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2015-04-13T20:59:23+05:30

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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  ] 


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