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A 1 kilogram mass is attached to a spring whose constant is 16 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 10 times instantaneous velocity. Determine the equations of motion if:

a) The mass is initially released from rest from a point 1 meter below the equilibrium position, and then

b) The mass initially released from a point 1 meter below the equilibrium position with an upward velocity of 12 m/s

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Answers

a) x = (4/3) e⁻²ᵗ - (1/3) e⁻⁸ᵗ

x = (1/3) [4e⁻²ᵗ - e⁻⁸ᵗ]

b) x = (-2/3) e⁻²ᵗ + (5/3) e⁻⁸ᵗ

x = (1/3) [-2e⁻²ᵗ + 5e⁻⁸ᵗ]

Explanation:

Let the position of the mass at any time be x.

The general formula for the elastic material is given as

mx" + cx' + kx = 0

m = mass of spring = 1 kg

c = damping constant = 10 kg/s

k = spring constant = 16 N/m

mx" + cx' + kx = 0

Becomes

x" + 10x' + 16x = 0

We then solve this differential equation

x" + 10x' + 16x = 0

Let x = eˢᵗ (s is an arbitrary number)

x' = seˢᵗ

x" = s²eˢᵗ

x" + 10x' + 16x = 0

Becomes

s²eˢᵗ + 10seˢᵗ + 16eˢᵗ = 0

s² + 10s + 16 = 0

s² + 2s + 8s + 16 = 0

s(s + 2) + 8(s + 2) = 0

(s + 2)(s + 8) = 0

s = -2 or -8

So,

x = Ae⁻²ᵗ + Be⁻⁸ᵗ

a) x(0) = 1 m, x'(0) = 0 (released from rest)

x = Ae⁻²ᵗ + Be⁻⁸ᵗ

x(0) = 1 = A + B

A + B = 1 (eqn 1)

x' = -2Ae⁻²ᵗ - 8 Be⁻⁸ᵗ

x'(0) = -2A - 8B = 0

2A = -8B

A = -4B

Recall that A + B = 1

-4B + B = 1

-3B = 1

B = (-1/3)

A = -4B = -4(-1/3) = (4/3)

x = Ae⁻²ᵗ + Be⁻⁸ᵗ

x = (4/3) e⁻²ᵗ - (1/3) e⁻⁸ᵗ

x = (1/3) [4e⁻²ᵗ - e⁻⁸ᵗ]

b) x(0) = 1 m, x'(0) = - 12 m/s

x = Ae⁻²ᵗ + Be⁻⁸ᵗ

x(0) = 1 = A + B

A + B = 1 (eqn 1)

x' = -2Ae⁻²ᵗ - 8 Be⁻⁸ᵗ

x'(0) = -2A - 8B = -12

A + 4B = 6

A + B = 1

A + 4B = 6

Solving the simultaneous equation

B = (5/3)

A = (-2/3)

x = Ae⁻²ᵗ + Be⁻⁸ᵗ

x = (-2/3) e⁻²ᵗ + (5/3) e⁻⁸ᵗ

x = (1/3) [-2e⁻²ᵗ + 5e⁻⁸ᵗ]

Hope this Helps!!!

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