**Statement: **A simply-supported beam with a
concentrated load acting on the midspan is shown in the figure below. If the
mass of the beam is negligible compared to the mass acting, find the natural frequency
of the system.

**Solution:**

Let us consider a beam AB of length L is simply supported at A and B as displayed in following figure. Let us thing one load P is acting at the midpoint of the beam.

Reaction force at point A, `R_A` = `\frac{P}{2}`

Reaction force at point B, `R_B` = `\frac{P}{2}`

The differential equation of elastic curve of a beam is expressed as,

`- M` = `E.I\frac{d^2y}{dx^2}`

`\Rightarrow E.I\frac{d^2y}{dx^2}` = `-\frac{Px}{2}`

`E.I\frac{dy}{dx}` = `-\frac{Px^2}{4}+C_1`

Slope will be zero at the centre
of the loaded beam. i.e dy/dx = 0, at x = L/2

Let us integrate equation (i) and
considering the boundary condition, we will have equation for deflection at a
section of beam.

Now, deflections at end support A
and B, x = 0,

Hence, the
deflection at mid span of a simply supported beam due to a concentrated load P
at the center of the beam is given by,

Where,

E
= Modulus of Elasticity

I
= Moment of Inertia

For small deflections,

`m\ddot{x}+c\dot{x}+kx` =` F(t)`

Here, c = 0 and F(t) =0

and, The natural Frequency, `\omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{48EI}{mL^3}}`