Derive the differential equation for damped oscillations and write its solution.

(N/A) The resistive force acting on an oscillator in a fluid medium depends upon the velocity of the oscillator.
In practice,for not very large velocities,the resistive force is directly proportional to the velocity of the oscillator.
$\therefore F_{d} \propto -v$
$\therefore F_{d} = -bv$ ... $(1)$
Where $b$ is a positive constant called the damping coefficient.
It depends on the characteristics of the medium (like viscosity) and the size and shape of the block.
The unit of the damping coefficient is $N \cdot s/m$ and its dimensional formula is $[M^1 L^0 T^{-1}]$.
When the oscillator has a displacement $x$ from the mean position,the restoring force $(F_{s})$ is $F_{s} = -kx$ ... $(2)$
Thus,the total force acting on the mass at any time $t$ is:
$F = F_{s} + F_{d}$
$\therefore F = -kx - bv$ ... $(3)$
If $a(t)$ is the acceleration of the mass at time $t$,then by Newton's second law of motion,$F = ma(t)$.
Substituting into equation $(3)$:
$m \frac{d^2x}{dt^2} = -kx - b \frac{dx}{dt}$
$\therefore m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$ ... $(4)$
This is the differential equation for damped oscillations.
The solution of equation $(4)$ is:
$x(t) = A e^{-\frac{bt}{2m}} \cos(\omega' t + \phi)$ ... $(5)$
Where $A e^{-\frac{bt}{2m}}$ is the time-dependent amplitude and $\omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$ is the angular frequency of the damped oscillator.