Establish the equation of motion of a damped oscillator and show that for a weakly
damped oscillator, the displacement is given by
( ) exp( )cos( ) x t = a0 −bt ωd
t + φ
where symbols have their usual meanings.
Answers
Answered by
2
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)
===========================
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 ]
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)
===========================
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 ]
Similar questions