Differential Equations 2 A spring with spring constant ANm i

Differential Equations

2. A spring with spring constant AN/m is attached to a 1kg mass with friction constant 4Ns/m. If the mass is displaced 1m to the left and has an initial velocity of 1 m/s to the right, determine if and when the mass will pass through equilibrium (t can be negative)

Solution

Data.

k = 4 N/m

m = 1 kg

b = 4 N s/m

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d = 1 m

v = 1 m/s

t = ??

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Answer.

We know that;

b² - 4mk = 4² - 4*1*4 = 0

so the spring mass system is critically damped. Roots for characteristic polynomial are;

r² + 4r + 4 = 0

r = -b/2m

friction will cause the mass to limit to equilibrium here. Replacing;

r = -4/2*1 = -2, -2

and the general spring mass system is given by,

y(t) = c₁e^rt + c₂e^rt = c₁e^-b/2mt + c₂e^-b/2mt = (c₁ + c₂t*)e^-b/2mt*

t* = - c₁/c₂

A critically damped spring mass system can pass through the equilibrium position at most once (or never is c2 = 0).

y(t) = c₁e^-2t + c₂e^-2t

Plugging in the initial conditions (c₁ = 1):

1 = y(0) = c₁ + c₂

0 = y\'(0) = 2c₁ + 2c₂

Note that since the mass initially located to the left of equilibrium and is moving toward equilibruim (right), y\'(0) is positive. Adding we have;

-1 = 3c₁ + 3c₂

c₂ = -4/3

Thus,

y(t) = e^-2t - 4/3e^-2t

0 = e^-2t - 4/3e^-2t

1/3e^-4t = -4

so the answer is;

t* = 1/3 ln(9) = 0.7324s

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A spring mass system that is critically damped is not physically a possible solution, since it\'s unlikely that exactly b² - 4mk = 0.