Let us take a spring mass system with a natural frequency ω0.

m = mass of the body oscillating

Spring restoration force F = - k x , k = spring constant.

Damping force F_d = - b v , where b = damping factor.

x = displacement.

v = velocity of mass.

ω₀ is the frequency of the spring mass when there is no damping. Only force is spring force.

so, m d² x / d t² = - k x has the solution:

x = A0 Sin (ω₀ + Ф₀) --- (0)

and ω₀ = √(k/m) or ω0² = k/m ---- (12)

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net force acting on mass m is: F = - k x - b v --- (1)

The minus sign is because spring restoration force is opposite to displacement direction and the damping force is in the direction opposite to velocity.

m a = m d² x / d t² = - k x - b v

d² x /d t² = -k/m * x - b/m * dx/dt --- (2)

This is the equation of motion in the form of differential equation. The general solution for the above differential equation is in the following form:

let x = A_t e^{-at} Sin (ω' t + Ф') --- (3)

we need to determine A_t, a, and ω'.

differentiate two times:

dx/dt = - a A_t e^{-at} Sin (ω't+Ф')+ + A_tω' e^{-a x} Cos (ω't+Ф') -- (4)

= A_t e^{-at} [ - a Sin ω't + ω' Cos ω't ]

d² x/d t² = a² A_t e^{-at} Sin (ω't+Ф') - a A0 ω' e^{-at} Cos (ω't+Ф')

- A_t a ω' e^{-at} Cos (ω't+Ф') - A_t ω'² e^{-at} Sin (ω't+Ф')

= (a² - ω'²) A_t e^{-at} Sin (ω't+Ф') - 2 A_t a ω'e^{-at} Cos (ω't+Ф') --(5)

Find the RHS of equation (2)

-kx/m - b/m * dx/dt

= -k/m A_t e^{-at} Sin(ω't+Ф') - b/m A_t e^{-at} [- a Sin(ω't+Ф') + ω' Cos(ω't+Ф')]

= A_t e^{-at} Sin(ω't+Ф') [ -k/m + ab/m] - bω'/m A_t e^{-at} Cos(ω't+Ф') --(6)

Compare equations (5) and (6), we get :

(ab - k) / m = (a² - ω'²) -- (7)

a = b/(2m) => b = 2 a m -- (8)

=> 2 a² - k / m = a² - ω'²

=> ω'² = ( k/m - a²) = (ω₀² - a²) = [ω₀² - b²/4m² ]

=> ω' = √[ω₀² - b²/4m² ] --- (9)

also the phase angle of the damped oscillation is :

Cos Ф' = a/ ω₀ = b/(2m ω₀)

Tan Ф' = ω'/a = √[ 4m² ω₀²/ b² - 1 ] = √[ ω₀²/a² - 1 ] --- (10)

in terms of initial value of displacement x = x₀ at t = 0sec and initial velocity v₀ = dx/dt at t = 0sec of the damped oscillations,

Tan Ф' = ω' x₀ / [ v₀ + a x₀ ]

x = A_t e^{ -bt /(2m) } Sin [ ω' t + Ф' ] -- (11)

A_t = x₀ Cosec Ф' from eq (3)

= √ [ x₀² + (v₀ + a x₀)² / ω'² ]