A spring with a mass of 2 kg has a damping constant 14 kg/s. A force of 3.6 N is required to keep the spring stretched 0.3 m beyond its natural length. The spring is stretched 0.8 m beyond its natural length and then released. Find the position of the mass at any time t. (Assume that movement to the right is the positive x-direction and the spring is attached to a wall at the left end.)

x = 0.96 e⁻ᵗ - 0.16 e⁻⁶ᵗ

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 = 2 kg

c = damping constant = 14 kg/s

k = spring constant

F = kx

3.6 = 0.3k

k = 12 N/m

mx" + cx' + kx = 0

Becomes

2x" + 14x' + 12x = 0

x" + 7x' + 6x = 0

x(0) = 0.8 m, x'(0) = 0 m/s (It was initially at rest)

We then solve this differential equation

x" + 7x' + 6x = 0

x = eˢᵗ (s is an arbitrary number)

x' = seˢᵗ

x" = s²eˢᵗ

x" + 7x' + 6x = 0

Becomes

s²eˢᵗ + 7seˢᵗ + 6eˢᵗ = 0

s² + 7s + 6 = 0

s² + s + 6s + 6 = 0

s(s + 1) + 6(s + 1) = 0

(s + 1)(s + 6) = 0

s = -1 or -6

So,

x = Ae⁻ᵗ + Be⁻⁶ᵗ

At t = 0, x = 0.8

0.8 = A + B (eqn 1)

At t = 0, x' = 0

x' = -Ae⁻ᵗ - 6Be⁻⁶ᵗ

0 = -A - 6B

A = -6B

From eqn 1

A + B = 0.8

-6B + B = 0.8

-5B = 0.8

B = -0.16

A = -6B = -6 × -0.16 = 0.96

x = 0.96 e⁻ᵗ - 0.16 e⁻⁶ᵗ

Hope this Helps!!!

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