Respuesta :
Answer:
u''(t) + 10*u'(t) + 98*u(t) = 2 sin(t/2) , u(0) = 0 , u'(0) = 0.03 m/s
Explanation:
Solution:
- The Force exerted F(t) = 10 sin(t/2)
- The viscous force at time t* is given by F(t*) = - 2N
- The speed of mass m at point t* is u'(t*) = 0.04 m/s
- The initial velocity u(0) = 0.03 m/s
- The IVP ODE of a damped system is given by a general form as follows:
        m*u''(t) + b*u'(t) + k*u(t) = F
- We will evaluate the system stiffness constant k and damping constant b for the system.
- We know that system stiffness k is given by Newton's equation of motion at equilibrium position we have:
                Fs = Fg
Where, Fs: spring force = k*x
      Fg: Weight of the object = m*g
                k*x = m*g
                k = m*g / x
- Given mass m = 5 kg stretches spring by x = 10 cm .... (Missing part of question)
- The stiffness constant k is:
                k = 5*9.81 / 0.1
                k = 490 N/m
- The damping force at time t* is given by:
                Fd(t*) = - b*u'(t*)
                b = - Fd(t*) / u'(t*)
                b = 2 / 0.04
                b = 50 Ns/m
- The IVP for the system is given as:
          m*u''(t) + b*u'(t) + k*u(t) = F
          5u''(t) + 50*u'(t) + 490*u(t) = 10 sin(t/2)
          u''(t) + 10*u'(t) + 98*u(t) = 2 sin(t/2) , u(0) = 0 , u'(0) = 0.03 m/s
