A spring with a 4 kg mass is resting on a horizontal frictionless table and is attached to a wall at the left end. It will stretch 0.5 m beyond its natural length when a force of 12 N is applied. If the spring begins at the equilibrium position and is pushed to the right with an initial velocity of 3 m/s, find the position of the mass after t seconds.

The position of the mass after t seconds is;

Explanation:

Let x(t) represent the position after t seconds

Apply the following differential equation;

where;

m is the mass of the spring = 4 kg

k is the spring constant, = F/L = 12 / 0.5 = 24 N/m

the, ω² = k/m = 24 / 4 = 6

ω = √6 = 2.45

The general solution is given by;

at x(0) = 0 , C₁ = 0    and

at x'(0) = 3 m/s , C₂ = 1.225

Therefore, the position of the mass after t seconds is given by;

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