#6. If a force of 400N stretches the spring 2 meters, then the spring constant is k = 400/2 = 200.

Since a mass of 50 kg is now attached to the spring, the DE is: 50 x'' + 200 x = 0, which equals x'' + 4x = 0.

The spring is then released from equilibrium (so x(0) = 0) with an upward velocity of 10 m/s (so x'(0) = -10).

The general solution is x(t) = c_{1}cos(2t) + c_{2}sin(2t).

The initial condition x(0) = 0 tells us that c_{1} = 0.

The initial condition x'(0) = -10 tells us that c_{2} = -5.

Thus our solution is: solution is: x(t) = -5 _{}sin(2t)

#23. a) We are given that the mass is 1 kg and the spring constant is 16, so ω^{2} = 16/1 = 16.

We are also given that the damping coefficient is β = 10, so 2λ = 10/1 = 10.

Thus our DE is: x'' + 10x' + 16x = 0.

Our characteristic equation is m^{2} + 10m + 16 = 0,

which factors to (m + 8)(m + 2) = 0, so m = -2 and -8.
Our general solution is x(t) = c_{1}e^{-2t} + c_{2}e^{-8t}

Plugging in the initial conditions x(0) = 1 and x'(0) = 0, we get x(t) = (4/3)_{}e^{-2t} + (-1/3)_{}e^{-8t}

b) With the initial condition x(0) = 1 and x'(0) = -12, we get x(t) = (-2/3)_{}e^{-2t} + (5/3)_{}e^{-8t}

#26. a) If a 10 lb force stretches the spring an extra 2 feet, that means the spring constant is k = 10/2 = 5.

We are told that a weight of 8 lbs is attached to the spring, which means we have a mass m = weight/g = 8/32 = 1/4

Thus ω^{2} = k/m = 5/(1/4) = 20.

We are also told that the damping coefficent is β = 1, so 2λ = β/m = 1/(1/4) = 4.

Thus our DE is: x'' + 4x' + 20x = 0.

Solving this (using the characteristic equation) gives x(t) = e^{-2t}(c_{1} cos(4t) + c_{2} sin(4t))

Plugging in the initial conditions x(0) = 1/2 and x'(0) = 1 gives: x(t) = e^{-2t}((1/2)_{} cos(4t) + (1/2) sin(4t))