A body of mass 0.2 kg is suspended from a spring of force constant 80 Nm−1
. A damping
force is acting on the system for which γ = 4 Nsm−1
. Write down the equation of motion
of the system and calculate the period of its oscillations. Now a harmonic force
F =10cos10t is applied. Calculate a and θ when the steady state response is given by
a cos(ωt − θ).

1

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2015-04-13T11:42:03+05:30

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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
===============================

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 = \frac{F_1/m}{\sqrt{(\omega-\omega_0)^2+(b\omega/m)^2}}=\frac{10/0.2}{\sqrt{(10-20)^2+(4*10/0.2)^2}}\\\\=\frac{50}{\sqrt{10^2+200^2}}= 0.2497 meters
          Ф = phase =

1 5 1
tan Ф = v₀ / (ω₀ x₀), where v₀ , ω₀ and x₀ are the velocity, frequency and displacement at t = 0, of the system, just before the external force is applied.
x₀ = A Sin (ω₀ t + Ф₀) = A SinФ₀
v₀ = A ω₀ Cos (ω₀ t + Ф₀) = A ω₀ Cos Ф₀
tan Ф = cot Ф₀ = tan (π/2 - Ф₀)
=> Ф = π/2 - Ф₀
thqq kvnmurty
Correction: In the formula for amplitude a, in the denominator, it is (ω² - ω₀²)² + (bω/m)² under square root. So answer is a = 50 / [sqroot{ (10² - 20²)² + 200² } ] = 1/2sqroot(13) = 0.1387 meters.
CORRECTION: For the phase of the steady state response: x = a Cos (ω₁ t - Ф):
tan Ф = b ω₁ /[ m(ω₀² - ω₁²) ]
= 4 * 10 /[ 0.2 (20² - 10² ] = 2/3
Ф = 33.69 deg.