M = 0.2 kg

Spring restoration force F = - k x , k = 80 N/m

Damping force F_d = -b v , where b = 4 N-sec/meter

F = - k x - b v --- (1)

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.

* let x = A e^{-a x} Sin (ωt+Ф) --- (3)** we need to determine A, a, and ω*.

differentiate two times:

dx/dt = - a A e^{-a x} Sin ωt + Aω e^{-a x} Cos ωt -- (4)

= A e^{-a x} [ - a Sin ωt + ω Cos ωt ]

d² x/d t² = a² A e^{-a x} Sin ωt - a A ω e^{-a x} Cos ωt

- Aaω e^{-a x} Cos ωt - Aω² e^{-a x} Sin ωt

= (a² - ω²) A e^{-a x} Sin ωt - 2 A a ωe^{-a x} Cos ωt ---- (5)

-kx/m - b/m * dx/dt = -k/m A e^{-a x} Sin ωt - b/m A e^{-ax} [- a Sin ωt + ω Cos ωt ]

= A e^{-ax} Sin ωt [ -k/m + ab/m ] - bω/m A e^{-ax} Cos ωt --- (6)

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

(ab-k)/m = (a² - ω²) -- (7)

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

=> 2 a² - k / m = a² - ω²

=> ω² = (k/m - a²)

* x = A e^{- a x } Sin [ √(k/m - a²) t + Ф ]*===========================

m = 0.2 kg , k = 80 N/m and b = 4 N-sec/m

b = 2 a m => a = 10 units

angular frequency ω = √ (k/m - a²) = √(400- 100) = 10√3 rad/sec

* frequency = 5√3/π Hz and time period = T = 1/f = π /(5√3) sec = 0.362 sec**This is the dampened frequency of the spring mass system.**Natural frequency = ω₀ = √(k/m)* = √(80/0.2) = 20 rad/sec

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when a force of

*F = 10 Cos 10 t = F₁ Cos (ω₁t)* s applied, The natural frequency of the force and the system are different. This is a case of forced oscillations.

net force = m a = - k x - b v + 10 Cos 10t --- (9)

m d²x/dt² = - k x - b dx/dt + 10 Cos 10t

steady state response :

* x = a Cos (ω₁ t - Ф), where,* ω₁ = frequency of the external force = 10 rad/sec.

so frequency =

* f₁ = 10/2π = 5/π Hz * and the

*time period T₁ = π/5 Sec.* ω₀ = natural freuency of the spring with out damping or external force = √(k/m)

= 20 rad /sec

a = amplitude =

* * Ф = phase =