Spring Differential Equation w/ Damping

1. Mar 31, 2013

AmagicalFishy

1. The problem statement, all variables and given/known data
A spring with a spring constant of 20 pounds per foot is loaded with a 10-pound weight and allowed to reach equilibrium. It is then displaced 1 foot downward and released. If the weight experiences a retarding force in pounds equal to four times the velocity at every point, find the equation of motion where t is time and y(t) is displacement with respect to time.

2. Relevant equations

$$10y'' + 4y' + 20y = 0 \\ y(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} \\ e^{i\theta} = cos{\theta} + isin{\theta} \\ y(0) = 1 \\ y'(0) = 0 \\$$

3. The attempt at a solution
The ODE's characteristic equation is:
$$10\lambda ^2 + 4\lambda + 20 = 0 \\ \text{or} \\ 5\lambda ^2 + 2\lambda + 10 = 0$$
Solving for lambda...
$$\lambda = \frac{-1 \pm 7i}{5}$$

Therefore, y(t)—the solution to the ODE—is...
$$y(t) = C_1e^{\frac{-1 + 7i}{5} t} + C_2e^{\frac{-1 - 7i}{5} t} \\ \\ = C_1e^{\frac{-1}{5}t}e^{\frac{7i}{5}t} + C_2e^{\frac{-1}{5}t}e^{\frac{-7i}{5}t} \\ = e^{\frac{-1}{5}t} (C_1e^{\frac{7i}{5}t} + C_2e^{\frac{-7i}{5}t}) \\ = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t}) \\ = e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{-7}{5}t} + C_2isin{\frac{-7}{5}t})$$
Since cos[x] is even and sin[x] is odd...
$$= e^{\frac{-1}{5}t} (C_1cos{\frac{7}{5}t} + C_1isin{\frac{7}{5}t} + C_2cos{\frac{7}{5}t} - C_2isin{\frac{7}{5}t}) \\ = e^{\frac{-1}{5}t} ( (C_1 + C_2)cos{\frac{7}{5}t} + (C_1 - C_2)isin{\frac{7}{5}t}) \\ \text{Let:} \\ A = (C_1 + C_2), B = i(C_1 - C_2) \\ \text{So...} \\ y(t) = e^{\frac{-1}{5}t} (Acos{\frac{7}{5}t} + Bsin{\frac{7}{5}t})$$

Now we use the initial conditions to solve for A and B.

$$y(0) = e^{\frac{-1}{5}(0)} (A\cos{\frac{7}{5}(0)} + B\sin{\frac{7}{5}(0)}) = 1 \\ = A\cos 0 + B\sin 0 = 1 \\ A = 1 \\ y'(t) = \frac{-1}{5}e^{\frac{-1}{5}t}(A\cos{\frac{7}{5}t} + B\sin{\frac{7}{5}t}) + e^{\frac{-1}{5}t} (\frac{-7}{5}A\sin{\frac{7}{5}t} + \frac{7}{5}B\cos{\frac{7}{5}t})$$

We end up w/: B = 1/7

So...
$$y(t) = e^{\frac{-1}{5}t} \left(\cos \frac{7}{5}t + \frac{1}{7}\sin \frac{7}{5}t \right)$$

(I'm not sure why the LaTeX stuff is showing up in boxes like that; I've tried a few things but it's always coming up).

Last edited by a moderator: Mar 31, 2013
2. Mar 31, 2013

Zondrina

$y(t) = c_1e^{\frac{-1 + 7i}{5} t} + c_2e^{\frac{-1 - 7i}{5} t} \\ = c_1e^{-\frac{t}{5}}(cos(\frac{7t}{5}) + isin(\frac{7t}{5})) + c_2e^{-\frac{t}{5}}(cos(\frac{7t}{5}) - isin(\frac{7t}{5}))$

Find y'.

Now use those initial conditions.

3. Mar 31, 2013

vela

Staff Emeritus
Wolfram Alpha's answer agrees with yours. It could be you're entering the answer incorrectly if it's graded by a computer. I expect, however, it's because the coefficient of y'' in your differential equation is wrong. It's suppose to be the mass, not the weight of the object.