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A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sin t 6 N (newtons) and moves in a medium that imparts a viscous force of 6 N when the speed of the mass is 3 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass. (Use g = 9.8 m/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in meters of the mass from its equilibrium position at time t seconds. Use up for u' and upp for u''.

Respuesta :

Answer:

u" + 40u' + 49u = 2 sin(t/6)        

upp + 40up + 49u = 2 sin(t/6)

Explanation:

Step 1: Data given

mass = 5 kg

L = 20 cm = 0.2 m

F = 10 sin(t/6)N

Fd(t) = - 6 N

u(0) = 0.03 m/s

u(0) = 0

u'(0) = 3 cm/s

Step 2:

ω =kL

k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²

Since Fd(t) = -γu'(t)  we know:

γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m

The initial value problem which describes the motion of the mass is given by

5u" + 200u' + 245u = 10 sin(t/6)   u(0) = 0  ;  u'(0) = 0.03

This is equivalent to:

u" + 40u' + 49u = 2 sin(t/6)        u(0) = 0  ;  u'(0) = 0.03

upp + 40up + 49u = 2 sin(t/6)

With u in m and t in s