Question

A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 meters. The mass is in a medium that exerts a damping force numerically equivalent to b multiplied by the instantaneous velocity. Write an inequality (using exact values) that describes the values of 'b' for which the system will be overdamped.

Answer 1

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

-kx - bv = ma

-kx -bdx/dt = md²x/dt².

Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

This is a second-order differential equation. The characteristic equation is thus,

D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

$D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}$

For an overdamped system,

$(b/m)^{2} - 4k/m} > 0$

$(b/m)^{2} > 4k/m}\\(b/m) > \sqrt{4k/m}} \\(b/m) > 2\sqrt{k/m}} \\b > 2\sqrt{km}}$

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s