Question

Industrial processes often require the damping of vibrations. Consider a workbench that uses four large ideal springs in place of traditional legs, with each spring supporting one corner of the workbench and exhibiting a force constant of 47500 N/m. This workbench is used when creating products that involve highly unstable chemicals. According to specifications, the acceleration of the manufacturing equipment must be held to less than 13.5 m/s2 for the manufacturing process to proceed safely. The amplitude of vibrations are expected to remain under 0.472 mm throughout. Given that the workbench and maufacturing equipment combine to 7.55 kg, find the maximum acceleration expected of the equipment and workbench. (Ignore the mass of the springs.)

Does the workbench meet specifications for acceleration?

Using shock absorbers, you decide to damp the oscillations applying one shock absorber to each spring. These shock absorbers apply a velocity-dependent drag force F = bv that opposes the motion of the bench. For proper damping, you decide that the motion of the bench should drop to half-amplitude within 0.479 seconds. (In other words, within 0.479 seconds the amplitude of motion should drop by a factor of one half.) When purchasing shock absorbers, what damping coefficient b should you specify to your industrial equipment supplier? (Assume that b^2 <4mk, that is, underdamping

PLEASE EXPLAIN QUALITATIVELY AS WELL

Answer 1

(a)

Angular speed,

w = sqrt [k / (m/4)]

w = sqrt [47500 / (7.55 / 4)] = 158.63 rad/s

maximum acceleration expected of the equipment and workbench,

a = Aw^2

a = 0.472*10^-3 * (158.63)²

a = 11.87 m/s^2

(b)

We know that,

A(t) = xm * e^{-(bt / 2m)}

xm / 2 = xm * e^{-[b*0.479 / 2*(7.55/4)]}

0.5 = e^{-[b*0.479 / 2*(7.55/4)]}

ln (0.5) = -b*0.49 / 2*(7.55 / 4)

-0.693 = -b*0.479 / 3.775

b = 5.46 kg/s