A 5 kg mass is suspended at the free end of an overhanging massless beam, having a pin support and a roller support, as shown in the figure below. Young's modulus of the material of the beam is 200 GPa and area moment of inertia of the beam is $10^{-8}$ m$^4$. The natural frequency of the beam in rad/s is
Step 1: Given data
$m = 5 \text{ kg}$
$E = 200 \times 10^9 \text{ N/m}^2$
$I = 10^{-8} \text{ m}^4$
$ EI = 200 \times 10^9 \times 10^{-8} = 2000 \text{ Nm}^2 $
Geometry:
Left support to roller = $1$ m
Roller to free end = $2$ m
Step 2: Deflection at free end (important result)
For this overhanging beam, deflection at free end due to load $P$ is:
$ \delta = \frac{P \cdot (2)^3}{3EI} + \frac{P \cdot (2)^2 \cdot 1}{2EI} $
$ = \frac{8P}{3EI} + \frac{4P}{2EI} = \frac{8P}{3EI} + \frac{2P}{EI} $
Take LCM:
$ \delta = \frac{8P + 6P}{3EI} = \frac{14P}{3EI} $
Step 3: Equivalent stiffness
$ k = \frac{P}{\delta} = \frac{3EI}{14} $
Substitute $EI = 2000$:
$ k = \frac{3 \times 2000}{14} = \frac{6000}{14} \approx 428.57 \text{ N/m} $
Step 4: Natural frequency
$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{428.57}{5}} = \sqrt{85.714} \approx 9.26 $
Step 5: Closest option
Closest value ≈ $10$
Important Questions from Vibration of Beam and Pendulum
The periodic time of one oscillation for a simple pendulum is:
The natural frequency of a simply supported beam of length l with mass M at its centre, flexural rigidity EI and negligible beam mass is
The fundamental natural frequency of the cantilever beam with point load P acting at the free end, in rad/sec is
