A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous

Question

A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping constant of 2 lb-sec/ft. If the mass is set in motion from its equilibrium position with a downward velocity of 3 in/sec, calculate the logarithmic decrement for this system.

Explanation / Answer

The equation set up is a force balance. Newton's law states that the force is mass times acceleration. If the displacement from equilibrium at time t is u(t), with u in feet and t in seconds, then acceleration is u ' ' (t). So F = m u '' = 1/2 u ' '(t) where 1/2 is the rounded slugs for 16 lbs This must balance the forces exerted by the spring and the damper. Hooke's law states that the spring force is proportional to displacement. F(spring) = ku. To find k we use that 16 lbs stretches the spring 1/4 ft (3in) 16 = k(1/4) ==> k=64. A typical law for damping is that damping is proportional to velocity, that is u ' (t). So the damping force is F(damping) = 2u ' (t). Thus 1/2 u ' '(t) = -2u ' (t) -64 u ==> 1/2 u '' +2 u' +64 u=0. If the mass starts at equilibrium, u(0)= 0. If the initial velocity is 3 in/ s u ' (0)= 1/4. What isn't clear is whether you intend downward to be the positive or the negative orientation. If you want it to be the negative orientation, you need to change that to u '(0)= -1/4. The problem is u' ' + 4 u ' +128 u = 0, u(0)=0, u'(0)= 1/4 The characteristic equation is r^2+4r+128=0 with roots r = -2 (+,-) 2sqrt(31)i. Set w=2sqrt(31) u(t) = exp(-2t)[C1 cos(wt) +C2 sin(wt)] u(0)=C1=0 u ' (0) = wC2 = 1/4 ==> C2=1/(4w) Thus u(t) = 1/(4w) exp(-2t) sin(wt) The mass returns to equilibrium when the sine function takes it first zero after t=0. That is when the argument is pi. That is, u(t)=0 next when wt=pi, so t=pi/w. | u(t) | < 1/(4w) exp(-2t) since the sine is bounded by 1. exp(-2tau) = (0.01)(4w) ==> tau = -1/2 ln(4w/100) is approximately 0.4 seconds

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