrt{\frac{K_{eff}}{I}}$
$\omega =\sqrt{\frac{2k{l}^2-mg\frac{l}{2}}{\frac{m{l}^2}{3}}}$
Putting $l=1.25$ m and $g=10{ms}^{-2}$
We get $\omega =6rad/s$