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''.

Question

Physics

Wolfgang

5 September, 22:49

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''.

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Aria Ellison · Answer 1

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