The system shown in the figure below consists of a cantilever beam (with flexural rigidity EI and negligible mass), a spring (with spring constant K and negligible mass) and a block of mass m. Assuming a lumped parameter model for the system, the fundamental natural frequency (ωn) of the system is

Correct Answer:

$\sqrt{\frac{\frac{3EI}{L^3}+K}{m}}$

Step-by-Step Explanation:

1. Analysis of the Physical System from the Image:
The schematic diagram illustrates a cantilever beam of length L fixed at support O on the left end. At the free end on the right, two restoring elements are connected to the lumped mass m:
- A vertical spring with a spring constant K attached to a rigid overhead support.
- The cantilever beam itself, which has a flexural rigidity EI and behaves as an elastic spring element under transverse loading at its tip.

2. Parallel Connection of Restoring Elements:
When the block of mass m is displaced vertically downward by a displacement x, both the spring and the free end of the cantilever beam must deflect by the exact same distance x. Because both elements share the same displacement and their restoring forces combine to oppose the motion, they are connected in parallel.
The equivalent spring stiffness (k_eq) for elements connected in parallel is the direct sum of their individual stiffnesses:
$k_{\text{eq}}=k_{\text{beam}}+k_{\text{spring}}$

3. Calculating the Stiffness of the Cantilever Beam:
For a cantilever beam of length L and flexural rigidity EI, the tip deflection δ due to a point load P applied at the free end is given by:
$\delta =\frac{P{L}^3}{3EI}$
Using the definition of stiffness, k = P / δ, the transverse stiffness of the cantilever beam at its free end is:
$k_{\text{beam}}=\frac{3EI}{L^3}$

4. Determining the Equivalent System Stiffness:
Substituting the beam stiffness and the spring constant (k_spring = K) into the parallel stiffness formula gives:
$k_{\text{eq}}=\frac{3EI}{L^3}+K$

5. Fundamental Natural Frequency:
For a single-degree-of-freedom lumped parameter system, the fundamental natural frequency (ω_n) is calculated as:
${\omega }_n=\sqrt{\frac{k_{\text{eq}}}{m}}$
Substituting the expression for k_eq, we arrive at:
${\omega }_n=\sqrt{\frac{\frac{3EI}{L^3}+K}{m}}$
This confirms the validity of the correct option.