kgSpring restoration force F = - k x ,    k = 80 N/mDamping 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 secThis is the dampened frequency of the spring mass system.Natural frequency = ω₀ = √(k/m) = √(80/0.2) = 20 rad/sec===============================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 =